enVisionmath 2.0: Grade 6, Volume 1

Chapter 4 Represent And Solve Equations And Inequalities

Section 4.3: Write And Solve Addition And Subtraction Equations

Page 189 Exercise 1 Answer

An equation can be thought of as a pan – a pan is balanced when both sides weigh the same, similarly the equation is balanced, that is true, when both sides have the same value.

When the same weight is added or subtracted from both sides of the pan, the pan stays balanced. Similarly, when the same value is added or subtracted from both sides of the equation, the equation stays balanced, that is true – this is known as the Addition and Subtraction Properties of Equality.

Result

When the same weight is added or subtracted from both sides of the pan, the pan stays balanced. Similarly, when the same value is added or subtracted from both sides of the equation, the equation stays balanced, that is true – this is known as the Addition and Subtraction Properties of Equality.

Page 189 Exercise 1 Answer

Let n mark the number of students on the school bus before the last stop.

Since at the last stop 16 students boarded the bus, there were n + 16 students after the last stop. That is the left side of the equation.

The right side of the equation is 25, since the bus arrived at the school with 25 students.

Thus, the equation is

n + 16 = 25.

To find how many students were on the bus before the last stop use the equation and find the value of n.

n + 16 = 25

n = 25 – 16 (Subtract 16 from both sides of the equation.)
​
n = 9

Result

There were 9 students on the bus before the last stop.

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Page 190 Exercise 1 Answer

To solve equations, we need to use inverse operations. To solve addition equations, we can then use subtraction to solve for the variable since addition and subtraction are inverses:

n + 12 = 16 Given equation.

n + 12 – 12 = 16 – 12 Subtract 12 on both sides.

n = 4 Simplify.
​
Cabrina then has 4 markers at the start.

Since 12 was subtracted on both sides of the equation, then the Subtraction Property of Equality was used to solve the equation. No other property could have been used since subtraction is the only inverse operation of addition.

Result

n + 12 = 16

n + 12 – 12 = 16 – 12

n = 4

Cabrina then has 4 markers at the start.

The Subtraction Property of Equality was used to solve the equation. No other property could have been used since subtraction is the only inverse operation of addition.

Page 191 Exercise 2 Answer

Both sides of the equation will represent how many pages she read. It is stated that she read 60 pages, so that will be one side of the equation. The other side will be p − 14 where p represents the number of pages she was assigned to read, and she read 14 pages fewer than she was assigned.

To find how many pages Vivian was assigned solve the equation – find the value of p which is the solution of the following equation.

p – 14 = 60

p – 14 + 14 = 60 + 14 (Use the Addition Property of Equality.)
​
p = 74

Vivian was assigned 74 pages to read.

Result

Vivian was assigned 74 pages to read.

Page 192 Exercise 1 Answer

To write an addition or subtraction equation, we determine which operation to use in the equation based on the given relationship. To solve the equation, we use an inverse operation to isolate the variable on one side of the equation. For an addition equation, the inverse operation is subtraction. For a subtraction equation, the inverse operation is addition.

Page 192 Exercise 2 Answer

To solve the equation n + 7 = 25 we need to find the value of n. To do this using the inverse relationship of addition and subtraction is to subtract 7 from both sides, since we need n on the left side, and everything else on the other.

That is, if n + 7 equals 25, then n + 7 − 7 must be equal to 25 − 7. Since n + 7 − 7 is n, then n = 25 − 7 = 18.

Result

Subtract 7 from both sides to get n on the left side and everything else on the other.

Page 192 Exercise 3 Answer

Both sides of the equation will represent how many book Clare had after she bought additional books. Since we know that is 24, that will be one side of the equation. However, we can think of that as t + 8 since she had t books and than bought 8 more books.

The equation is

24 = t + 8.

To find how many books she started with solve the equation – find the value of t. Use the Subtraction Property of Equality.

24 = t + 8

24 – 8 = t + 8 – 8 (Subtract 8 from both sides.)
​
16 = t

Clare started with 16 books.

Result

Clare started with 16 books.

Page 192 Exercise 4 Answer

Both sides of the equation will represent the outside temperature at the time Arianna ate dinner, t will represent the outside temperature when Arianna ate breakfast. When she ate dinner the temperature was 35°F – that will be one side of the equation. The other side will be t – 20° since the temperature dropped 20°F from the time she ate breakfast until the dinner.

To find the outside temperature when Arianna ate breakfast solve the equation – find the value of t. Use the Addition Property of Equality since when 20 is added we know the value of t.

35° = t – 20°

35° + 20 = t – 20° + 20 (Add 20 to both sides)

55° = t

The outside temperature at the time Arianna ate breakfast was 55°F.

Result

55°F

Page 192 Exercise 5 Answer

24 + m – 24 = 49 – 24 (Subtract 24 from both sides.)

m = 25 (Evaluate to find the value of m.)
​
Result

m = 25

Page 192 Exercise 6 Answer

12 + 11 = y – 11 + 11 (Add 11 to both sides.)

23 = y (Evaluate to find the value of y.)
​
Result

y = 23

Page 192 Exercise 7 Answer

22 – 13 = 13 + a – 13 (Subtract 13 from both sides.)

9 = a (Evaluate to find the value of a.)
​
Result

a = 9

Page 192 Exercise 8 Answer

t – 40 + 40 = 3 + 40 (Add 40 to both sides.)

t = 43 (Evaluate to find the value of t.)
​
Result

t = 43

Page 192 Exercise 9 Answer

d + 11 – 11 = 15 – 11 (Subtract 11 from both sides.)

d = 4 (Evaluate to find the value of d.)

Result

d = 4

Page 192 Exercise 10 Answer

32 + 19 = s – 19 + 19 (Add 19 to both sides.)

51 = s (Evaluate to find the value of s.)
​
Result

s = 51

Page 193 Exercise 11 Answer

y – 12 + 12 = 89 + 12 (Add 12 to the left side of the equation.)

y = 101 (Evaluate to find the value of y.)
​
Result

y = 101

Page 193 Exercise 12 Answer

80 + r – 80 = 160 – 80 (Subtract 80 from both sides.)

r = 80 (Evaluate to find the value of r.)
​
Result

r = 80

Page 193 Exercise 13 Answer

60 + 16 = x – 16 + 16 (Add 16 to both sides.)

76 = x (Evaluate to find the value of x.)

Result

x = 76

Page 193 Exercise 14 Answer

20 – 12 = y + 12 – 12 (Subtract 12 from both sides.)
​
8 = y

Result

y = 8

Page 193 Exercise 15 Answer

x + 2 − 2 = 19 − 2 (Subtract 2 from both sides.)

x = 17
​
Result

x = 17

Page 193 Exercise 16 Answer

z – 313 + 313 = 176 + 313 (Add 313 to both sides.)

z = 489 (Evaluate to find the value of z.)
​
Result

z = 489

Page 193 Exercise 17 Answer

Both sides of the equation will represent how many cards you have after the trade – that is one side of the equation is 9. The other side is t − 21, since you had t cards but you gave 21 cards to your friend.

The equation is

9 = t − 21.
​
To find how many cards were in the original deck, that is how many cards you had before the trade, solve the eqution – find the value of t. Use the Addition Property of Equality.

9 = t − 21

9 + 21 = t − 21 + 21

30 = t

There were 30 cards in the original deck.

Result

There were 30 cards in the original deck.

Page 193 Exercise 18 Answer

Both sides of the equation will represent the number of contacts Joy has in her phone list – one side of the equation is 100. If c represents the number of contacts she had in her phone list before she updated it, and we know 26 new contacts are added, the other side is c + 26.

The equation is

100 = c + 26.

To find the number of contacts Joy had in her phone list before she updated it solve the equation – find the value of c. Use the Subtraction Property of Equality.

100 = c + 26

100 – 26 = c + 26 – 26 (Subtract 26 from both sides.)
​
74 = c

Joy had 74 contacts in her phone list before she updated it.

Result

Joy had 74 contacts in her phone list before she updated it.

Page 193 Exercise 19 Answer

Both sides of the given equation show how much money Jeremy spent on a sandwwich and a drink. It is stated that he spent $7 and the price of the drink is known – $1.75. Solve the equation to find the cost of the sandwich he bought. Use the Subtraction Property of Equality.

7 = s + 1.75

7 – 1.75 = s + 1.75 – 1.75 (Subtract 1.75 from both sides.)
​
5.25 = s

The cost of Jeremy’s sandwich is $5.25.

Result

The cost of Jeremy’s sandwich is $5.25.

Page 194 Exercise 20 Answer

Solve the equation using the Subtraction Property of Equality.

42 + d = 51

42 + d – 42 = 51 – 42 (Subtract 42 from both sides.)
​
d = 9

The distance of the third leg of the race is 9 kilometers.

Result

d = 9

Page 194 Exercise 21 Answer

To solve the equation subtract 45 from both sides, that is use the Subtraction Property of Equality.

153 = g + 45

153 – 45 = g + 45 – 45 (Subtract 45 from both sides.)
​
108 = g

Result

g = 108

Page 194 Exercise 22 Answer

Use substitution to check if either of these values, y = 2,3,4, or 5, is the solution.

6 + 3y = 4y + 2

6 + 3(2) = 4(2) + 2

6 + 6 = 8 + 2

12 ≠ 10

6 + 3y = 4y + 2

6 + 3(3) = 4(3) + 2

6 + 9 = 12 + 2

15 ≠ 14

6 + 3y = 4y + 2

6 + 3(4) = 4(4) + 2

6 + 12 = 16 + 2

18 = 18

6 + 3y = 4y + 2

6 + 3(5) = 4(5) + 2

6 + 15 = 20 + 2

21 ≠ 22

The equation is balanced only if y = 4, thus that is the solution.

Result

y = 4

Page 194 Exercise 23 Answer

Both sides of the equation will represent the height of the helicopter after the descend which is equal to 477 – that is one side of the equation. The other side is the original height minus 127 meters for which it descended}$.

The equation which represents the problem is

477 = h − 127.
​
Solve the equation to find the original height of the helicopter.

477 = h − 127

477 + 127 = h − 127 + 127

604 = h

Result

477 = h − 127,h = 604.

Page 194 Exercise 24 Answer

In the three days of their annual production the drama club sold 826 tickets. Since on the first day they sold 143, on the second day 295, and on the third day t number of tickets, the sum of these numbers is 826.

Since 143 + 295 equals 438, the equation which represents the total amount of thickets sold is

438 + t = 826.

Solve the equation – find the value of t to answer the question. Use the Subtraction Property of Equality.

438 + t = 826

438 + t – 438 = 826 – 438 (Subtract 438 from both sides.)
​
t = 388

388 tickets were sold the third day.

Result

388 tickets were sold the third day.

Page 194 Exercise 25 Answer

Since there are 35 almonds, 34 hazelnuts, and 32 walnuts, the sum 35 + 34 + 32 represents the total number of almonds, hazelnuts, and walnuts.

35 + 34 + 32 = 101
​
Both sides of the equation will represent the total number of nuts – almonds, hazelnuts, walnuts, and pistachios. One side will thus be equal to 134. The other side is the sum of the total number of nuts – almonds, hazelnuts, walnuts, and pistachios. Since we have alread calculated the total number of almonds, hazelnuts, and walnuts, which is 101, the other side is equal to 101 + p, where p is the number of pistachios.

The equation is

134 = 101 + p.

Solve the equation – find the value of p to answer the question. Use the Subtraction Property of Equality.

134 = 101 + p

134 – 101 = 101 + p – 101 (Subtract 101 from both sides.)
​
33 = p

There are 33 pistachios in the bag.

Result

The total number of almonds, hazelnuts, and walnuts is 101. There are 33 pistachios in the bag.

Page 194 Exercise 26 Answer

To find which equations have g = 6 as the solution, substitute the variable with its value, that is substitute g with 6, and evaluate to see if the equation is balanced. If and only if the equation is balanced, g = 6 is the solution.

g + 2 = 10

6 + 2 = 10 (Substitute g with 6.)

8 ≠ 10 (Evaluate and compare.)

g – 1 = 10

6 – 1 = 10 (Substitute g with 6.)

5 ≠ 10 (Evaluate and compare.)

g – 2 = 4

6 – 2 = 4 (Substitute g with 6.)

4 = 4 (Evaluate and compare.)
​
58 + g = 60

58 + 6 = 60 (Substitute g with 6.)

64 ≠ 60 (Evaluate and compare.)

44 – g = 38

44 – 6 = 38 (Substitute g with 6.)

38 = 38 (Evaluate and compare.)
​
g = 6 the solution of equations: g − 2 = 4 and 44 − g = 38. g = 6 is not the solution of other equations since, when g is substituted with its value, the equations are not balanced.

Result

g − 2 = 4, 44 − g = 38.

Page 194 Exercise 27 Answer

To find which equations have x = 4 as the solution, substitute the variable with its value, that is substitute x with 4, and evaluate to see if the equation is balanced. If and only if the equation is balanced, x = 4 is the solution.

42 = 38 + x

42 = 38 + 4 (Substitute x with 4.)

42 = 42 (Evaluate and compare.)

x + 15 = 19

4 + 15 = 19 (Substitute x with 4.)

19 = 19 (Evaluate and compare.)

18 = x – 2

18 = 4 – 2 (Substitute x with 4.)

18 ≠ 2 (Evaluate and compare.)
​
36 = x + 32

36 = 4 + 32 (Substitute x with 4.)

36 = 36 (Evaluate and compare.)

52 – x = 46

52 – 4 = 46 (Substitute x with 4.)

48 ≠ 46 (Evaluate and compare.)
​
x = 4 is not the solution of other equations since, when x is substituted with its value, the equations are not balanced.

Result

42 = 38 + x, x + 15 = 19, and 36 = x + 32.

Chapter 4 Represent And Solve Equations And Inequalities

Section 4.4: Write And Solve Multiplication And Division Equations

Page 195 Exercise 1 Answer

Since the 29 people also equally share the cost of the hotel bill, we can use the same strategy to find each person’s share of the hotel bill that we used in the Solve & Discuss It! problem.

From the Solve & Discuss It! problem, we know that the total cost will be 29 ⋅ s if s represents each person’s share.

From the receipt, the total hotel bill cost is $10,034 so the equation representing this situation is 29 ⋅ s = 10,034.

To solve this equation, we will also divide both sides by 29:

29s ÷ 29 = 10,034 ÷ 29

s = 346

Each person’s share is then s = $346.

Result

Since the 29 people also equally share the cost of the hotel bill, we can use the same strategy to find each person’s share of the hotel bill that we used in the Solve & Discuss It! problem. The equation representing this situation is 29 ⋅ s = 10,034 and each person’s share is s = $346.

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Page 195 Exercise 1 Answer

It is given that each person’s share is s dollars so if 29 people are equally sharing the cost, the total cost is:

(number of people)⋅(each person’s share) = 29 ⋅ s

From the receipt, we know the total cost is $19,111 so the equation that represents the situation is 29 ⋅ s = 19,111.

To solve equations, we need to use inverse operations. Since we have a multiplication equation and the inverse operation of multiplication is division, we need to divide both sides of the equation by 29 so solve for s:

29s ÷ 29 = 19,111 ÷ 29

s = 659

Each person’s share is then s = $659.

Result

s = $659

Page 196 Exercise 1 Answer

To solve equations, we need to use inverse operations to isolate the variable. Since we have a multiplication equation and the inverse operation of multiplication is division, we need to to divide both sides to isolate the variable:

4n = 52 Given equation.

4n ÷ 4 = 52 ÷ 4 Divide both sides by 4.

n = 13 Simplify.

Theresia then picked 13 tomatoes each day.

To solve the equation we needed to divide both sides by 4 so the property of equality that was used was the Division Property of Equality.

Result

4n = 52

4n ÷ 4 = 52 ÷ 4

n = 13
​
Theresia picked 13 tomatoes each day.

To solve the equation we needed to divide both sides by 4 so the property of equality that was used was the Division Property of Equality.

Page 197 Exercise 2 Answer

We know that:

total number of pages = (number of days) ⋅ (number of pages per day)

This relationship can also be written as:

If d is the number of days, the book has 630 pages, and she reads 18 pages per day, then the division equation representing this situation is:

d = \(\frac{630}{18}\)

Evaluating the right side of the equation gives \(\frac{630}{18}\) so d = 35. It will then take her 35 days to finish her book.

Result

35 days

Page 198 Exercise 1 Answer

To write a multiplication or division equation, we determine which operation to use in the equation based on the given relationship. To solve the equation, we use an inverse operation to isolate the variable on one side of the equation. For a multiplication equation, the inverse operation is division. For a division equation, the inverse operation is multiplication.

Page 198 Exercise 2 Answer

For the equation 8n = 16, the left side has 8 times the variable. To isolate the variable, we must then use the inverse operation of multiplication, which is division.

The equation is then solved by using the Division Property of Equality to divide both sides of the equation by 8.

Result

Division Property of Equality

Page 198 Exercise 3 Answer

For the equation a ÷ 9 = 2, the left side has the variable divided by 9. To isolate the variable, we must then use the inverse operation of division, which is multiplication.

The equation is then solved by using the Multiplication Property of Equality to multiply both sides of the equation by 9.

Result

Multiplication Property of Equality

Page 198 Exercise 4 Answer

Since each van will carry an equal number of students, then:

total of students = (of vans) . (of students per van)

We know that there are 30 total students, 5 vans, and s represents the number of students per van. The multiplication equation representing this situation is then:

30 = 5s

To solve a multiplication equation, we need to use the inverse operation of multiplication, which is division. Dividing both sides of the equation by gives:

30 ÷ 5 = 5s ÷ 5

6 = s

Each van then has s = 6 students.

Result

30 = 5s

s = 6 students

Page 198 Exercise 5 Answer

18m = 36 is a multiplication equation since the left side is 18 times the variable.

To solve a multiplication equation, we need to use the inverse operation of multiplication, which is division.

To solve 18m = 36, we must then divide both sides by 18:

18m ÷ 18 = 36 ÷ 18

m = 2
​
Result

Divide both sides by 18 to get m = 2.

Page 198 Exercise 6 Answer

t ÷ 3 = 10 is a division equation since the left side has the variable divided by 3.

To solve a division equation, we need to use the inverse operation of division, which is multiplication.

To solve t ÷ 3 = 10, we must then multiply both sides by 3:

t ÷ 3 ⋅ 3 = 10 ⋅ 3

t = 30

Result

Multiply both sides by 3 to get t = 30.

Page 198 Exercise 7 Answer

12 = 2y is a multiplication equation since the right side is 2 times the variable.

To solve a multiplication equation, we need to use the inverse operation of multiplication, which is division.

To solve 12 = 2y, we must then divide both sides by 2:

12 ÷ 2 = 2y ÷ 2

6 = y
​
Result

Divide both sides by 2 to get 6 = y.

Page 198 Exercise 8 Answer

22 = a ÷ 5 is a division equation since the right side has the variable divided by 5.

To solve a division equation, we need to use the inverse operation of division, which is multiplication.

To solve 22 = a ÷ 5, we must then multiply both sides by 5:

22 . 5 = a ÷ 5 ⋅ 5

110 = a
​
Result

Multiply both sides by 5 to get 110 = a.

Page 198 Exercise 9 Answer

23d ÷ 23 = 2,392 ÷ 23 (Divide both sides by 23.)

d = 104 (Evaluate.)

Result

d = 104

Page 198 Exercise 10 Answer

74f ÷ 74 = 6,179 ÷ 74 (Divide both sides by 74.)

f = 83.5 (Evaluate.)
​
Result

f = 83.5

Page 198 Exercise 11 Answer

y ÷ 11 × 11 = 987 × 11 (Multiply both sides by 11.)

y = 10857 (Evaluate.)

Result

y = 10857

Page 198 Exercise 12 Answer

r ÷ 187 × 187 = 9 × 187 (Multiply both sides by 187.)

r = 1683 (Evaluate.)
​
Result

r = 1683

Page 199 Exercise 13 Answer

8y = 56 is a multiplication equation since the left side has 8 times the variable.

To solve a multiplication equation, we need to use the inverse operation of multiplication, which is division.

To solve 8y = 56, we must then divide both sides by 8:

8y ÷ 8 = 56 ÷ 8

y = 7

Result

Divide both sides by 8 to get y = 7.

Page 199 Exercise 14 Answer

t ÷ 15 = 3 is a division equation since the left side has the variable divided by 15.

To solve a division equation, we need to use the inverse operation of division, which is multiplication.

To solve t ÷ 15 = 3, we must then multiply both sides by 15:

t ÷ 15 ⋅ 15 = 3 ⋅ 15

t = 45

Result

Multiply both sides by 15 to get t = 45.

Page 199 Exercise 15 Answer

u ÷ 8 = 12 is a division equation since the left side has the variable divided by 8.

To solve a division equation, we need to use the inverse operation of division, which is multiplication.

To solve u ÷ 8 = 12, we must then multiply both sides by 8:

u ÷ 8 ⋅ 8 = 12 ⋅ 8

u = 96

Result

Multiply both sides by 8 to get u = 96.

Page 199 Exercise 16 Answer

31y = 310 is a multiplication equation since the left side has 31 times the variable.

To solve a multiplication equation, we need to use the inverse operation of multiplication, which is division.

To solve 31y = 310, we must then divide both sides by 31:

31y ÷ 31 = 310 ÷ 31

y = 10

Result

Divide both sides by 31 to get y = 10.

Page 199 Exercise 17 Answer

d ÷ 2 = 108 Given equation.

d ÷ 2 . 2 = 108 . 2 Multiply both sides by 2.

d = 216 Simplify.
​
Result

d = 216

Page 199 Exercise 18 Answer

7,200 = 800s Given equation.

7,200 ÷ 800 = 800s ÷ 800 Divide both sides by 800.

9 = s Simplify.

Result

s = 9

Page 199 Exercise 19 Answer

x ÷ 3 = 294 Given equation.

x ÷ 3 . 3 = 294 . 3 Multiply both sides by 3.

x = 882 Simplify.
​
Result

x = 882

Page 199 Exercise 20 Answer

99 = 3x Given equation.

99 ÷ 3 = 3x ÷ 3 Divide both sides by 3.

33 = x Simplify.

Result

x = 33

Page 199 Exercise 21 Answer

We need to write a division equation and a multiplication equation to represent the problem. Note that we are not being asked to solve the equations, we only need to write them.

We know that:

total of words typed = (of minutes) . (of words per minute)

In Lolo types a total of 1,125 words in 15 minutes at a rate of w words per minute, then the multiplication equation representing this situation is:

1,125 = 15w

To write the division equation, we can divide both sides of the equation by 15 to get:

w = \(\frac{1125}{15}\)

Result

15w = 1,125

w = \(\frac{1125}{15}\)

Page 199 Exercise 22 Answer

We need to write a division equation and a multiplication equation to represent the problem. Note that we are not being asked to solve the equations, we only need to write them.

We know that:

total amount earned = (of weeks) . (amount earned each week)

If Felipe earns a total of $4,500 in 12 weeks when he earns m dollars each week, then the multiplication equation is:

4,500 = 12 m

To write the division equation, we can divide both sides of the equation by 12 to get:

m = \(\frac{4500}{12}\)

Result

12m = 4,500

m = \(\frac{4500}{12}\)

Page 199 Exercise 23 Answer

From the bar diagram, we know that the 3,330 toothpicks are being divided equally into 18 rows with t toothpicks in each row. The division equation is then:

3,330 ÷ 18 = t

Evaluating the left side of the equation gives 3,330 ÷ 18 = 185 so each row has t = 185 toothpicks.

Result

185 toothpicks

Page 199 Exercise 24 Answer

To write the equation, we can use the following relationship:

distance = rate . time

From the ticket, we know the distance was 2,184 miles and then travel time was 12 hours. If m represents the rate (the number of miles flown each hour), then the multiplication equation is 2,184 = m . 12, which can be written more simply as 2,184 = 12 m.

Note that the directions only asked you to write an equation so you do not need to solve this equation for m.

Result

12m = 2,184

Page 200 Exercise 25 Answer

To find the height of the triangle, we need to solve the given equation for h:

\(\frac{1}{2}\)(8h) = 44 Given equation.

(\(\frac{1}{2}\) . 8)h = 44 Use the Associative Property.

4h = 44 Multiply \(\frac{1}{2}\) and 8.

4h ÷ 4 = 44 ÷ 4 Divide both sides by 4.

h = 11 Simplify.
​
The height of the triangle is then 11 cm.

Result

11 cm

Page 200 Exercise 26 Answer

Let s be the lengths of the two congruent sides of the triangle.

The perimeter of a triangle is the sum of its side lengths so the perimeter is s + s + 8, which simplifies to 2s + 8.

It is given that the perimeter is 32 cm so 2s + 8 = 32.

The equation has two operations of multiplication and addition. To solve a two-step equation, first isolate the variable term by adding or subtracting and then isolate the variable by multiplying or dividing.

Solving the equation for s gives:

2s + 8 = 32 Perimeter equation.

2s + 8 – 8 = 32 – 8 Subtract 8 on both sides.

2s = 24 Simplify.

2s ÷ 2 = 24 ÷ 2 Divide both sides by 2.

s = 12 Simplify.
​
The length of each of the two congruent sides is then 12 cm.

Result

2s + 8 = 32 12 cm

Page 200 Exercise 27 Answer

We know that:

hours each girl spent cleaning = (total time cleaning) ÷ (number of girls)

It the total time they spent cleaning is c hours, there are 5 girls, and each girl spent 3 hours cleaning, then the division equation is:

3 = c ÷ 5

To solve this equation for c, we need to multiply both sides by 5:

3 . 5 = c ÷ 5 . 5

15 = c

Therefore, the total number of hours the girls spent cleaning is 15 hours.

Result

3 = c ÷ 5 15 hours

Page 200 Exercise 28 Answer

Let d be the number of days she drove 85 miles per day.

Since distance = (rate)(time), then the distance she drove for these d days is 85d miles.

She also drove a distance of 52 miles on the last day of her trip so the total distance she drove is 85d + 52 miles.

It is given that she traveled a total distance of 562 miles so the equation is:

85d + 52 = 562

This is a two-step equation so to solve for the variable, we must first isolate the variable term by adding or subtracting and then isolate the variable by multiplying or dividing:

85d + 52 – 52 = 562 – 52 Subtract 52 on both sides.

85d = 510 Simplify.

85d ÷ 85 = 510 ÷ 85 Divide both sides by 85.

d = 6 Simplify.
​
Therefore, Veronia drove 85 miles per day for 6 days and drove 52 miles for 1 additional day. The total number of days she traveled was 6 + 1 = 7 days.

Result

85d + 52 = 562 7 days

Page 200 Exercise 29 Answer

We know that:

of people per showing = (total of tickets) ÷ (of showings)

Let p be the number of people per showing. If the total number of tickets is 11,150 and the number of showings is 50, then the division equation is:

p = 11,550 ÷ 50

Evaluating the right side of the equation gives p = 231 so 231 people bought tickets for each showing.

Result

p = 11,550 ÷ 50 p = 231 people

Page 200 Exercise 30a Answer

We know that:

total amount of rock salt = (pounds per bag).(number of bags)

We know the school used a total of 4,920 pounds of rock salt and that each bag has 40 pounds of rock salt. If the school used b bags, the multiplication equation is:

4,920 = 40b

Comparing this equation with the given equation of 4,920b = 40, we can see that the equations are not equivalent since they don’t have the variable multiplied by the same number. We can then say that No, the equation cannot be used to find b.

Result

No

Page 200 Exercise 30b Answer

We know that:

total amount of rock salt = (pounds per bag).(number of bags)

We know the school used a total of 4,920 pounds of rock salt and that each bag has 40 pounds of rock salt. If the school used b bags, the multiplication equation is:

4,920 = 40b

Therefore, yes the given equation of 4,920 ÷ b = 40 can be used to find b.

Result

Yes

Page 200 Exercise 30c Answer

We know that:

total amount of rock salt = (pounds per bag).(number of bags)

We know the school used a total of 4,920 pounds of rock salt and that each bag has 40 pounds of rock salt. If the school used b bags, the multiplication equation is:

4,920 = 40b

Dividing both sides by b gives the following division equation:

4,920 ÷ b = 40

Therefore, yes the given equation of 4,920 ÷ b = 40 can be used to find b.

Result

Yes

Page 200 Exercise 30d Answer

We know that:

total amount of rock salt = (pounds per bag).(number of bags)

We know the school used a total of 4,920 pounds of rock salt and that each bag has 40 pounds of rock salt. If the school used b bags, the multiplication equation is:

4,920 = 40b

Dividing both sides by b gives the following division equation:

4,920 ÷ b = 40

Therefore, yes the given equation of 4,920 ÷ b = 40 can be used to find b.

Result

Yes

Chapter 4 Represent And Solve Equations And Inequalities

Section 4.5: Write And Solve Equations With Rational Numbers

Page 201 Exercise 1 Answer

When solving for unknowns involving money, we use the same process of using inverse operations to isolate the variable that we use when solving for unknowns involving whole numbers.

Page 201 Exercise 1a Answer

We know that:

cost of Lorna’s shirt + cost of Ben’s shirt = total cost

Since Lorna is on Team A, then the cost of Lorna’s shirt is $12.50. Since we don’t know what team Ben is on, let b represent the cost of Ben’s shirt. The total cost is $21.25 so the equation is:

12.50 + b = 21.25

Result

12.50 + b = 21.25 where b is the cost of Ben’s shirt.

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Page 201 Exercise 1b Answer

If Dario is correct, then the price of Ben’s T-shirt must be $11.95 if Ben is on Team B.

From Part A, the equation we need to solve to find the cost of Ben’s shirt is 12.50 + b = 21.25. To solve for b, we need to subtract 12.50 on both sides:

12.50 + b − 12.50 = 21.25 − 12.50

b = 8.75
​
Since Ben’s T-shirt costs $8.75, then Ben is on Team C, not Team B, so Dario is incorrect.

Result

Dario is incorrect. Ben is on Team C since his T-shirt costs $8.75.

Page 202 Exercise 1 Answer

\(1 \frac{3}{8}+x-1 \frac{3}{8}=2 \frac{1}{4}-1 \frac{3}{8}\)
​
\(x=1 \frac{5}{4}-1 \frac{3}{8}\)

x = \(1 \frac{10}{8}-1 \frac{3}{8}\)

x = \(\frac{10}{8}-\frac{3}{8}\)
​
x = \(\frac{7}{8}\)

The length of the shorter piece is \(\frac{7}{8}\) foot.

If the situation is different, that is we know the lenght of the shorter piece is \(\frac{7}{8}\) foot. Than the equation is:

where x is now the length of the longer of the two pieces.

Result

x = \(\frac{7}{8}\)

Page 203 Exercise 2 Answer

When solving a multiplication equation with a rational coefficient, we isolate the variable by multiplying both sides of the equation by the reciprocal of the coefficient.

For the equation \(\frac{5}{9}\)y = 25, the coefficient is \(\frac{5}{9}\), which has a reciprocal of \(\frac{9}{5}\) so solving the equation gives:

\(\frac{5}{9}\)y = 25 Given equation.

\(\frac{9}{5} \cdot \frac{5}{9} y=\frac{9}{5} \cdot 25\) Multiply both sides by \(\frac{9}{5}\).

\(y=\frac{225}{5}=45\) Simplify.

Result

y = 45

Page 203 Exercise 3 Answer

We know that:

(number of apples) . (price per apple) = total cost

Molly buys 8 apples for a total cost of $3.60. Let m be the cost of each apple. The equation is then:

8m = 3.60

To solve this equation we need to divide both sides by 8:

8m ÷ 8 = 3.60 ÷ 8

m = 0.45

Molly then paid $.45 for each apple.

Result

$0.45

Page 204 Exercise 4 Answer

Since Carmen spent $12.50 in total, the sum $6.35 + c equals $12.50.

The equation is:

$6.35 + c = $12.50.

Solve the equation to find the value of c, that is the cost of the compass.

$6.35 + c = $12.50

$6.35 + c – $6.35 = $12.50 – $6.35 (Subtract $6.35 from both sides.)
​
c = $6.15

To cost of the compass is $6.15.

Result

To cost of the compass is $6.15.

Page 205 Exercise 1 Answer

To write equations involving rational numbers, we use the same process as when writing equations involving whole numbers, which is using the given relationship to determine what operation is needed for the equation. We also use the same process when solving, which is using properties of equality and inverse relationships.

Page 205 Exercise 2 Answer

When solving equations, we need to isolate the variable by performing an operation that will undo” the operation we already have. Since inverse operations undo” each other, then inverse relationships are important for solving equations.

Result

When solving equations, we need to isolate the variable by performing an operation that will undo” the operation we already have. Since inverse operations undo” each other, then inverse relationships are important for solving equations.

Page 205 Exercise 3 Answer

When solving equations, you must always perform an operation on both sides of the equation to keep it balanced.

To solve the equation, 3.5 must be added to both sides of the equation. Johnny is then incorrect because he said he only added 3.5 to the left side, not both sides.

Result

Johhny is incorrect since 3.5 must be added to both sides of the equation, not just the left side.

Page 205 Exercise 4 Answer

When adding or subtracting fractions and mixed numbers, we always need to get a common denominator. This means that when solving addition and subtraction equations with mixed numbers, we need to rewrite the mixed number and fraction to have a common denominator before we can add or subtract.

For example, to solve y + \(\frac{3}{4}\) = \(4 \frac{1}{2}\), we must rewrite the mixed number \(4 \frac{1}{2}\) as \(4 \frac{2}{4}\) so it will have a common denominator with \(\frac{3}{4}\).

When we multiply fractions, we must always convert mixed numbers to improper fractions. This means that when solving multiplication division equations with mixed numbers, we need to rewrite the mixed number as an improper fraction before we can multiply or divide.

For example, to solve \(\frac{3}{4}\)y = \(4 \frac{1}{2}\), we must rewrite the mixed number \(4 \frac{1}{2}\) as \(\frac{9}{2}\).

Result

When solving addition and subtraction equations, you need to write the mixed number and fraction with a common denominator. When solving multiplication and division equations, you need to write the mixed number as an improper fraction.

Page 205 Exercise 5 Answer

To solve equations, we isolate the variable using inverse relationships and the properties of equality. To solve addition equations, we use the Subtraction Property of Equality, and to solve subtraction equations, we use the Addition Property of Equality since addition and subtraction are inverse operations. To solve multiplication equations, we use the Division Property of Equality, and to solve division equations, we use the Multiplication Property of Equality since multiplication and division are inverse operations.

Page 205 Exercise 6 Answer

\(t-\frac{2}{3}=25 \frac{3}{4}\) Given equation.

\(t-\frac{2}{3}+\frac{2}{3}=25 \frac{3}{4}+\frac{2}{3}\) Add \(\frac{2}{3}\) on both sides.

\(t=25 \frac{9}{12}+\frac{8}{12}\) Get a common denominator.

t = \(25 \frac{17}{12}\) Add.

t = \(26 \frac{5}{12}\) Regroup.

Result

t = \(26 \frac{5}{12}\)

Page 205 Exercise 7 Answer

\(\frac{f}{2} \times 2=\frac{5}{8} \times 2\) (Multiply both side by 2.)

\(\frac{10}{8}\) (Multiply the fractions.)

f = \(1 \frac{1}{4}\) (Simplify.)
​
Result

f = \(1 \frac{1}{4}\)

Page 205 Exercise 8 Answer

13.27 + 24.45 = t − 24.45 + 24.45 (Add 24.45 to both sides.)

37.72 = t (Evaluate.)
​
Result

t = 37.72

Page 205 Exercise 9 Answer

r ÷ 5.5 × 5.5 = 18.2 × 5.5 (Multiply both sides by 5.5.)

r = 100.1 (Evaluate.)

Result

r = 100.1

Page 205 Exercise 10 Answer

\(\frac{7}{10}+\frac{3}{5}=x-\frac{3}{5}+\frac{3}{5}\) (Add \(\frac{3}{5}\) to both sides.)

\(\frac{7}{10}\) + \(\frac{6}{10}\) = x (Add the fractions.)

\(\frac{10}{10}\) + \(\frac{3}{10}\) = x
​
\(1 \frac{3}{10}\) = x

Result

x = \(1 \frac{3}{10}\)

Page 205 Exercise 11 Answer

1.8x ÷ 1.8 = 40.14 ÷ 1.8 (Divide both sides by 1.8.)

x = 22.3 (Evaluate.)
​
Result

x = 22.3

Page 205 Exercise 12 Answer

17.3 + v − 17.3 = 22.32 − 17.3 (Subtract 17.3 from both sides.)

v = 5.02 (Evaluate.)

Result

v = 5.02

Page 205 Exercise 13 Answer

\(9 \times \frac{8}{3}=\frac{3}{8} y \times \frac{8}{3}\) (Multiply both sides by \(\frac{8}{3}\).)

\(\frac{9}{1} \times \frac{8}{3}=y\) (Multiply the fractions.)

\(\frac{72}{3}\) = y

24 = y (Evaluate.)

Result

y = 24

Page 205 Exercise 14 Answer

\(\frac{7}{4}+z=\frac{8}{3}\) (Rewrite the mixed numbers as fractions.)

\(\frac{7}{4}+z-\frac{7}{4}=\frac{8}{3}-\frac{7}{4}\) (Subtract \(\frac{7}{4}\) from both sides.)

\(z=\frac{8 \times 4}{3 \times 4}-\frac{7 \times 3}{4 \times 3}\)
​
z = \(\frac{32}{12}\) – \(\frac{21}{12}\)

z = \(\frac{11}{12}\)

Result

z = \(\frac{11}{12}\)

Page 206 Exercise 15 Answer

w − 3.2 + 3.2 = 5.6 + 3.2

w = 8.8 (Add 3.2 to both sides.)
​
Result

w = 8.8

Page 206 Exercise 16 Answer

9.6 ÷ 1.6 = 1.6y ÷ 1.6 (Divide both sides by 1.6.)

6 = y (Evaluate.)

Result

y = 6

Page 206 Exercise 17 Answer

48.55 + k − 48.55 = 61.77 − 48.55 (Subtract 48.55 to both sides.)

k = 13.22 (Evaluate.)

Result

k = 13.22

Page 206 Exercise 18 Answer

m ÷ 3.54 × 3.54 = 1.5 × 3.54 (Multiply both sides by 3.54.)

m = 5.31 (Evaluate.)
​
Result

m = 5.31

Page 206 Exercise 19 Answer

\(7 \frac{1}{9}=2 \frac{4}{5}+m\) Given equation.

\(7 \frac{1}{9}-2 \frac{4}{5}=2 \frac{4}{5}+m-2 \frac{4}{5}\) Subtract \(2 \frac{4}{5}\) on both sides.

\(7 \frac{5}{45}-2 \frac{36}{45}=m\) Get a common denominator.

\(6 \frac{50}{45}-2 \frac{36}{45}=m\) Regroup 7 \(7 \frac{5}{45}\).

\(4 \frac{14}{45}\) = m Subtract.

Result

m = \(4 \frac{14}{45}\)

Page 206 Exercise 20 Answer

\(a+3 \frac{1}{4}=5 \frac{2}{9}\) Given equation.

\(a+3 \frac{1}{4}-3 \frac{1}{4}=5 \frac{2}{9}-3 \frac{1}{4}\) Subtract \(3 \frac{1}{4}\) on both sides.

\(a=5 \frac{8}{36}-3 \frac{9}{36}\) Get a common denominator.

\(a=4 \frac{44}{36}-3 \frac{9}{36}\) Regroup \(5 \frac{8}{36}\)

a = \(1 \frac{35}{36}\) Subtract.
​
Result

a = \(1 \frac{35}{36}\)

Page 206 Exercise 21 Answer

\(\frac{1}{8}\) x y x 8 = 4 x 8 (Multiply both sides by 8.)

y = 32 (Evaluate.)

Result

y = 32

Page 206 Exercise 22 Answer

\(k-6 \frac{3}{8}=4 \frac{6}{7}\) Given equation.

\(k-6 \frac{3}{8}+6 \frac{3}{8}=4 \frac{6}{7}+6 \frac{3}{8}\) Add \(6 \frac{3}{8}\) on both sides.

\(k=4 \frac{48}{56}+6 \frac{21}{56}\) Get a common denominator.

k = \(10 \frac{69}{56}\) Add.

k = \(11 \frac{13}{56}\) Regroup.
​
Result

k = \(11 \frac{13}{56}\)

Page 206 Exercise 23 Answer

Mr. Marlon bought four tickets. Let’s mark the cost of one ticket x. Since the total cost is $210. Thus, the equation which represents this situation is:

4x = 210.
​
To answer the question find the solution of the equation.

4x = 210

x = 210 ÷ 4 (Divide both sides by 4.)
​
x = \(\frac{105}{2}\)

The cost of one ticket is $52.50.

Result

$52.50

Page 206 Exercise 24 Answer

The right sides of the given equations are equal.

However, the solution of the equation \(\frac{5}{9}\)m = \(2 \frac{3}{4}\) must be greater than the solution of the equation \(\frac{5}{8}\)m = \(2 \frac{3}{4}\).

To solve the equations, we multiply both sides by the reciprocals of the coefficients, which are \(\frac{8}{5}\) and \(\frac{9}{5}\).

\(\frac{9}{5}\) is greater than \(\frac{8}{5}\) which means that \(2 \frac{3}{4}\) x \(\frac{9}{5}\), which is the solution of the second equation, is greater than the \(2 \frac{3}{4}\) x \(\frac{8}{5}\) which is the solution of the first equation.

Result

The solution of the equation \(\frac{5}{9}\)m = \(2 \frac{3}{4}\) must be greater than the solution of the equation \(\frac{5}{8}\) m =​ \(2 \frac{3}{4}\) since the reciprocal of \(\frac{5}{9}\) is greater than the reciprocal of \(\frac{5}{8}\).

Page 206 Exercise 25 Answer

It is given that the record is 21 ft \(2 \frac{1}{4}\) in and that Tim’s best long jump is 20 ft \(9 \frac{1}{2}\) in.

Since the long jumps are measured to the nearest quarter inch, Tim must jump one-quarter inch farther than the current record in order to beat it. Tim must then jump a new record distance of:

To find how much farther Tim must jump to break the record, we then need to subtract the new record and Tim’s current long jump distance.

Before we subtract, we should convert the two lengths to inches. Since there are 12 inches in 1 foot, we can multiply the number of feet by 12 to convert it to inches:

new record:

current distance:

​\(20 \mathrm{ft} 9 \frac{1}{2} \text { in }=\left(20 \times 12+9 \frac{1}{2}\right) \text { in }=\left(240+9 \frac{1}{2}\right) \text { in }=249 \frac{1}{2} \text { in }\)

The difference of the new record and his current long jump distance is then:

Tim must then jump 5 inches farther to break the record.

Result

5 inches farther

Page 207 Exercise 26 Answer

It takes one gallon of fuel to move the space shuttle 40 feet.

To calculate how many gallons it takes to move it 3 miles, first rewrite 3 miles in feet. Since 1 mile is 5,280 feet, 3 miles is 3 times that.

3 gal = 3 x 5,280 ft

= 15,840 ft

Since it take 1 gallon per 40 feet, divide 15,840 by 4.

15,840 ÷ 40 = 396

It takes 396 gallons of fuel to move the space shuttle 3 miles.

Result

396 gallons.

Page 207 Exercise 27 Answer

The solution of b × \(\frac{5}{6}\) = 25 must be greater than 25 since the solution is b multiplied by a number less than one – that is, we can take less than the whole number, specifically \(\frac{5}{6}\) of the number b, and it will be equal to 25 so it must be greater.

Solve and compare to check the answer.

b x \(\frac{5}{6}\) = 25

\(b \times \frac{5}{6} \times \frac{6}{5}=25 \times \frac{6}{5}\) (Multiply by a reciprocal of \(\frac{5}{6}\).)

\(b=\frac{25}{1} \times \frac{6}{5}\)
​
\(b=\frac{25 \times 6}{1 \times 5}\)

b = \(\frac{150}{5}\)

b = 30 > 25

The solution of the given equation is 30 which is greater than 25.

Result

The solution is 30 which is greater than 25.

Page 207 Exercise 28 Answer

Area of a rectangle is given as a product of its width and length, that is if w is its width and l is its length, the area is A = w × l.

Since the length and the area are given, substitute and solve for w to answer the question.

w x l = A

​\(w \times \frac{3}{7} \mathrm{ft}=2 \mathrm{ft}^2\)

\(w \times \frac{3}{7} \times \frac{7}{3}=2 \times \frac{7}{3}\) (Multiply by a reciprocal of \(\frac{3}{7}\).)

w = \(\frac{2}{1}\) x \(\frac{7}{3}\) (Rewrite the whole number as a fraction.)

w = \(\frac{14}{3}\)(Multiply the fractions.)

w = \(\frac{12}{3}+\frac{2}{3}\)

w = \(4 \frac{2}{3}\)ft (Rewrite as a mixed number.)

The width of the rectangle is \(4 \frac{2}{3}\)ft.

Result

\(4 \frac{2}{3}\) ft.

Page 207 Exercise 29 Answer

Since Helen needs to fill the pool with \(10 \frac{1}{2}\) gallons of water that is one side of the equation. The other side is the product of how much she can carry in one trip – that is \(1 \frac{7}{8}\) of a gallon, and x – the number of trips she must make.

To answer the question solve the following equation:

\(1 \frac{7}{8}\)x = \(10 \frac{1}{2}\).

\(1 \frac{7}{8}\)x = \(10 \frac{1}{2}\)

\(\frac{15}{8}\)x = \(\frac{21}{2}\) (Rewrite mixed numbers as fractions)

x = \(\frac{168}{30}\)

x = \(\frac{28}{5}\)

x = \(5 \frac{3}{5}\)

Helen needs to make \(5 \frac{3}{5}\) trips, since that is not a whole number, she must make at least 6 trips to fill the whole pool.

Result

6 trips

Page 207 Exercise 30 Answer

When the pool was full there was \(10 \frac{1}{2}\) gallons of water in it, that is one side of the equation. The other side is the sum of the water which was spashed, g gallons, and the water still left in the pool, that is \(7 \frac{7}{8}\)

The equation which describes the situation is:
​
\(g+7 \frac{7}{8}=10 \frac{1}{2}\)

To answer the question solve the equation.

​\(g+7 \frac{7}{8}=10 \frac{1}{2}\)

\(g+\frac{63}{8}=\frac{21}{2}\) (Rewrite mixed numbers as fractions.)

\(g+\frac{63}{8}-\frac{63}{8}=\frac{21}{2}-\frac{63}{8}\) (Subtract \(\frac{63}{8}\) from both sides.)

g = \(\frac{84}{8}\) – \(\frac{63}{8}\)

g = \(\frac{84}{8}\) – \(\frac{63}{8}\)

g = \(\frac{84-63}{8}\)
​
g = \(\frac{21}{8}\)

g = \(\frac{16}{8}\) + \(\frac{5}{8}\)

g = \(2 \frac{5}{8}\)
​
\(2 \frac{5}{8}\) gallons was splashed out of the pool.

Result

\(2 \frac{5}{8}\) gallons.

Page 207 Exercise 31 Answer

\(2 \frac{1}{2}\)y = \(\frac{5}{8}\) (The first step is obviously the equation itself.)

\(\frac{5}{2}\)y = \(\frac{5}{8}\) (The second step is rewriting mixed numbers as fractions.)

y = \(\frac{5}{8}\) . \(\frac{2}{5}\) (The third step is multiplying by a reciprocal of \(\frac{5}{2}\).)

y = \(\frac{10}{40}\) or \(\frac{1}{4}\) (The last step is the value of y.)

Result

\(2 \frac{1}{2}\)y = \(\frac{5}{8}\)

\(\frac{5}{2}\)y = \(\frac{5}{8}\)

y = \(\frac{5}{8}\) . \(\frac{2}{5}\)
​
y = \(\frac{10}{40}\) or \(\frac{1}{4}\)

Page 208 Exercise 32 Answer

Solve the given equation, start by dividing both sides by 4.5.

4.5n ÷ 4.5 = 8,640 ÷ 4.5

n = 1920

Nontasters have 1,920 taste buds.

Result

n = 1,920

Page 208 Exercise 33 Answer

Since there are 2.25 times more women supertasters than men, and m represents the number of men classified as supertasters, the product 2.25m is the number of women supertasters, that is 72.

2.25m = 72

2.25m ÷ 2.25 = 72 ÷ 2.25(Divide both sides by 2.25.)
​
m = 32

There are 32 men classified as supertasters.

Result

32

Page 208 Exercise 34 Answer

Five times the fraction f equals \(\frac{1}{8}\). That is 5f = \(\frac{1}{8}\).

​5f = \(\frac{1}{8}\)

5f ÷ 5 = \(\frac{1}{8}\) ÷ 5 (Divide both sides by 5.)

f = \(\frac{1}{8}\) x \(\frac{1}{5}\) (Instead of dividing by 5, multiply by its reciprocal.)
​
f = \(\frac{1}{40}\)

To solve the equation divide both sides by five. Since dividing by a number is the same as multiplying by the reciprocal of that number, we can multiply \(\frac{1}{8}\) by \(\frac{1}{5}\), which is a reciprocal of 5, and get the same result.

Result

Page 208 Exercise 35 Answer

To answer the question solve the equation, start by rewriting the mixed numbers as fractions.

​\(\frac{43}{8}\) + m = 8

m = 8 – \(\frac{43}{8}\) (Subtract \(\frac{43}{8}\) from both sides.)

m = \(\frac{64}{8}\) – \(\frac{43}{8}\) (Rewrite 8 as a fraction with 8 as a denominator.)

m = \(\frac{64-43}{8}\) (Subtract the fractions.)
​
m = \(\frac{21}{8}\)

m = \(\frac{16}{8}+\frac{5}{8}\)

m = \(2 \frac{5}{8}\)

Yelena needs to swim \(2 \frac{5}{8}\) more miles.

Result

\(2 \frac{5}{8}\) miles.

Page 208 Exercise 36 Answer

Any equation written using addition can be rewritten as an equivalent equation using subtraction. One way is to subtract the same number from both sides, for example:

m + 4.56 = 55.7

m + 4.56 − 55.7 = 55.7 − 55.7

m − 51.14 = 0.

Another way is to subtract the opposite number of the one you are adding. Since −4.56 is the opposite of 4.56, the equation above would look like:

m + 4.56 = 55.7

m − (−4.56) = 55.7

Result

Yes because you can subtract the same number from both sides to get a subtraction equation or you can rewrite the addition as subtracting the opposite number.

Page 208 Exercise 37 Answer

Oscar’s expression is true now, when he is 12 years old and his sister is 6, however it will not be true in a year.

If Oscar’s age is a and his sister’s age is s, using Oscar’s expression the equation showing the relationship between their age would be:

a ÷ 2 = s.

This is true when a = 12 and s = 6, that is, it is true now.

In a year Oscar will be 13 and his sister 7, that is a = 13 and s = 7, than the eqaution

13 ÷ 2 = 7
​
not be true.

Result

No.

Page 208 Exercise 38 Answer

Solve the equation to answer the question, start by dividing both sides by 0.26.

0.26y ÷ 0.26 = 0.676 ÷ 0.26

y = 2.6

The correct answer is marked C.

Result

C 2.6

Page 208 Exercise 39 Answer

Solve the equation to answer the question – start by subtracting 0.435 from both sides.

0.435 + x − 0.435 = 0.92 − 0.435

x = 0.485

The correct answer is marked D.

Result

D 0.485

Chapter 4 Represent And Solve Equations And Inequalities

Section 4.7: Solve Inequalities

Page 217 Exercise 1 Answer

Let n be the number of which Henry is thinking. It is less than 17. The inequality which shows this is:

n < 17.
​
To show all the numbers that are less than 17 on a number line, mark 17 on the number line with an open circle since it is not included, and mark the line on the left.

Result

n < 17

Page 217 Exercise 1 Answer

Henry could not be thinking of 17, since it clearly states he is thinking of a number less than 17.

Result

No.

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Page 218 Exercise 1 Answer

To graph x < 8, start by drawing an open circle at 8 on the number line since 8 is not included as a solution.

7 and 4 are two of the many possible solutions of the inequality and both of these numbers lie to the left of 8 on the number line. To complete the graph, you then need to shade the solutions to the left of the open circle you drew at 8.

If the less than sign is changed to a greater than sign, then the only part of the graph that changes is the part you shade. Instead of shading to the left of 8, you would have to shade to the right of 8. The graph would still have an open circle at 8 since 8 would still not be included as a solution.

Result

Draw an open circle at 8 on the number line. Shade to the left of the open circle you drew at 8.

If the less than sign is changed to a greater than sign, then the only part of the graph that changes is the part you shade. Instead of shading to the left of 8, you would have to shade to the right of 8. The graph would still have an open circle at 8 since 8 would still not be included as a solution.

Page 219 Exercise 2 Answer

All the possible costs of the items on the children’s menu are marked x and for them the inequality x ≤ $8.50, is true. Also, they are obviously greater than zero.

To mark this on a number line, first mark the point 8.5 and than all the points to the left of it.

Result

x ≤ $8.50

Page 219 Exercise 3 Answer

First, write an inequality which represents the situation.

Let y mark how far a jumper has jumped. Since it states that to qualify for the finals one must jump at least \(20 \frac{1}{2}\) feet, all that qualify must have jumped either exacatly \(20 \frac{1}{2}\) feet or more.

y ≥ \(20 \frac{1}{2}\)

Check for which jumpers the inequality is true. Those are the one who would qualify.

Amir: \(22 \frac{1}{3} \geq 20 \frac{1}{2}\)

Jake: \(16 \nsucceq 20 \frac{1}{2}\)

Tyrell: \(18 \frac{1}{2} \nsucceq 20 \frac{1}{2}\)

Ryan: \(20 \frac{1}{2} \geq 20 \frac{1}{2}\)
​
Amir and Ryan would qualify for finals.

Result

Amir and Ryan would qualify.

Page 220 Exercise 1 Answer

You can represent the solutions of an inequality by making a graph on a number line. If the inequality symbol is < or >, plot an open circle on the number line and if it is ≤ or ≥, plot a closed circle on the number line. Then shade to the left if the symbol is < or ≤ and shade to the right if the symbol is > or ≥.

Page 220 Exercise 2 Answer

The inequality x > 5 means that x can be any number that is larger than 5. Since 9 is larger than 5, then 9 is a solution of x > 5.

Page 220 Exercise 3 Answer

The inequality x > 5 means that x can be any number that is larger than 5. Since 2 is not larger than 5, then 2 is not a solution of x > 5.

Page 220 Exercise 4 Answer

The inequality x > 12 means that x can be any number that is greater than 12, such as 12.1, \(15 \frac{1}{2}\), and 50. Since there are infinitely many numbers that are greater than 12, then the inequality has infinitely many solutions.

Result

infinitely many solutions

Page 220 Exercise 5 Answer

When the inequality symbol is < or >, we use an open circle and when the symbol is ≤ or ≥, we use a closed circle. We shade to the right when the symbol is > or ≥ and shade to the left when the symbol is < or ≤.

The graphs of the solutions of inequalities involving > and ≥ will then both be shaded to the right, but the > graph will have an open circle and the ≥ graph will have a closed circle.

Result

The graphs of the solutions of inequalities involving > and ≥ will both be shaded to the right, but the > graph will have an open circle and the ≥ graph will have a closed circle.

Page 220 Exercise 6 Answer

To determine which inequality symbol to choose, we need to look at what type of circle there is on the number line and what direction the graph is shaded.

The graph has an open circle at 14 so the inequality symbol will be either > or <.

The graph is shaded to the left of the open circle which means z must be smaller than 14. The inequality is then z < 14.

Result

z < 14

Page 220 Exercise 7 Answer

To determine which inequality symbol to choose, we need to look at what type of circle there is on the number line and what direction the graph is shaded.

The graph has a closed circle at 18 so the inequality symbol will be either ≥ or ≤.

The graph is shaded to the right of the closed circle which means d could be greater than 18. The inequality is then d ≥ 18.

Result

d ≥ 18

Page 220 Exercise 8 Answer

Use substitution to answer the question.

4.3 < 8

5.3 < 8

8.3 > 8

9 > 8
​
The solution to the inequality are w = 4.3 and w = 5.3, however w = 8.3 and w = 9 are not solutions.

Result

w = 4.3, w = 5.3

Page 220 Exercise 9 Answer

Use substitution to answer the question.

24 < 25

25 = 25

25.1 > 25

27 > 25
​
The solution to the given inequality are t = 25.1 and t = 27, however t = 24 and t = 25 are not solutions.

Result

t = 25.1, t = 27

Page 220 Exercise 10 Answer

Use substitution to find the answer the question.

0 ≤ 4

4 ≤ 4

5 > 4

6 > 4
​
The solution to the give inequality are g = 0 and g = 4, however g = 5 and g = 6 are not solutions.

Result

g = 0, g = 4

Page 220 Exercise 11 Answer

Use substitution to find the solution of the inequality.

4 < 8

5 < 8

6 < 8

7 < 8
​
Neither of the given values of the variable is the solution.

Result

Neither of the given values of the variable is the solution.

Page 221 Exercise 12 Answer

Number 7 beginning of the marked segment on the number line, however the point 7 itself is not marked. The marked segment is to the left of 7. Thus, the inequality that the graph represents is

y < 7.
​
Result

y < 7

Page 221 Exercise 13 Answer

Number 0 is the beginning of the marked segment on the number line, however the point 0 itself is not marked. The marked segment is to the right of 0. Thus, the inequality that the graph represents is

b > 0.
​
Result

b > 0

Page 221 Exercise 14 Answer

Number 3 is the beginning of the marked segment on the number line, however the point 3 itself is not marked, so x ≠ 3. The marked segment is to the right of 3, so all x are greater than 3. Thus, the inequality that the graph represents is

x > 3.

Result

x > 3

Page 221 Exercise 15 Answer

Number 5 is the beginning of the marked segment on the number line, and since the point 5 itself is is marked, t = 5. The marked segment is to the left of the point 5, so the all t are less than 5. Thus, the inequality that the graph represents is

t ≤ 5.

Result

t ≤ 5

Page 221 Exercise 16 Answer

To graph the inequality on a number line, first mark the point 9 with a closed circle. Since the inequality states that h are all numbers equal to or greater than 9, mark the line segment to the right of the point 9.

Result

Mark the point 9 with a closed circle and mark the line segment to the right of the point 9.

Page 221 Exercise 17 Answer

The inequality states that p are all numbers less than 3. First mark the point 3, however, since p ≠ 3, mark it only with an open circle. Mark the line to the left of the point 3.

Result

Mark the point 3 with an open circle and mark the line to the left of the point 3.

Page 221 Exercise 18 Answer

The inequality states that t are all the numbers equal to or less than 6, thus mark the point 6 with a closed circle and the line to the left of the point 6.

Result

Mark the point 6 with a closed circle and mark the line to the left of the point 6.

Page 221 Exercise 19 Answer

The inequality states that the s are all the numbers greater than 1, thus mark the point 1 with an open circle and mark the line to the right of the point 1.

Result

Mark the point 1 with an open circle and mark the line to the right of the point 1.

Page 221 Exercise 20 Answer

The solutions to the inequality are all numbers greater than 10.5, for example: 11, 321, or 10.9.

Result

11, 321, 10.9

Page 221 Exercise 21 Answer

The solutions to the equation are all the numbers less than 19. For example: 18, 1.3, and −9.

Result

18, 1.3, −9

Page 221 Exercise 22 Answer

The solutions to the inequality are all the numbers equal to or greater than 200. For example: 540, 201.34, and 200.

Result

540, 201.34, 200

Page 221 Exercise 23 Answer

The solutions to the inequality are all the numbers equal to or less than 82. For example: 82, −4, and 81.4.

Result

82, −4, 81.4

Page 221 Exercise 24 Answer

The solutions to the inequality are all the numbers equal to or greater than 12. For example: 12, 321, and 12.23.

Result

12, 321, 12.23

Page 221 Exercise 25 Answer

The solutions to the inequality are all the numbers equal to or less than 3.5. For example: 2, −3.4, and 3.5.

Result

2, −3.4, 3.5

Page 221 Exercise 26 Answer

The solutions to the inequality are all the numbers grater than 35. For example: 38, 35.4, and 342.

Result

38, 35.4, 342

Page 221 Exercise 27 Answer

The solutions to the inequality are all the numbers which are less than 2.5. For example: −2, 2, and 2.4.

Result

−2, 2, 2.4

Page 221 Exercise 28 Answer

Substitute w with 1,505 and 1,600 in the inequality and check wheather it is true.

w ≤ 1,500

1,505 ≥ 1,500

1,600 ≥ 1,500
​
Since neither of the inequalities are true when w is substituted, 1,505 pounds and 1,600 pounds are not allowed in a freight elevator.

Result

No.

Page 222 Exercise 29 Answer

In the inequality x > 2, the solution are all numbers greater than 2. In the inequality x < 2, the solution are all the numbers less than 2. Notice, 2 is not a solution to either of the inequalities.

Result

2 is not a solution to either of the inequalities.

Page 222 Exercise 30 Answer

Let t mark the temperature at the Death Valley.

Since the highest recorded temperature was 134° F, the inequality showing this is:

t ≤ 134°F.

The lowest recorded temperature was 15°F, thus the inequality showing this is:

t ≥ 15°F.
​
Result

t ≤ 134° F, t ≥ 15°F.

Page 222 Exercise 31 Answer

The solutions of the inequality x > 7 are all the numbers greater than 7, thus both 7.1 and 7.01 are solutions since both of them are greater, even though by a small difference, but still they are greater. That is, if we look at the number line they are to the right of the point 7 and are thus solutions.

Result

Both 7.1 and 7.01 are solutions of the inequality.

Page 222 Exercise 32 Answer

Since the temperature should be either exactly 65 degrees or higher the inequality describing the situation is:

t ≥ 65°,
​
where t represents the allowable temperature in the greenhouse.

Result

t ≥ 65°

Page 222 Exercise 33 Answer

Since the gift card’s value is enough to buy any of the apps thus it must be equal to or greater than the most expensive app.

$9.50 < $10.50 < $12.00
​
The most expensive app is the Remote Desktop which costs $12.00. Thus, the value of the gift card must be at least $12.00. If the value of the gift card is markedv, the inequality which represents this situation is:

v ≥ $12.00.

Result

v ≥ $12.00

Page 222 Exercise 34 Answer

Since 400 pounds is the maximum weight, w, the weight on the plane must be either equal to or less than 400 pounds.

w ≤ 400.
​
Result

w ≤ 400

Page 222 Exercise 35 Answer

Inequality n > 21 is true for all numbers greater than 21, that is, n = 22, 23, 24, 25, …

Result

n = 22, 23, 24, 25, …

Page 222 Exercise 36 Answer

The first two tick marks on the number line are 3.0 and 3.1 so we know the number line has an interval of 0.1. Finish labeling the number line with 3.2 through 3.9.

To graph the inequality y < 3.7, we need to start by plotting an open circle at 3.7 since the inequality symbol doesn’t have an equal sign. Since y must be a number smaller than 3.7, we then need to shade to the left of 3.7 on the number line:

Result

Finish labeling the number line with 3.2 through 3.9. Plot an open circle at 3.7 and then shade to the left.

Page 222 Exercise 37 Answer

The first two tick marks on the number line are 20 and 21 so we know the number line has an interval of 1. Finish labeling the number line with 23 through 28.

To graph the inequality x ≤ 25, we need to start by plotting a closed circle at 25 since the inequality symbol has an equal sign. Since x can be a number smaller than 25, we then need to shade to the left of 25 on the number line:

Result

Finish labeling the number line with 23 through 28. Plot a closed circle at 25 and then shade to the left.

Chapter 4 Represent And Solve Equations And Inequalities

Section 4.6: Understand And Write Inequalities

Page 211 Exercise 1 Answer

Since Fran won the contest, his was the heaviest pumpkin. Every other pumpkin than must have weighed less. The mathematical statement describing this is an inequality:

217 > w,

where w is the weight of any other pumpkin.

Result

217 > w

Page 211 Exercise 1 Answer

Since the record is 24.49 seconds, to break the record Camilla has to be faster, that is her time must be less than the current record.

Some possible times are: 24.40, 23.50; anything less than 24.49.

The answer can be described by an inequality

24.49 > t,

where t is the time in which she need to swim 50 meters to break the record.

Result

24.49 > t

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Page 212 Exercise 1 Answer

From Example 1, we know that any child who is less than 8 years old has to be accompanied by an adult.

This would mean that any child who is at least 8 years old does not need to be accompanied by an adult. The phrase at least” is equivalent to the phrase greater than or equal to” so the inequality that represents the ages of people, n, who do not need to be accompanied by an adult is:

n ≥ 8

To show some ages of people who do not need to be accompanied by an adult on the number, plot some points at values that are 8 or bigger:

For the inequality n ≥ 8, an age of 8 years old is included so we know that an 8-year-old belongs in the group of those who do not need to be accompanied by an adult.

Result

n ≥ 8

For the inequality n ≥ 8, an age of 8 years old is included so we know that an 8-year-old belongs in the group of those who do not need to be accompanied by an adult.

Page 213 Exercise 2a Answer

Let h represent Harry’s height.

Since Harry is taller than 60 inches, then his height must be greater than 60 inches. The inequality is then h > 60.

Result

h > 60

Page 213 Exercise 2b Answer

Let s represents Sherry’s age.

Since Sherry is not 4 years old, then her age is not equal to 4. The inequality is then s ≠ 4.

Result

s ≠ 4

Page 213 Exercise 2c Answer

Let m be the amount of money that Hank has.

Since Hank has at least $7.50, then the amount of money he has is greater than or equal to to $7.50. The inequality is then m ≥ 7.50.

Result

m ≥ 7.50

Page 214 Exercise 1 Answer

To write an inequality to describe a situation, we start by choosing a variable to represent the unknown quantity. We then use the given relationship to determine what inequality symbol we need to use. It can help to show some of the possible values of the variable on a number line to determine which inequality symbol to choose.

Page 214 Exercise 2 Answer

An equation with a variable has an equal sign: =. An inequality with a variable has an inequality symbol: <, ≤, >, ≥, or ≠.

For example, x + 2 = 5 is an equation with a variable while x + 2 < 5 is an inequality with a variable.

Result

An equation with a variable has an equal sign: =. An inequality with a variable has an inequality symbol: <, ≤, >, ≥, or ≠.

Page 214 Exercise 3 Answer

It is more efficient to use an inequality to give all the quantities less than 6. This is because there is an infinite number of quantities that are less than 6 so it is impossible to list them all.

For example, if you tried to list them as 5, 4, 3, 2, 1, 0, −1, −2 then your list would be missing all of the fractions and decimals that lie between each pair of integers, such as 4.5, \(3 \frac{1}{4}\), and −2.1, and would be missing all of the integers that are smaller than −2.

Result

It is more efficient to use an inequality to give all the quantities less than 6. This is because there is an infinite number of quantities that are less than 6 so it is impossible to list them all.

Page 214 Exercise 4 Answer

The greater than symbol, >, and the greater than or equal to symbol, ≥, are related in that they both mean the variable can be bigger than some value.

For example x > 2 and x ≥ 2 both mean that x can be bigger than 2. The difference is that ≥ also includes the value so x ≥ 2 means that x could also be equal to 2 while x > 2 only means x is bigger than 2.

Result

The greater than symbol, >, and the greater than or equal to symbol, ≥, both mean the variable can be bigger than some value.

Page 214 Exercise 5 Answer

Write an inequality which states that n is greater than 22.

n > 22

Result

n > 22

Page 214 Exercise 6 Answer

Write an inequality which states that v is not equal to \(2 \frac{1}{2}\).
​
v ≠ \(2 \frac{1}{2}\)

Result

v ≠ \(2 \frac{1}{2}\)

Page 214 Exercise 7 Answer

Write an inequality which states that Sally’s age, which is marked a, is at most 15.

a ≤ 15

Result

a ≤ 15

Page 214 Exercise 8 Answer

Write an inequality that states that the width of the picture, which is marked w, is shorter than 8.5 inches.

w < 8.5

Result

w < 8.5

Page 214 Exercise 9 Answer

Write an inequality that states that Steve’s height, which is marked h, is at least 48 inches.

h ≥ 48

Result

h ≥ 48

Page 214 Exercise 10 Answer

Write an inequality that states that the baby’s age, which is marked b, is not 24 months.

b ≠ 24

Result

b ≠ 24

Page 214 Exercise 11 Answer

Write an inequality that states that the number of quarters in the jar, which is marked q, is less than 75.

q < 75

Result

q < 75

Page 214 Exercise 12 Answer

Write an inequality that states that the length of the fish, which is marked f, is at least 10 inches.

f ≥ 10
​
Result

f ≥ 10

Page 215 Exercise 13 Answer

It is given that p represents the number of people that can ride in the van.

If the number of people is up to 12, then the number of people must be less than or equal to 12. The inequality is then p ≤ 12.
​
Result

p ≤ 12

Page 215 Exercise 14 Answer

Write an inequality which states that the number of days of sunshine, which are marked d, is not 28.

d ≠ 28

Result

d ≠ 28

Page 215 Exercise 15 Answer

Write an inequality which represents that the distance of the race, which is marked r, is farther than 6.2 miles.

r > 6.2
​
Result

r > 6.2

Page 215 Exercise 16 Answer

Write an inequality which states that the value of the bracelet, which is marked v, is less than $85.25.

v < $85.25
​
Result

v < $85.25

Page 215 Exercise 17 Answer

Write an inequality which states that the number of people that can be seated in the restaurant, which is marked p, is not greater than 171.

p ≤ 171
​
Result

p ≤ 171

Page 215 Exercise 18 Answer

Write an inequality which states that the time the customer has left, which is marked t, is at least 25 minutes.

t ≥ 25
​
Result

t ≥ 25

Page 215 Exercise 19 Answer

Write an inequality which states that the bill, which is marked b, is less than $45.

b < $45

Result

b < $45

Page 215 Exercise 20 Answer

Write an inequality which states that the girls do not live \(7 \frac{1}{2}\) block apart, where b represents the number of blocks.

b ≠ \(7 \frac{1}{2}\)

Result

b ≠ \(7 \frac{1}{2}\)

Page 215 Exercise 21 Answer

Write an inequality which states that the speed must be no less than 34 miles per hour, where s represents the speed.

s ≥ 34 mph

Result

s ≥ 34

Page 215 Exercise 22 Answer

Write an inequality which states that the number of baseball games is more than 5, where x marks the number of baseball games.

x > 5
​
Result

x > 5

Page 215 Exercise 23 Answer

Since Mia is taller than Gage and m represents Mia’s height and g Gage’s, the inequality is m > g.

Result

m > g

Page 216 Exercise 24 Answer

If d represents the amount of money Taryn earned and she sold at least 15 rolls, $8 per roll, than

d ≥ 15 x $8,

d ≥ $ 120

Result

d ≥ $120

Page 216 Exercise 25 Answer

In one day it fell 19.7 inches of snow. To beat the record it would have to fall more than 19.7 inches. The inequality showing this is:

19.7 < x,

where x represents the number of inches of snow that would have to fall in one day.

Result

19.7 < x

Page 216 Exercise 26 Answer

Since the first bookcase can hold one book less than the second bookcase, the inequality representing this relationship is:

a < 2,492.

Result

a < 2,492

Page 216 Exercise 27 Answer

Since the airplane can carry precisely 134 passengers, thus any more would not be allowed. The inequality which represents that is:

134 < p.
​
Result

134 < p

Page 216 Exercise 28 Answer

Since a rider must be at least 42 inches tall, if he or she would be exactly 42 inches tall they would be allowed to ride. Thus, Elias is correct.

h ≥ 42

Result

Elias.

Page 216 Exercise 29 Answer

To answer the question, first calculate the total number of hours Miguel will work.

7 + 9 + 5 + 7 = 28

Miguel will work 28 hours total, however Heather did not count the number of hours he will work on Saturday the second weekend. Thus, her solution is not correct.

To calculate how much per hour Miguel needs to earn, divide the cost of the bike by the number of hours he will work.

$140 ÷ 28 = 5

Miguel needs to earn at least 5.00 per hour. The inequality which shows this is: m ≥ $5, where m$ is the amount of money Miguel needs to earn.

Result

Heather did not count the number of hours Miguel will work on saturday the second weekend so her solution is not correct. The inequality m ≥ $5 represents the amount of money Miguel needs to earn.

Chapter 4 Represent And Solve Equations And Inequalities

Section 4.8: Understand Dependent And Independent Variables

Page 227 Exercise 1a Answer

Three factors that will affect the weight of the box are:

The dimensions of the box. Larger dimensions will weigh more than smaller dimensions.

The contents of the box. Heavier contents will give a larger total weight than lighter contents. The weight of the contents of the box will change based on what present is in the box and how much/what kind of packing material is used to protect the fragile item.

The material used to make the box. Thicker cardboard will weigh more than thinner cardboard.

Three factors that will not affect the weight of the box are:

The orientation of the box. Rotating the box so a different side is on the bottom will not change its weight.

The location of the stickers on the box.

The placement of the contents inside of the box. Placing the item in the middle or closer to one side will not change the overall weight of the box, it will just make the box unbalanced.

Result

Three factors that will affect the weight of the box are the dimensions of the box, the contents of the box, and the material used to make the box. Three factors that will not affect the weight of the box are the orientation of the box, the location of the stickers on the box, and the placement of the contents inside of the box.

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Page 227 Exercise 1b Answer

The size of the box affects the weight of the box because a larger box means more cardboard and more tape for the box. The contents of the box affect the weight because a heavier present will increase the total weight of the box. Also, the packing material used inside the box to protect the fragile item will affect the weight. Some packing materials weigh more than others and a larger box would also need more packing materials than a smaller box.

Page 227 Exercise 1 Answer

Possible answer:

Another situation in which changing one factor results in changed to another factor could be wrapping a present. In this situation, the contents of the box would not affect the amount of wrapping paper needed but the size of the box would. Increasing the dimensions of the box would mean the amount of wrapping paper needed would increase.

Result

Possible answer: When wrapping a present, changing the size of the box will affect how much wrapping paper is needed.

Page 228 Exercise 1 Answer

When the baker is making the pancakes, the amount of batter used affects the number of pancakes that can be made. Using more batter means the baker can make more pancakes. Decreasing the amount of batter decreases the number of pancakes that can be made.

The variable p pancakes is then the dependent variable since it changes in response to the number of cups of batter b that are used.

If the baker doubled the number of cups of batter used, I would expect the number of pancakes to double since the number of cups of batter used determines how many pancakes are used.

Result

The variable p pancakes is the dependent variable since it changes in response to the number of cups of batter b that are used. If the baker doubled the number of cups of batter used, I would expect the number of pancakes to double since the number of cups of batter used determines how many pancakes are used.

Page 229 Exercise 2 Answer

From Example 2, we know the independent variables are the length of the rental l, the price per day d, and the price per week w.

Result

The independent variables are the length of the rental l, the price per day d, and the price per week w.

Page 230 Exercise 1 Answer

When a variable is dependent on another variable, that is the dependent variable depends on the independent variable, when the independent variable changes, the dependent variable also changes according to a particular rule.

Result

When the independent variable changes, teh dependent variable also changes.

Page 230 Exercise 2 Answer

The more miles they bike, the more calories they burn. That is, when m, the number of miles cycled, changes, so does c, the number of calories burnt.

Thus, Viola is right. The number of calories, c, depends on the number of miles cycled, m.

Result

Yes.

Page 230 Exercise 3 Answer

In this problem, the dependent variable is the number of calories burnt in an hour, c.

It could depend on how fast they cycle. If they cycle faster, they would burn more calories.

Result

Possible answer: The number of calories burnt could also depend on how fast they cycle. If they cycle faster, they would burn more calories.

Page 230 Exercise 4 Answer

The dependent variable is the amount of money, m, since it depends on the amount of tickets which are sold, t – the independent variable.

Result

m is the dependent variable and t the independent.

Page 230 Exercise 5 Answer

Since the amount of money is greater the more hours one works, the amount of money, m, is the dependent variable and the number of hours worked, h, is the independent variable.

Result

m is the dependent variable and h the independent.

Page 230 Exercise 6 Answer

The more shelves the bookcase has, the more books it can hold. Thus, the number of books, b, the bookcase can hold is the dependent variable and the number of shelves, s, is the independent variable.

Result

b is the dependent variable and s the independent.

Page 230 Exercise 7 Answer

The more hours you read, the more pages you will read. Thus, the number of pages read, p, is the dependent variable and the hours you spent reading, h, is the independent variable.

Result

p is the dependent variable and h is the independent.

Page 230 Exercise 8 Answer

The longer the water runs, the more gallons the hose will produce. Thus, the number of gallons, g, is the dependent variable since it depends on the number of the minutes, m, the water is running, which is the independent variable.

Result

g is the dependent variable and m is the independent.

Page 230 Exercise 9 Answer

The number of peaches is greater the more bushes there are. Thus, the number of peaches, y, is the dependent variable and the number of bushes, x, is the independent variable.

Result

y is the dependent variable and x is the independent.

Page 230 Exercise 10 Answer

The faster you drive the faster you will arrive, that is the number of hours will be less. Thus, the number of hours, h, is the dependent variable and the speed, r, is the independent variable.

Result

h is the dependent variable and r is the independent.

Page 230 Exercise 11 Answer

The more hours the lights are on, the greater the electric bill will be. Also, the greater the price of the electricity, the greater the bill will be.

Thus, one of the independent variable that can change the monthly electric bill is the number of hours one spends with the lights on and the other such variable is the price of electricity.

Result

The number of hours one spends with the lights on and the price of electricity result in a change in a monthly electric bill.

Page 231 Exercise 12 Answer

The more pages the book has, the more it weighs. Thus, the weight, w, is the dependent variable, and the number of pages, p, in a book is the independent variable.

Result

w is the dependent variable and p is the independent.

Page 231 Exercise 13 Answer

The more hamburgers are sold, the greater the amount of money earned. Thus, the amount of sales, s, is the dependent variable and the number of hamburgers, h, is the independent variable.

Result

s is the dependent variable and h is the independent.

Page 231 Exercise 14 Answer

It is implied that you already know the number of bread loaves you want to make and than go buy flour. The more bread loaves you want to make, the more pounds of flour you will have to buy.

Thus, the number of pounds of flour you buy, p, is the dependent variable and the number of bread loaves, b, is the independent variable.

Result

p is the dependent variable and b is the independent.

Page 231 Exercise 15 Answer

The longer the water is in the freezer, the lower its temperature is.

Thus, the temperature, t, is the dependent variable, and the number of minutes the water is in the freezer, m, is the independent variable.

Result

t is the dependent variable and m is the independent.

Page 231 Exercise 16 Answer

Possible answer:

The more you study for a test, the better your test score will be. This means your test score depends on how many hours you study. The independent variable is then the number of hours, h, that you study and the dependent variable is your test score, s.

Result

Possible answer: The more you study for a test, the better your test score will be so the independent variable is the number of hours, h, that you study and the dependent variable is your test score, s.

Page 231 Exercise 17 Answer

The more apples there are in a basket, the more the basket of apples will cost.

The greater the price of one apple, the greater the price of a basket of apples.

Independent variables that could result in a change of the price of a basket of apples are: the number of apples in a basket and the price of one apple.

Result

The number of apples in a basket and the price of one apple.

Page 231 Exercise 18 Answer

The total amount you spend on pants depends on the number of pants you purchase. This means the cost c dollars is the dependent variable and the number of pairs of pants p that you buy is the independent variable. Your friend is then wrong.

Result

Your friend is wrong. The total cost depends on the number of pants purchased so the cost c dollars is the dependent variable and the number of pairs of pants p is the independent variable.

Page 231 Exercise 19 Answer

The more oranges there are in a bag, the more the bag of oranges will cost. Thus, the independent variable is the number of oranges in a bag.

Result

The number of oranges in a bag.

Page 231 Exercise 20 Answer

The growth of a plant depends on the amount of light the plant gets and the amount of nutrients and water the plant gets.

The growth of a plant could also be tracked in terms of time, such as daily or weekly.

Thus, the independent variables in this situation are the amount of light, the amount of nutrients and water, and the amount of time.

Result

The independent variables in this situation are the amount of light, the amount of nutrients and water, and the amount of time.

Page 232 Exercise 21 Answer

We know that distance = (rate)(time). The total distance they drive each day is then affected by their rate (the speed they are driving) and the time (the number of hours they drive).

Possible independent variables that would affect the total distance they drove are then their speed and the number of hours they drive per day.

Result

Their speed and the number of hours they drive per day are two possible independent variables that would affect the total distance they drove each day.

Page 232 Exercise 22 Answer

Some of the factors that would affect the amount of money the Williams family spends on meals during their vacation are the number of people in the family, the total cost for the parents, the total cost for the children, the number of days of the trip, and the cost for each meal.

The cost for the children is affected by factors such as how many children are in the family and the average cost per meal for each child. Since the cost for the children depends on another quantity, then the total cost for the children is a dependent variable.

The cost of each meal is affected by factors such as where the family eats, the sizes of the meals, whether it’s from the regular menu or a childrens’ menu, etc. Since the cost of each meal depends on another quantity, then the cost of each meal is a dependent variable.

Result

Two possible dependent variables are the total cost for the children and the cost of each meal.

Page 232 Exercise 23 Answer

Some of the factors that affect the cost of a salad are its size, the total cost of its ingredients, the amount of a specific ingredient used, and what restaurant it is from.

Two possible independent variables that would affect the cost of a salad are then its size and how much lettuce is used.

Result

Two possible independent variables that would affect the cost of a salad are its size and how much lettuce is used.

Page 232 Exercise 24 Answer

We know that distance = (rate)(time) so the distance Julian drives is affected by both rate and time.

Two possible independent variables that could affect the number of days Julian took to make the trip are then his speed (which is the rate in the distance equation) and the number of hours per day that he drove (which is the time in the distance equation).

Result

Two possible independent variables that could affect the number of days Julian took to make the trip are his speed and the number of hours per day that he drove.

Page 232 Exercise 25 Answer

The more incorrect answers a person gives on a math test, the lowest their test score is. The person’s test score then depends on the number of incorrect answers. Therefore, the dependent variable is the math test score.

Result

The dependent variable is the math test score.

Page 232 Exercise 26 Answer

Possible answers:

If you drive a total distance of d miles in t hours at an average speed of 60 miles per hour, then the total distance d you drive depends on the number of hours t. Therefore, t is the independent variable in this situation.

If you drove a total distance of 500 miles at an average speed s for t hours, then the number of hours your trip takes depends on your speed s. Therefore, t is the dependent variable in this situation.

Result

Possible answers:

If you drive a total distance of d miles in t hours at an average speed of 60 miles per hour, then t is the independent variable.

If you drove a total distance of 500 miles at an average speed s for t hours, then t is the dependent variable.

Page 232 Exercise 27a Answer

We are given the following table:

As we can see from the table, the newer machines use fewer gallons of water. One possible independent variable that could affect the amount of water Jonas uses to do the laundry could then be the age of the washing machine.

Also from the table, we can see the more energy-efficient model uses fewer gallons of water so a second possible independent variable could be the amount of energy used to run the machine.

Larger washing machines will require more gallons of water so a third possible independent variable could be the size of the washing machine.

Result

Possible answers: age of the washing machine, the amount of energy used, the size of the washing machine

Page 232 Exercise 27b Answer

The total cost, c, depends on the number of gallons of water, g, that he uses. This is because an increase in the number of gallons will cause the cost to increase.

The dependent variable is then the total cost, c, and the independent variable is the number of gallons of water, g.

Result

The dependent variable is the total cost, c, and the independent variable is the number of gallons of water, g.

enVisionmath 2.0: Grade 6, Volume 1 Chapter 4 Represent And Solve Equations And Inequalities

Section 4.9: Use Patterns To Write And Solve Equations

Page 233 Exercise 1 Answer

From the solve & Discuss It! problem, we know the number of candles is equal to 4 times the number of boxes.

If b is the number of boxes and c is the number of candles, then this relationship gives us the equation:

of candles = 4 x of boxes

c = 4b

Result

c = 4b

Page 233 Exercise 1 Answer

We are given the following table:

From the table, we can see that the number of candles is equal to 4 times the number of boxes since 2 x 4 = 8, 3 x 4 = 12 and 4 x 4 = 16.

To find the number of candles for 10 boxes, we must then multiply this number by 4:

4 x 10 = 40

Therefore, 10 boxes has 40 candles.

Result

The number of candles is equal to 4 times the number of boxes.

10 boxes has 40 candles.

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Page 234 Exercise 1 Answer

24.4 ÷ 2 = 12.2

36.6 ÷ 3 = 12.2

61 ÷ 5 = 12.2

73.2 ÷ 6 = 12.2
​
Notice, when y is divided by s the result is 12.2, for all given values of s and y. In other words, y is equal to 12.2 times the value of s.

Thus, the rule can be written as:

y = 12.2s.
​
If the cyclist maintains this speed and rides for 8 seconds, that is s = 8,

y = 12.2s

= 12.2(8)

= 97.6,

he will cross 97.6 yards.

The equation y = 12.2s describes the pattern in the table because it is true for all the values of s and y that are given.

Result

97.6 yards.

Page 235 Exercise 2 Answer

From Example 2, we know the equation a = $75 − 5w represents the amount still owed, a, after Ethan has made $5 payments each week to his mother for w weeks.

Also from the Example, we also know that after 12 weeks, Ethan still owes $15. Since he makes payments of $5 each week and 15 ÷ 5 = 3, then Ethan will still need to make payments to his mother for 3 more weeks.

Result

After 12 weeks he still owes $15 so if he pays her $5 each week, he will still need to make payments to his mother for 3 more weeks.

Page 236 Exercise 1 Answer

We can use a pattern to write and solve an equation by comparing the values of the independent and dependent variables. Using our comparison, we can write the relationship between the variables in words. Using the relationship in words, we can then write the equation by replacing each quantity in the verbal description with its variable.

For example, consider the following relationship given in words:

The cost is 5 times the number of tickets.

Replacing the cost” with the variable $c$ and replacing the number of tickets” with the variable t then gives the equation c = 5t.

Result

We can use a pattern to write and solve an equation by comparing the values of the independent and dependent variables. Using our comparison, we can write the relationship between the variables in words. Using the relationship in words, we can then write the equation by replacing each quantity in the verbal description with its variable.

Page 236 Exercise 2 Answer

To find a pattern that relates the values in a table, we need to compare the values of the independent variable and the dependent variable. Using our comparison, we can determine what operation(s) and what number (s) must be used with the values of the independent variable to get the values of the dependent variable.

For example, consider the following table:

Comparing the values of the variables, we can see that the cost is always 5 times the number of tickets since 1 x 5 = 5, 2 x 5 = 10, 3 x 5 = 15, and 4 x5 = 20.

Result

To find a pattern that relates the values in a table, we need to compare the values of the independent variable and the dependent variable. Using our comparison, we can determine what operation(s) and what number(s) must be used with the values of the independent variable to get the values of the dependent variable.

Page 236 Exercise 3 Answer

In Example 2, the following table was made:

From this table, we can see that each time the number of weeks w increases by 1, the amount owed decreases by $5, which makes sense since he pays his mom $5 each week so the amount he still owes should also decrease by $5.

Result

As the number of weeks w increases by 1, the amount owed a decreases by $5.

Page 236 Exercise 4 Answer

We age given the following table:

We need to find a pattern for this table so we can write an equation. The y-values are bigger than the x-values so the operation might be addition or multiplication.

We know that 1 + 6 = 7 but 2 + 6 = 8 ≠ 12 so addition is not the pattern.

We know that 1 x 7 = 7 but 2 x 7 = 14 ≠ 12 so multiplication is also not the pattern.

The pattern will then need two operations in it.

To find one of those operations, we can determine how the value of y changes when the value of x increases by 1. From the table, the values of y increase by 5 each time x increases by 1 since 7 + 5 = 12, 12 + 5 = 17, and 17 + 5 = 22. This means that the pattern will have 5 times x in it.

Knowing that we need 5 times the value of x in the pattern, we can rewrite the values of y as:

7 = 5 + 2 = 5(1) + 2

12 = 10 + 2 = 5(2) + 2

17 = 15 + 2 = 5(3) + 2

22 = 20 + 2 = 5(4) + 2

y = 5(x) + 2

The pattern as an equation is then y = 5x + 2

Result

y = 5x + 2

Page 236 Exercise 5 Answer

We are given the following table:

From the table, we can see that Brenda’s age, b, is equal to Talia’s age, t, minus 5 since 7 – 5 = 2, 9 – 5 = 4, and 10 – 5 = 5. The pattern as an equation is then b = t – 5.

If Talia is 12, then t = 12. Using the equation gives:

b = t – 5 = 12 – 5 = 7

Therefore, Brenda is 7 when Talia is 12.

Result

Brenda’s age, b, is equal to Talia’s age, t, minus 5.

b = t − 5

Brenda is 7 when Talia is 12.

Page 236 Exercise 6 Answer

Use substitution – substitute x with its value, that is with 7 and 8, and calculate the value of y which depends on the value of x.

y = 2x − 7

= 2(7) − 7

= 14 − 7

= 7
​
If x = 7, then y = 7.

= 2x − 7

= 2(8) − 7

= 16 − 7

= 9

If x = 8, then y = 9.

Result

If x = 7, then y = 7. If x = 8, then y = 9.

Page 236 Exercise 7 Answer

The rule for the pattern as an equation is y = 2x − 7 so the rule for the pattern in words is:

The variable y is equal to the value of 2 times x minus 7.

Result

The variable y is equal to the value of 2 times x minus 7.

Page 237 Exercise 8 Answer

We are given the following table:

From the table, we can see that value of y is always 32 more than the value of x since:

33 = 1 + 32

34 = 2 + 32

35 = 3 + 32

36 = 4 + 32

37 = 5 + 32

The rule is then: The value of y is equal to the value of x plus 32.

As an equation, this rule is: y = x + 32

Result

Rule: The value of y is equal to the value of x plus 32.

Equation: y = x + 32

Page 237 Exercise 9 Answer

We are given the following table:

From the table, we can see that value of n is always 3 times the value of m since:

0 = 3 x 0

3 = 3 x 1

6 = 3 x 2

9 = 3 x 3

12 = 3 x 4

The rule is then: The value of n is equal to 3 times the value of m.

As an equation, this rule is: n = 3m.

Result

Rule: The value of n is equal to 3 times the value of m.

Equation: n = 3m

Page 237 Exercise 10 Answer

We are given the following table:

From the table, we can see that the value of k is always 15 less than the value of g:

17 = 32 – 15

22 = 37 – 15

27 = 42 – 15

The rule is then: The value of k is equal to the value of g minus 15.

The rule as an equation is: k = g – 15

To complete the table, we can substitute g = 47 and g = 52 into the equation and then simplify:

g = 47 : k = g – 15 = 47 – 15 = 32

g = 52 : k = g – 15 = 52 – 15 = 37

The completed table is then:

Result

Rule: The value of k is equal to the value of g minus 15.

Equation: k = g – 15

Page 237 Exercise 11 Answer

We are give the following table:

From the table, we can see that the value of y is always the value of x divided by 9:

0 = 0 ÷ 9

1 = 9 ÷ 9

2 = 18 ÷ 9

The rule is then: The value of y is equal to the value of x divided by 9.

The rule as an equation is: y = x ÷ 9

To complete the table, we can substitute x = 27 and x = 36 into the equation and then simplify:

x = 27 : y = x ÷ 9 = 27 ÷ 9 = 3

x = 36 : y = x ÷ 9 = 36 ÷ 9 = 4

The completed table is then:

Result

Rule: The value of y is equal to the value of x divided by 9.

Equation: y = x ÷ 9

Page 237 Exercise 12 Answer

We are given the following table:

To write the equation, it would help to find the cost for each teddy bear. To find the cost for 1 teddy bear, we can divide one of the costs from the table by its corresponding number of teddy bears:

$188,000 ÷ 4 = $47,000

Each teddy bear then costs $47,000 which means the cost is equal to 47,000 times the number of teddy bears:

188,000 = 47,000(4)

329,000 = 47,000(7)

517,000 = 47,000(11)

c = 47,000(n)

The equation is then c = 47,000n

Result

c = 47,000n

Page 237 Exercise 13a Answer

We are given the following table:

We are also given to the right of the table the following information:

General Admission: $8.00

Rides: $2.50

Since the cost of each ride is $2.50, then the cost for 5 rides is $2.50 more than the cost for 4 rides. The cost for 5 rides is then $18.00 + $2.50 = $20.50.

Since the cost of 1 ride is $2.50, then the cost of 2 rides is 2 x $2.50 = $5.00. The cost for 8 rides is then $5.00 more than the cost for 6 tides so the cost is $23.00 + $5.00 = $28.00.

The completed table is then:

Since the total cost is the cost for the rides plus the general admission of $8.00 and the cost for the rides is equal to 2.50 times the number of rides, we can rewrite the costs in the table as follows:

$15.50 = 7.50 + 8.00 = 2.50(3) + 8.00

$18.00 = 10.00 + 8.00 = 2.50(4) + 8.00

$20.50 = 12.50 + 8.00 = 2.50(5) + 8.00

$23.00 = 15.00 + 8.00 = 2.50(6) + 8.00

$28.00 = 20.00 + 8.00 = 2.50(8) + 8.00

c = 2.50(r) + 8.00

The pattern as an equation is then: c = 2.5r + 8

Result

c = 2.5r + 8

Page 237 Exercise 13b Answer

From part (a), we know the total cost, c, is represented by the equation c = 2.5r + 8 where r is the number of rides.

To find the cost for 12 rides, we can then substitute r = 12 into the equation and simplify:

c = 2.5r + 8 = 2.5(12) + 8 = 30 + 8 = 38

The cost for 12 rides is then $38.

Result

$38

Page 238 Exercise 14 Answer

Notice, for each w and z, the value of z is two less than the value of w. That is:

2 − 2 = 0

4 − 2 = 2

6 − 2 = 4

8 − 2 = 6

10 − 2 = 8.
​
Thus, the pattern is:

z = w − 2.
​
Result

z = w − 2

Page 238 Exercise 15 Answer

Comparing the values of x and y, we see that:

4 x 0 = 0

4 x 1 = 4

4 x 2 = 8

Thus, the value of y is 4 times the value of x so the pattern is y = 4x

Result

y = 4x

Page 238 Exercise 16 Answer

Use substitution – substitute d with its value, that is with 3 and 4, and calculate the value of t.

t = 5d + 5

= 5(3) + 5

= 15 + 5

= 20
​
If d = 3, then t = 20.

t = 5d + 5

= 5(4) + 5

= 20 + 5

= 25
​
If d = 4, then t = 25.

Result

If d = 3, then t = 20. If d = 4, then t = 25.

Page 238 Exercise 17 Answer

Use substitution – substitute x with its value, that is with 8 and 10, and calculate the value of y.

y = \(\frac{1}{2}\)x − 1
​
= \(\frac{1}{2}\)(8)−1

= 4 − 1

= 3
​
If x = 8, then y = 3.

y = \(\frac{1}{2}\)x − 1

= \(\frac{1}{2}\)(10)−1

= 5 − 1

= 4

If x = 10, then y = 4.

Result

If x = 8, then y = 3. If x = 10, then y = 4.

Page 238 Exercise 18 Answer

Use substitution – substitute x with its value, that is with 2 and 3, and calculate the value of y.

y = 2x + 1

= 2(2) + 1

= 4 + 1

= 5
​
If x = 2, then y = 5.

y = 2x + 1

= 2(3) + 1

= 6 + 1

= 7
​
If x = 3, then y = 7.

Result

If x = 2, then y = 5. If x = 3, then y = 7.

Page 238 Exercise 19 Answer

Use substitution – substitute a with its value and calculate the value of b.

b = \(\frac{a}{2}\) – 2

= \(\frac{17}{2}\) – 2

= \(\frac{17}{2}\) – \(\frac{4}{2}\)

= \(\frac{17-4}{2}\)

= \(\frac{13}{2}\)

= \(\frac{12}{2}\) + \(\frac{1}{2}\)

= \(6 \frac{1}{2}\)

If a = 17, then b = \(6 \frac{1}{2}\)

b = \(\frac{a}{2}\) – 2

= \(\frac{14}{2}\) – 2

= 7 -2

= 5

If a = 14, then b = 5

b = \(\frac{a}{2}\) – 2

= \(\frac{11}{2}\) – 2

= \(\frac{11}{2}\) – \(\frac{4}{2}\)

= \(\frac{11-4}{2}\)

= \(\frac{7}{2}\)

= \(\frac{6}{2}\) + \(\frac{1}{2}\)

= \(3 \frac{1}{2}\)

If a = 11, then b = \(3 \frac{1}{2}\)

b = \(\frac{a}{2}\) – 2

= \(\frac{8}{2}\) – 2

= 4 – 2

=2
​
If a = 8, then b = 2

b = \(b=\frac{a}{2}-2\)

= \(\frac{5}{2}\) – 2

= \(\frac{5}{2}\) – \(\frac{4}{2}\)

= \(\frac{5-4}{2}\)

= \(\frac{1}{2}\)

If a = 5, then b = \(\frac{1}{2}\)

Result

If a = 17, then b =​ \(6 \frac{1}{2}\). If a = 14, then b = 5. If a = 11, then b = \(3 \frac{1}{2}\). If a = 8, then b = 2. If a = 5, then b = \(\frac{1}{2}\).

Page 238 Exercise 20 Answer

Substitute h and d with corresponding values from the table to check if Maya is right.

h = d + 22

3 ≠ 33 + 22 = 55

5 ≠ 55 + 22 = 77

7 ≠ 77 + 22 = 99

9 ≠ 99 + 22 = 121
​
Maya is not correct. h = d + 22 is not true for any of the corresponding values of h and d.

d = 11h

33 = 11(3)

55 = 11(5)

77 = 11(7)

99 = 11(9)
​
The correct equation is d = 11h.

Result

Maya is not correct.

Page 238 Exercise 21 Answer

Find an equation which is true for the given values of t and c, than use it to answer the question.

26.25 ÷ 3 = 8.75

61.25 ÷ 7 = 8.75

78.75 ÷ 9 = 8.75
​
That is, for each value of t and c it is true c ÷ t = $8.75, or written differently c = $8.75t.

Use c = $8.75t to find the cost of 5 tickets, that is find the value of c if t = 5.

c = $8.75(5)

= $43.75

Result

The cost of 5 tickets is $43.75.

Chapter 4 Represent And Solve Equations And Inequalities

Section 4.10: Relate Tables, Graphs And Equations

Page 239 Exercise 1 Answer

Draw a line through the three given points on the graph.

Remember that each ordered pair on the line is of the form (m,n) so the ordered pair on the line that includes m = 5 is the ordered pair (5,9).

Since m is the number of blocks that Maria has walked and n is the number of blocks that Nancy has walked, then the ordered pair (5,9) means

Nancy has walked 9 blocks when Maria has walked 5 blocks

Result

The ordered pair is (5,9) and means Nancy has walked 9 blocks when Maria has walked 5 blocks.

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

enVisionmath 2.0: Grade 6 Page 239 Exercise 1 Answer

Since Nancy walked 4 blocks to Maria’s house and then they continued walking together, then Nancy walked 4 more blocks than Maria. The number of blocks that Nancy walked will then always be equal to the number of blocks Maria walked plus 4.

This relationship is represented in the given equation n = m + 4 since n represents the number of blocks Nancy walked, m represents the number of blocks Maria walked, and the equation represents the pattern of n equalling 4 more than m.

We are also given the following table:

Since 5 = 1 + 4, 6 = 2 + 4, and 7 = 3 + 4, then the values of n in the table are equal to the value of m plus 4.

The given graph has the ordered pairs from the table so the graph also represents the value of n being equal to the value of m plus 4.

Result

The equation, table, and graph all have values of n that equal the values of m plus 4 so they all represent Nancy walking 4 more blocks than Maria.

Page 240 Exercise 1 Answer

It is given that the booster club raises $0.41 For each pom pom they sell so the total money raised, r is equal to $0.41 times the number of pom poms, n.

That is r = $0.41 x n so the equation for this situation is r = 0.41n

To complete the table, we need to find the value of r for n = 150. Substituting n = 150 into the equation and simplifying gives:

r = 0.41n = 0.41(150) = 61.50

The table is then completed as:

To complete the graph, plot the three ordered pairs from the table and then draw a line passing through the points. Since n and r can’t be negative, your line should start at the origin:

To find how many pom poms they must sell to raise $50, we can substitute r = 50 into the equation and then solve for n:

r = 0.41n

50 = 0.41n Substitute r = 50.

50 ÷ 0.41 = 0.41n ÷ 0.41 Divide both sides by 0.41.

121.95 ≈ n Simplify.
​
The number of pom poms must be a whole number so they need to sell 122 pom poms to raise $50.

Finding three values for x and y helps us represent the relationship between x and y because it gives us three ordered pairs for the relationship. These ordered pairs can be used to create a table and a graph. The ordered pairs can also be used to write the equation if it wasn’t given.

Result

They need to sell 122 pom poms to raise $50.

Finding three values for x and y helps us represent the relationship between x and y because it gives us three ordered pairs for the relationship. These ordered pairs can be used to create a table and a graph. The ordered pairs can also be used to write the equation if it wasn’t given.

Page 241 Exercise 2 Answer

It is given that all of the supplies have a cost of $5 and that the company sells the decorations for $2 a piece.

Profit is equal to revenue minus cost so to find the profit, y, we need to multiply the number of decorations sold, x by $2 to get the revenue, and then subtract the cost of $5.

If the number of decorations sold are x = 3, x = 6, and x = 9, then the profits are:

x = 3 : y = $2(3) – $5 = $6 – $5 = $1

x = 6 : y = $2(6) – $5 = $12 – $5 = $7

x = 9 : y = $2(9) – $5 = $18- $5 = $13

y = $2(x) – $5

A possible way to complete the table is then:

Note that you can choose any two values of x for the last two rows of the table so your table may vary from mine if you chose different values for x.

From completing the table, we determined that the equation is y = 2x – 5.

To complete the graph, plot the three ordered pairs from your table and then draw a line passing through the points. Remember that profit y, can be negative but the number of decoration, x, can’t be negative so your line should start on the y-axis at (0,-5).

If the company must have a profit of $15, then y = 15. Substituting this into the equation and solving for x gives:

y = 2x − 5

15 = 2x − 5 Substitute y = 15.

15 + 5 = 2x − 5 + 5 Add 5 on both sides.

20 = 2x Simplify.

10 ÷ 2 = 2x ÷ 2 Divide both sides by 2.

10 = x Simplify.

The company must then sell 10 decorations to have a profit of $15. Notice this matches with our graph since the line passes through the point (10,15).

Result

Possible table:

y = 2x – 5

The company must sell 10 decorations to have a profit of $15.

Page 242 Exercise 1 Answer

The independent variable is represented on the x-axis and the dependent variable is on the y-axis.

An equation describes the relationship between the independent and the dependent variable by telling us what numbers and operations relate the values of the two variables. The equation can be made into a table, one column show the values of the independent and the other the values of the dependent variable. The table then gives coordinate pairs which can be plotted and a graph can be drawn.

Result

An equation describes the relationship between the independent and the dependent variable by telling us what numbers and operations relate the values of the two variables. The equation can be made into a table which gives coordinate pairs that can be plotted and a graph can be drawn.

Page 242 Exercise 2 Answer

To raise $75 for charity they need to sell x pom poms for the cost of $0.45 each. Thus, the answer is the solution of the equation:$ x × 0.45 = 75.$

x × 0.45 = 75

x × 0.45 ÷ 0.45 = 75 ÷ 0.45 (Divide both sides by 0.45.)
​
x ≈166.67

They must sell 167 pom poms to raise $75.

Result

167

Page 242 Exercise 3 Answer

It is given that for every 4 bananas a grocery store sells, it sells 2 apples.

This means that if the grocery store sells b = 4 bananas, it would also sell a = 2 apples so the number of apples sold, a, is half the number of bananas sold, b. That is, a = \(\frac{1}{2}\)b.

Therefore, Maria’s equation of 4b × 2 = a does not correctly represent the relationship. We can verify this by substituting a = 2 and b = 4 into her equation and showing both sides are not equal:

4b x 2 = a

4(4) x 2 = 2

16 x 2 = 2
​
32 ≠ 2

Result

Mary’s equation does not correctly represent the relationship. The correct equation is a = \(\frac{1}{2}\)b.

Page 242 Exercise 4 Answer

It is given that d = 4t. Substituting each of the given values of t into this equation and simplifying gives:

d = 4t

t = 1 : d = 4(1) = 4

t = 2 : d = 4(2) = 8

t = 3 : d = 4(3) = 12

The completed table is then:

Result

Page 242 Exercise 5 Answer

In Exercise 4, we made the following table:

From the table, we know that three ordered pairs found on the line are then (1,4), (2,8) and (3,12)

To find a fourth ordered pair, we need to substitute a fourth value of t into the equation. Substituting t = 4 and simplifying gives:

d = 4t = 4(4) = 16

A fourth ordered pair on the line is then (4,16)

Result

(1,4), (2,8), (3,12), and (4,16)

Page 242 Exercise 6 Answer

We know the equation d = 4t where d is the distance and t is the time represents the relationship between the variables.

This relationship described in words is:

The distance, d, is equal to 4 times the time, t.

Result

The distance, d, is equal to 4 times the time, t.

Page 242 Exercise 7 Answer

We are given the equation d = 5 + 5t where d is the distance and t is the time.

First, we need to complete the table for t = 0, t = 2, and one additional value of t. I have chosen to find d for t = 4. To find the values of d, substitute each value of t into the equation and then simplify:

d = 5 + 5t

t = 0 : d = 5 + 5(0) = 5 + 0 = 5

t = 2 : d = 5 + 5(2) = 5 + 10 = 15

t = 4 : d = 5 + 5(4) = 5 + 20 = 25

The table can then be completed as follows:

To complete the graph, plot the ordered pairs from your table and then draw a line passing through the points. Since d and t can’t be negative, your line should start on the y-axis at (0, 5):

Result

Page 243 Exercise 8 Answer

We are given the equation l = w + \(\frac{1}{2}\)

We need to complete the table for w = 1, w = 2, and one additional value of w. I have chosen to find l when w = 3. To find the values of l, substitute each of the given values of w into the equation and then simplify:

l = w + \(\frac{1}{2}\)

The table can then be completed as:

To complete the graph, plot the ordered pairs from your table and then draw a line passing through the points. Since l and w can’t be negative, your line should start on the y-axis at (0, \(\frac{1}{2}\)):

Result

Page 243 Exercise 9 Answer

We are given the equation s = r – 5.

We need to complete the table for r = 10, r = 20, and one additional value of r. I have chosen to find s when r = 30. To find the values of s, substitute each of the given values of r into the equation and then simplify:

s = r – 5

r = 10 : s = 10 – 5 = 5

r = 20 : s = 20 – 5 = 15

r = 30 : s = 30 – 5 = 25

The table can then be completed as:

To complete the graph, plot the ordered pairs from your table and then draw a line passing through the points. Since r and s can’t be negative, your line should start on the x-axis at (5, 0):

Result

Page 243 Exercise 10 Answer

Since both points, (2,4) and (−2,−4), are on the graph of y = a × x, when the coordinates of the points are substituted as the values of the equation, the equation must be true.

Use substitution to find the value of a.

y = a x x

​4 = a x 2

4 ÷ 2 = a (Divide both sides by 2.)

2 = a

−4 = a × (−2)

−4 ÷ (−2) = a (Divide both sides by −2.)
​
2 = a

The value of a is 2.

Result

a = 2

Page 243 Exercise 11 Answer

If a point is on the graph, its coordinates, when substituted as values of the equation, make the equation true.

The equation is y = 2 × x. A point on the graph is (x,y).

Let’s say x = 1, then y = 2 × 1 = 2. A point (1,2) is on the graph.

Let’s say x = 3, then y = 2 × 3 = 6. A point (3,6) is on the graph.

Let’s say x = 4, then y = 2 × 4 = 8. A point (4,8) is on the graph.

Result

(1,2), (3,6), (4,8)

Page 243 Exercise 12 Answer

It cost $40 and then an additional $0.10 per mile. Since they will drive for m miles, the product m × $0.10 represents how much they will have to pay for the miles. Add to that the $40 they must pay in the beginning.

The equation which represents the cost for renting a car is:

c = $40 + m × $0.10.
​
Result

c = $40 + m × $0.10

Page 244 Exercise 13 Answer

Let x be the number of weeks. We know the puppy gains \(\frac{1}{2}\) pound each week so after x weeks, the number of pounds the puppy has gained is:

(amount gained each week) . (number of weeks) = \(\frac{1}{2}\)x

The puppy initially weighs 1 pound so the puppy’s weight after x weeks is:

initial weight + amount of weight gained = 1 + \(\frac{1}{2}\)x

Letting y be the puppy’s weight then gives the equation y = 1 + \(\frac{1}{2}\)x

To complete the table, we need to find the corresponding y-values for x = 0, x = 2, and one additional value of x. I have chosen to find the y-value for y = 4. To find the y-values, substitute each of the x-values into the equation and then simplify:

y = 1 + \(\frac{1}{2}\)x

The completed table is then:

From the table, the puppy weighs 3 pounds after 4 weeks.

To complete the graph, plot the points from your table and then draw a line passing through them. Remember that the number of weeks and the puppies weight can’t be negative so your line should start at the y-axis at (0, 1):

Result

y = 1 + \(\frac{1}{2}\)x-values

The puppy weighs 3 pounds after 4 weeks.

Page 244 Exercise 14 Answer

To write the equation, we need to use the following relationship:

total amount refunded = (matinee ticket price) x (number of people)

It is given that the matinee ticket price is $5.00. If we let r be the total amount refunded and n be the total number of people, then we get:

r = $5.00 x n

The equation representing the amount of money refunded is then r = 5n.

Result

r = 5n

Page 244 Exercise 15 Answer

We are given the following table:

In the table, we can see that the values of y increase by 3 each time the values of x increase by 1. This means the equation will have 3 times the value of x. Each value of y can then be written as the sum of a multiple of 3 and some other number. We can then rewrite the values of y as:

8 = 3 + 5 = 3(1) + 5

11 = 6 + 5 = 3(2) + 5

14 = 9 + 5 = 3(3) + 5

17 = 12 + 5 = 3(4) + 5

y = 3(x) + 5

The equation that matches with the table is then y = 3x + 5.

Result

The equation is y = 3x + 5. I found this equation by finding how much the values of y changed each time x increased by 1 and then used this to rewrite each y-value as a sum.

Page 244 Exercise 16a Answer

To find which equation represents the graph choose a point, for example (1,2), which is on the graph and substitute variables in both equations with their values.

If the equation is true for selected values, than it represents the graph.

y = 2x

2 = 2(1)

2 = 2

y = x + 2

2 = 1 + 2

2 ≠ 3
​
The equation y = 2x represents the graph.

Result

y = 2x

Page 244 Exercise 16b Answer

From PART A, we know the equation for the graph is y = 2x.

This relationship in words is: The value of y is 2 times the value of x.

Result

The value of y is 2 times the value of x.

enVisionmath 2.0: Grade 6, Volume 1 Chapter 4 Represent And Solve Equations And Inequalities

Page 245 Exercise 1 Answer

In mathematics, a variable is a symbol, for example, an alphabetic character, which represents a number. A variable can be either independent or dependent.

For example, in y = x + 9, the variable x is independent variable since it doesn’t depend on, that is it doesn’t change, as the variable y changes.

Result

Independent.

enVisionmath 2.0: Grade 6, Volume 1 Chapter 4 Page 245 Exercise 2 Answer

As with equations, a value of the variable for which the inequality is true is a solution of the inequality, and the set of all such solutions is the solution of the inequality. This set contains an infinite number of solutions.

Thus, an inequality has an infinite number of solutions.

Result

Inequality.

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Page 245 Exercise 3 Answer

In mathematics, a variable is a symbol, for example, an alphabetic character, which represents a number. A variable can be either independent or dependent.

For example, in y = x + 9, the variable y is dependent variable since it depends on, that is it changes as the variable x changes.

Result

Dependent.

enVisionmath 2.0: Grade 6, Volume 1 Chapter 4 Page 245 Exercise 4 Answer

The given equation represents the Subtraction Property of Equality, since we have the equation

6 + 3 = 9,

and than 3 is subtracted from both sides, which gives us

(6+3) − 3 = 9 − 3.

Result

Subtraction Property of Equality.

Page 245 Exercise 5 Answer

The given equation represents the Multiplication Property of Equality, since we have the equation

6 + 3 = 9,

and than both sides are multiplied by 3, which gives us

(6+3) × 3 = 9 × 3.

Result

Multiplication Property of Equality.

Page 245 Exercise 6 Answer

The given equation represents the Addition Property of Equality, since we have the equation

6 + 3 = 9,

and than 3 is added to both sides, which gives us

(6+3) + 3 = 9 + 3.

Result

Addition Property of Equality.

enVisionmath 2.0: Grade 6, Volume 1 Chapter 4 Page 245 Exercise 7 Answer

The given equation represents the Division Property of Equality, since we have the equation

6 + 3 = 9,

and than both sides are divided by 3, which gives us

(6+3) ÷ 3 = 9 ÷ 3.

Result

Division Property of Equality.

Page 246 Exercise 1 Answer

We are given the equation 6 + 2 = 8 and need to determine if 6 + 2 + 3 = 8 + 3 is also true.

The Addition Property of Equality allows us to add the same number on both sides of the equation.

Compare the two sides of the equation:

6 + 2 + 3 = 8 + 3

Since 3 was added to both sides of the equation, then 6 + 2 + 3 = 8 + 3 by the Addition Property of Equality.

Result

Yes, by the Addition Property of Equality.

Page 246 Exercise 2 Answer

We are given the equation 8 − 1 = 7 and need to determine if the equation 8 − 1 − 2 = 7 − 3 is also true.

The Subtraction Property of Equality allows us to subtract the same number on both sides of the equation.

Compare the two sides of the equation:

8 − 1 − 2 = 7 − 3

The left side has subtraction by 2 while the right side has subtraction by 3. Since the same number wasn’t subtracted on both sides, then 8 − 1 − 2 ≠ 7 − 3.

Result

No, since the same number wasn’t subtracted on both sides.

enVisionmath 2.0: Grade 6, Volume 1 Chapter 4 Page 246 Exercise 3 Answer

We are given the equation 4 + 6 = 10 and need to determine if the equation (4+6) × 3 = 10 × 3 is also true.

The Multiplication Property of Equality allows us to multiply both sides of the equation by the same number.

Compare the two sides of the equation:

(4+6) × 3 = 10 × 3

Since 3 was multiplied on both sides of the equation, then (4+6) × 3 = 10 × 3 by the Multiplication Property of Equality.

Result

Yes, by the Multiplication Property of Equality.

Page 246 Exercise 4 Answer

We are given the equation 5 + 4 = 9 and need to determine if the equation (5+4) ÷ 3 = 9 ÷ 4 is also true.

The Division Property of Equality allows us to divide both sides of the equation by the same number.

Compare the two sides of the equation:

(5+4) ÷ 3 = 9 ÷ 4

The left side has division by 3 while the right side has division by 4. Since the two sides were not divided by the same number, then (5+4) ÷ 3 ≠ 9 ÷ 4.

Result

No, the two sides were not divided by the same number.

enVisionmath 2.0: Grade 6, Volume 1 Chapter 4 Page 247 Exercise 1 Answer

8x = 64 Given equation.

8x ÷ 8 = 64 ÷ 8 Divide both sides by 8.

x = 8 Simplify.
​
Result

x = 8

Page 247 Exercise 2 Answer

x + 2 = 11 Given equation.

x + 2 – 2 = 11 – 2 Subtract 2 on both sides.

x = 9 Simplify.
​
Result

x = 9

Page 247 Exercise 3 Answer

x ÷ 20 = 120 Given equation.

x ÷ 20 x 20 = 120 x 20 Multiply both sides by 20.

x = 2,400 Simplify.
​
Result

x = 2,400

Page 247 Exercise 4 Answer

x – 17 = 13 Given equation.

x – 17 + 17 = 13 + 17 Add 17 on both sides.

x = 30 Simplify.
​
Result

x = 30

Page 247 Exercise 5 Answer

x ÷ 12 = 2 Given equation.

x ÷ 12 x 12 = 2 x 12 Multiply both sides by 12.

x = 24 Simplify.
​
Result

x = 24

enVisionmath 2.0: Grade 6, Volume 1 Chapter 4 Page 247 Exercise 6 Answer

8 + x = 25 Given equation.

8 + x – 8 = 25 – 8 Subtract 8 on both sides.

x = 17 Simplify.
​
Result

x = 17

Page 247 Exercise 7 Answer

7x = 77 Given equation.

7x ÷ 7 = 77 ÷ 7 Divide both sides by 7.

x = 11 Simplify.

Result

x = 11

Page 247 Exercise 8 Answer

x – 236 = 450 Given equation.

x – 236 + 236 = 450 + 236 Add 236 on both sides.

x = 686 Simplify.
​
Result

x = 686

Page 247 Exercise 9 Answer

26 = 13x Given equation.

26 ÷ 13 = 13x ÷ 13 Divide both sides by 13.

2 = x Simplify.
​
Result

x = 2

Page 247 Exercise 10 Answer

x + 21.9 = 27.1 Given equation.

x + 21.9 – 21.9 = 27.1 – 21.9 Subtract 21.9 on both sides.

x = 5.2 Simplify.
​
Result

x = 5.2

Page 247 Exercise 11 Answer

2,448 ÷ 48 = x Given equation.

51 = x Divide 2,448 and 48.
​
Result

x = 51

Page 247 Exercise 12 Answer

x + 15 = 31 Given equation.

x + 15 – 15 = 31 – 15 Subtract 15 on both sides.

x = 16 Simplify.
​
Result

x = 16

Page 247 Exercise 1 Answer

\(x+3 \frac{5}{8}=7 \frac{1}{4}\) Given equation.

\(x+3 \frac{5}{8}-3 \frac{5}{8}=7 \frac{1}{4}-3 \frac{5}{8}\) Subtract \(3 \frac{5}{8}\) on both sides.

\(x=7 \frac{2}{8}-3 \frac{5}{8}\) Get a common denominator.

\(x=6 \frac{10}{8}-3 \frac{5}{8}\) Regroup \(7 \frac{2}{8}\) as \(6 \frac{10}{8}\).

x = \(3 \frac{5}{8}\)Subtract.

Result

x = \(3 \frac{5}{8}\)

Page 247 Exercise 2 Answer

\(x-\frac{4}{8}=4 \frac{1}{4}\) Given equation.

\(x-\frac{4}{8}+\frac{4}{8}=4 \frac{1}{4}+\frac{4}{8}\) Add \(\frac{4}{8}\) on both sides.

\(x=4 \frac{1}{4}+\frac{2}{4}\) Get a common denominator.

x = \(4 \frac{3}{4}\) Add.
​
Result

x = \(4 \frac{3}{4}\)

enVisionmath 2.0: Grade 6, Volume 1 Chapter 4 Page 247 Exercise 3 Answer

x ÷ 15 = \(8 \frac{1}{3}\) Given equation.

\(x \div 15 \times 15=8 \frac{1}{3} \times 15\) Multiply both sides by 15.

x = \(\frac{25}{3} \times \frac{15}{1}\) Rewrite as improper fractions.

x = \(\frac{375}{3}\) Multiply.

x = 125 Reduce the fraction.
​
Result

x = 125

Page 247 Exercise 4 Answer

\(\frac{4}{2}\)x = 6 Given equation.

2x = 6 Reduce the fraction.

2x ÷ 2 = 6 ÷ 2 Divide both sides by 2.

x = 3 Simplify.
​
Result

x = 3

Page 247 Exercise 5 Answer

\(\frac{x}{3}\) = 9 Given equation.

\(\frac{x}{3}\) x 3 = 9 x 3Multiply both sides by 3.

x = 27 Simplify.
​
Result

x = 27

Page 247 Exercise 6 Answer

14x = 73.5 Given equation.

14x ÷ 14 = 73.5 ÷ 14 Divide both sides by 14.

x = 5.25 Simplify.
​
Result

x = 5.25

Page 247 Exercise 7 Answer

12x = 19.2 Given equation.

12x ÷ 12 = 19.2 ÷ 12 Divide both sides by 12.

x = 1.6 Simplify.
​
Result

x = 1.6

enVisionmath 2.0: Grade 6, Volume 1 Chapter 4 Page 247 Exercise 8 Answer

17.9 − x = 12.8 Given equation.

17.9 − x + x = 12.8 + x Add x on both sides.

17.9 = 12.8 + x Simplify.

17.9 − 12.8 = 12.8 + x − 12.8 Subtract 12.8 on both sides.

5.1 = x Simplify.
​
Result

x = 5.1

Page 247 Exercise 9 Answer

To write the equation, we need to use the following relationship:

total cost = (number of peaches) x (cost of each peach)

We know the total cost is $3.55, that he buys 5 peaches, and m represents the cost of each peach.

The equation is then:

3.55 = 5 x m

Dividing both sides of the equation by 5 to solve for m gives:

3.55 ÷ 5 = 5m ÷ 5

0.71 = m

The cost of each peach is then $0.71.

Result

5m = 3.55 $0.71

Page 247 Exercise 10 Answer

To write the equation, we need to use the following relationship:

amount left = (original amount) – (amount spent)

If m represents the amount she has left, she started with $1.54, and she spent $0.76, then the equation is:

x = 1.54 – 0.76

Subtracting gives x = 0.778 so she has $0.78 left.

Result

x = 1.54 – 0.76

$0.78 left

Page 248 Exercise 1 Answer

When graphing inequalities, we use an open circle for < or > and a closed circle for ≤ or ≥. We shade to the left for inequalities with x < or x ≤ and we shade to the right for inequalities with x > or x ≥.

The given graph has an open circle at 10 and is shaded to the left. The inequality is then x < 10.

Result

x < 10

Page 248 Exercise 2 Answer

When graphing inequalities, we use an open circle for < or > and a closed circle for ≤ or ≥. We shade to the left for inequalities with x < or x ≤ and we shade to the right for inequalities with x > or x ≥.

The given graph has an open circle at 5 and is shaded to the right. The inequality is then x > 5.

Result

x > 5

Page 248 Exercise 3 Answer

When graphing inequalities, we use an open circle for < or > and a closed circle for ≤ or ≥. We shade to the left for inequalities with x < or x ≤ and we shade to the right for inequalities with x > or x ≥.

The given graph has a closed circle at 3 and is shaded to the right. The inequality is then x ≥ 3.

Result

x ≥ 3

Page 248 Exercise 4 Answer

When graphing inequalities, we use an open circle for < or > and a closed circle for ≤ or ≥. We shade to the left for inequalities with x < or x ≤ and we shade to the right for inequalities with x > or x ≥.

The given graph has a closed circle at 7 and is shaded to the left. The inequality is then x ≤ 7.

Result

x ≤ 7

Page 248 Exercise 1 Answer

It is given that p represents the number of people Mary visited today.

If Mary visited up to 5 people today, then the number of people is less than or equal to 5. The inequality is then p ≤ 5.

Result

p ≤ 5

Page 248 Exercise 2 Answer

It is given that v is the value of the hat.

If the value of the hat is less than $9, then the inequality is v < 9.

Result

v < 9

Page 248 Exercise 3 Answer

It is given that g represents the number of guests.

If the number of guests is not 8, then the inequality is g ≠ 8.

Result

g ≠ 8

Page 248 Exercise 4 Answer

It is given that d represents the distance of the race.

If the distance is at least 6 miles, then the distance is greater than or equal to 6. The inequality is then d ≥ 6.

Result

d ≥ 6

Page 248 Exercise 5 Answer

It is given that t represents the time it takes to get to Grandma’s house.

If the time is longer than 2 hours, in other words, the time is greater than 2 hours, then the inequality is t > 2.

Result

t > 2

Page 249 Exercise 1 Answer

The dependent variable is the quantity that depends on the independent variable. That is, if the independent variable changes, it causes the dependent variable to change.

We are given the variables d and s where d represents the distance traveled and s represents the speed.

If you increase your speed, then the distance you travel will increase and if you decrease your speed, then the distance you travel will decrease. This means the distance traveled depends on the speed so the dependent variable is the distance traveled, d, and the independent variable is the speed, s.

Result

The dependent variable is the distance traveled, d, and the independent variable is the speed, s.

Page 249 Exercise 2 Answer

The dependent variable is the quantity that depends on the independent variable. That is, if the independent variable changes, it causes the dependent variable to change.

We are given the variables c and a where c represents the number of calories in a snack and a represents the amount of the snack.

If you eat more of the snack, then the number of calories you eat will increase and if you eat less of the snack, then the number of calories you eat will decrease. This means the number of calories depends on the amount of the snack so the dependent variable is the number of calories, c, and the independent variable is the amount of the snack, a.

Result

The dependent variable is the number of calories, c, and the independent variable is the amount of the snack, a.

Page 249 Exercise 3 Answer

The dependent variable is the quantity that depends on the independent variable. That is, if the independent variable changes, it causes the dependent variable to change.

We are given the variables s and m where s represents the amount of money you have spent and m represents how much money you have left.

The more money you spend, the less money you have left. This means the amount you have left depends on the amount you have spent so the dependent variable is how much money you have left, m, and the independent variable is the amount of money you have spent, s.

Result

The dependent variable is how much money you have left, m, and the independent variable is the amount of money you have spent, s.

Page 249 Exercise 4 Answer

The dependent variable is the quantity that depends on the independent variable. That is, if the independent variable changes, it causes the dependent variable to change.

We are given the variables r and e where r represents the number of apple slices remaining and e represents the number of apple slices eaten.

The more apple slices you eat, the fewer you have remaining. This means the number of slices remaining depends on the number of slices eaten so the dependent variable is the number of apple slices remaining, r, and the independent variable is the number of apple slices eaten, e.

Result

The dependent variable is the number of apple slices remaining, r, and the independent variable is the number of apple slices eaten, e.

Page 249 Exercise 1 Answer

We are given the following table:

From the table, we can see that the rule is the value of y is equal to the value of x divided by 2 since 0 ÷ 2 = 0, 2 ÷ 2 = 1, and 10 ÷ 2 = 5. The equation is then y = x ÷ 2.

To complete the table, we need to evaluate the equation for x = 16 and x = 20:

x = 16 : y = x ÷ 2 = 16 ÷ 2 = 8

x = 20 : y = x ÷ 2 = 20 ÷ 2 = 10

The completed table is then:

Result

Rule: The value of y is equal to the value of x divided by 2.

Equation: y = x ÷ 2

Page 249 Exercise 2 Answer

We are given the equation y = 6x + 1 and need to complete the table for x = 1, 2, 3, 4, and 5.

To find the y-values, substitute each of the given x-values into the equation and then simplify:

y = 6x + 1

x = 1 : y = 6(1) + 1 = 6 + 1 = 7

x = 2 : y = 6(2) + 1 = 12 + 1 = 13

x = 3 : y = 6(3) + 1 = 18 + 1 = 19

x = 4 : y = 6(4) + 1 = 24 + 1 = 25

x = 5 : y = 6(5) + 1 = 30 + 1 = 31

The complete table is then:

Result

Page 250 Exercise 1a Answer

To complete the table, we can use the following relationship:

distance = (rate)(time)

It is given that the average jogging rate is 6 miles per hour and x is the number of hours jogged. To complete the table, we can then multiply each of the given x-values by 6:

x = 1 : 1 x 6 = 6

x = 2 : 2 x 6 = 12

x = 3 : 3 x 6 = 18

The completed table is then:

Result

Page 250 Exercise 1b Answer

In part (a), we made the following table:

To complete the graph, plot the ordered pairs from the table and then draw a line passing through the points. Since x and y can’t be negative, your line should start at the origin:

Result

To complete the graph, plot the ordered pairs from the table made in part (a) and then draw a line passing through the points. Since x and y can’t be negative, your line should start at the origin.

Page 250 Exercise 1c Answer

When completing the table in part (a), we determined that the rule was:

The value of y is equal to 6 times the value of x.

The rule written as an equation is then y = 6x.

If one member jogged 3.5 hours, then x = 3.5. Substituting this into the equation and simplifying gives:

y = 6x = 6(3.5) = 21

The team member then jogged 21 miles.

Result

y = 6x 21 miles

Page 250 Exercise 2 Answer

Let x be the number of puppets and y be the total cost.

The cost of the string is $125 and the cost for the remaining material is $18 per puppet. For x puppets, the cost of the remaining material is then 18x dollars.

Adding the cost for the string gives a total cost of y = 18x + 125 dollars.

If he makes 50 puppets, then x = 50 so the total cost is:

y = 18x + 125 = 18(50) + 125 = 900 + 125 = $1025
​
Result

$1025

Chapter 2 Integers And Rational Numbers

Section 2.6: Represent Polygons On The Coordinate Plane

Page 101 Exercise 1 Answer

Since we know the vertices of the polygon we can graph them in a coordinate plane and then calculate the distance from one vertice to another to find the perimeter of the polygon.

The distance from point A to point B is the difference of absolute values of their x-coordinates, since they have the same y-coordinates and are both in Quadrant II.

∣−7∣ − ∣−1∣ = 7 − 1 = 6

The distance from point A to point B is 6 units.

The distance from point B to point C is the sum of absolute values of their y-coordinates, since they have the same x-coordinates and are not in the same quadrant.

∣6∣ + ∣−3∣ = 6 + 3 = 9

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

The distance from point B to point C is 9 units.

We can see from the graph that the polygon is a rectangle, so side AB is of equal length as side CD, and side BC is of equal length as side AD. However, we can check.

The distance from point C to point D is the difference of absolute values of their x-coordinates, since they have the same y-coordinates and are both in Quadrant III, thus the distance from point C to point D is 6 units.

The distance from point A to point D is the sum of absolute values of their y-coordinates, since they have the same x-coordinates and are not in the same quadrant, thus the distance from point B to point C is 9 units.

The perimeter is the sum of the lengths of all sides. 6 + 9 + 6 + 9 = 30

Result

The perimeter of the polygon is 30 units.

Page 101 Exercise 1 Answer

The polygon drawn in the Solve & Discuss It! exercise is a rectangle since it is a quadrilateral with four right angles.

Points A(−1,6), B(−7,6), C(−7,−3), and D(−1,−3) are vertices of this polygon. Since lines segments AB and CD are both parallel to the x-axis, and line segments BC and AD are both parallel to the y-axis, angles ∠ABC, ∠BCD, ∠CDA, and ∠DAB are all right angles which means this polygon is a rectangle.

Result

Rectangle

Page 102 Exercise 1 Answer

First we need to find the length of each side of the new rectangle.

The vertices of the new rectangle are A(−4,6), B(2,6), Working Tent, and Food Tent. To find the coordinates of the Working Tent and the Food Tent follow the grid lines directly to the x-axis to find the x-coordinates, follow the grid lines directly to the y-axis to find the y-coordinates. The Working Tent is at (−4,−2) and the Food Tent is at (2,−2).

A to B: ∣−4∣ + ∣2∣ = 4 + 2 = 6

B to food tent: ∣6∣ + ∣−2∣ = 6 + 2 = 8

Food tent to working tent: ∣2∣ + ∣−4∣ = 2 + 4 = 6

Working tent to A: ∣−2∣ + ∣6∣ = 2 + 6 = 8

6 + 8 + 6 + 8 = 28

The sum of the lengths of all sides is 28, so the archaeologist needs 28 meters of rope.

To find the perimeter of the larger rectangle we could have used just the distance from A to B and the distance from A to the Working Tent. Since points A and the Working Tent, and points B and the Food Tent have the same x-coordinates, and points A and B, and points the Working Tent and the Food tent have the same y-coordinates, the distance from A to the Working Tent and the distance from B to the Food Tent are the same, so is the distance from A to B, and from the Working Tent to the Food Tent, are the same as well.

Result

A to B = 6 m Food tent to working tent = 6 m

B to food tent: 8 m Working tent to A = 28 m

The archeologist needs 28 meters of rope.

Page 103 Exercise 2 Answer

Find the side lengths. For all the points which have the same x-coordinates and are in the same quadrant, find the difference of the absolute values of their y-coordinates. For all the points which have the same y-coordinates and are in the same quadrant, find the difference of the absolute values of their x-coordinates. If they are in different quadrants, find the sum instead of the difference.

AB = ∣15∣ − ∣4.5∣ = 15 − 4.5 = 10.5

BC = ∣14∣ − ∣8∣ = 14 − 8 = 6

DC = ∣15∣ − ∣10∣ = 15 − 10 = 5

DE = ∣8∣ − ∣3∣ = 8 − 3 = 5

EF = ∣15∣ − ∣10∣ = 15 − 10 = 5

FG = ∣3∣ + ∣−2.25∣ = 3 + 2.25 = 5.25

HG = ∣15∣ − ∣4.5∣ = 15 − 4.5 = 10.5

AH = ∣14∣ + ∣−2.25∣ = 14 + 2.25 = 16.25

10.5 + 6 + 5 + 5 + 5 + 5.25 + 10.5 + 16.25 = 63.5

The perimeter is the sum of all the lengths which is 63.5.

Result

The rancher needs 63.5 yards of fencing.

Page 103 Exercise 3 Answer

If quadrilateral ADEF is a square, then all of its sides have an equal length. Let’s check.

The length of side AD = ∣−5∣ + ∣7∣ = 5 + 7 = 12 units.

The length of side AF = ∣2∣ + ∣−8∣ = 2 + 8 = 10 units.

The length of side DE = ∣2∣ + ∣−8∣ = 2 + 8 = 10 units.

The length of side EF = ∣−5∣ + ∣7∣ = 5 + 7 = 12 units.

Not all the sides have an equal length, so quadrilateral ADEF is not a square.

Result

Quadrilateral ADEF is not a square since not all the sides have an equal length.

Page 104 Exercise 1 Answer

When a polygon is placed in a coordinate plane, its vertices are all assigned specific unique coordinates.

The lengths of the sides can be calculated by adding or subtracting the absolute values of the coordinates to find the distances between the vertices. When we know the lengths of sides of polygons we can further calculate its perimeter and its area.

For example, if we have a rectangle whose vertices are A(−2,7), B(5,7), C(5,−3), and D(−2,−3). The lengths of its side are then:

AB = ∣−2∣ + ∣5∣ = 2 + 5 = 7

BC = ∣7∣ + ∣−3∣ = 7 + 3 = 10

CD = ∣5∣ + ∣−2∣ = 5 + 2 = 7

AD = ∣7∣ + ∣−3∣ = 7 + 3 = 10

Its area is 7 × 10 = 70 units, and its perimeter is 7 + 10 + 7 + 10 = 34 units.

Result

When a polygon is placed in a coordinate plane, its vertices are all assigned specific unique coordinates.

The lengths of the sides can be calculated by adding or subtracting the absolute values of the coordinates to find the distances between the vertices. When we know the lengths of sides of polygons we can further calculate its perimeter and its area.

Page 104 Exercise 2 Answer

In Example 1, points A and B are in different quadrants. If we imagine walking the distance straight from A to B, first we must cross four units to get to the y-axis, since A is at (-4, 6), and then another two units to get to point B, which is at (2, 6).

Thus, to find the distance from A to B we need to add the units, to add the distance from A to the y-axis and the distance from B to the y-axis, which are actually the x-coordinates of the points.

However, points B and C are in the same quadrant, Quadrant I. If we imagin wlaking a straight line from B to C, we won’t cross either of the axes. Since B is at (2, 6) and C at (2, 1), we need to cross five units to get from B to C.

Thus, to find the distance from B to C we need to subtract the units, to subtract the distance from B to the x-axis and the distance from C to the y-axis, which are actually the y-coordinates of the points.

Result

A and B are in different quadrants so we need to add to find the distance. However, B and C are in the same quadrant so we need to subtract to find the distance.

Page 104 Exercise 3 Answer

To find the length of the diagonal AC we can neither add nor subtract the absolute values of coordinates since points A and C have different both the x-coordinates and the y-coordinates.

Result

To find the length of the diagonal AC we can neither add nor subtract the absolute values of coordinates since points A and C have different both the x-coordinates and the y-coordinates.

Page 104 Exercise 4 Answer

The vertices of rectangle MNOP are M(−2,5), N(−2,−4), O(3,−4), and P(3,5). To find the lengths of sides we need to calculate the distances from one vertice to another.

Points M and N are in different quadrants and have the same x-coordinates, so to find the distance from M to N we need to add the absolute values of their y-coordinates. The same is true for points O and P.

MN = ∣5∣ + ∣−4∣ = 5 + 4 = 9

OP = ∣5∣ + ∣−4∣ = 5 + 4 = 9

Points N and O are in different quadrants and have the same y-coordinates, so to find the distance from N to O we need to add the absolute values of their x-coordinates. The same is true for points P and M.

NO = ∣−2∣ + ∣3∣ = 2 + 3 = 5

MP = ∣−2∣ + ∣3∣ = 2 + 3 = 5

The perimeter is equal to the sum of the lengths of all four sides of rectangle MNOP.

9 + 9 + 5 + 5 = 28

Result

The perimeter of rectangle MNOP is 28 units.

Page 104 Exercise 5 Answer

If a polygon EFGH is a square, then all of its sides are equal length. To check if they are equal length, we need to check if the distance between E and F, F and G, G and H, and the distance between E and H, are same.

Points E and F are in different quadrants and have the same y-coordinates. To find the distance from E to F we need to add the absolute values of their x-coordinates. The same is true for points G and H.

EF = ∣−2∣ + ∣3∣ = 2 + 3 = 5

GH = ∣−2∣ + ∣3∣ = 2 + 3 = 5

Points F and G are in different quadrants and have the same x-coordinates. To find the distance from F to G we need to add the absolute values of their y-coordinates. The same is true for points E and H.

FG = ∣3.5∣ + ∣−1.5∣ = 3.5 + 1.5 = 5

EH = ∣3.5∣ + ∣−1.5∣ = 3.5 + 1.5 = 5

All the side of a rectangle EFGH are the same length, which means it is a square.

Result

EFGH is a square.

Page 104 Exercise 6 Answer

To find the area of square ABCD we must first find the length of one of its side.

Points A and B have the same y-coordinates and are in different quadrants, to find the distance from A to B we need to calculate the sum of absolute values of their x-coordinates.

AB = ∣−4.5∣ + ∣3.5∣ = 4.5 + 3.5 = 8

The area of square ABCD is equal to 8 × 8 = 64.

Result

The area of square ABCD is 64 units.

Page 105 Exercise 7 Answer

Points J and K have the same x-coordinates and are in different quadrants, so to find the distance between J and K we need to calculate the sum of absolute values of their y-coordinates.

JK = ∣8∣ + ∣−1∣ = 8 + 1 = 9

Points K and L have the same y-coordinates and are in different quadrants, so to find the distance between K and L we need to calculate the sum of absolute values of their x-coordinates.

KL = ∣−3∣ + ∣4∣ = 3 + 4 = 7

The rectangle has two sides with length 9 and two side with length 7.

The perimeter is equal to 2 × 9 + 2 × 7 = 18 + 14 = 32 units.

Result

The perimeter is equal to 32 units.

Page 105 Exercise 8 Answer

Points W and X have the same y-coordinates and are in different quadrants, so to find the distance between W and X we need to calculate the sum of absolute values of their x-coordinates.

WX = ∣−3∣ + ∣4∣ = 3 + 4 = 7

Points X and Y have the same x-coordinates and are in the same quadrant, Quadrant IV, so to find the distance between X and Y we need to calculate the difference of absolute values of their y-coordinates.

XY = ∣−5∣ − ∣−2∣ = 5 − 2 = 3

The rectangle has two sides with length 7 and two side with length 3.

The perimeter is equal to 2 × 7 + 2 × 3 = 14 + 6 = 20 units.

Result

The perimeter is 20 units.

Page 105 Exercise 9 Answer

If triangle JKL is equilateral, then the lengths of all of its sides are equal. To find the length of its side we need to calculate the distance from one vertice to another.

Since point J is the origin and point K is a point on the x-axis, the distance from J to K is equal to the absolute value of the x-coordinate of point K. Thus, the distance from J to K is 5 units.

Again, point J is the origin and point L is a point on the y-axis, the distance from J to L is equal to the absolute value of the y-coordinate of point L. Thus, the distance from J to L is 3 units.

Sides JK and JL are not of equal length, so triangle JKL is not equilateral.

Result

Triangle JKL is not equilateral.

Page 105 Exercise 10 Answer

Polygon WXYZ with vertices W(−1.5,1.5), X(6,1.5), Y(6,−4.5), and Z(−1.5,−4.5) is a rectangle if it is a quadrilateral with four right angles. Graph the vertices and then connect them with segments to draw the polygon.

From the graph, the polygon has four sides so it is a quadrilateral.

W and Z have the same x-coordinates and X and Y have the same x-coordinates. Sides WZ and XY are then both vertical.

W and X have the same y-coordinates and Z and Y have the same y-coordinates. Sides WX and ZY are then both horizontal.

Since horizontal lines and vertical lines intersect to form right angles, then quadrilateral WXYZ has four right angles. The polygon is then a rectangle.

Result

WXYZ is a rectangle since it has four sides and four right angles.

Page 105 Exercise 11 Answer

First we need to find the coordinates of the vertices of rectangle ABCD. For each point, follow the grid lines directly to the x-axis to find the x-coordinates, and follow the grid lines directly to the y-axis to find the y-coordinates.

The coordinates of the vertices are as follows: A(−3,−7), B(4,−7), C(4,6), D(−3,6).

To find the perimeter of rectangle ABCD, we need to calculate the sum of the lengths of all of its sides.

Points A and B have the same y-coordinates and are in different quadrants, so the distance between them is the sum of absolute values of their x-coordinates. The same is true for points D and C.

AB = ∣−3∣ + ∣4∣ = 3 + 4 = 7

DC = ∣−3∣ + ∣4∣ = 3 + 4 = 7

Points A and D have the same x-coordinates and are in different quadrants, so the distance between them is the sum of absolute values of their y-coordinates. The same is true for points B and C.

AD = ∣−7∣ + ∣6∣ = 7 + 6 = 13

BC = ∣−7∣ + ∣6∣ = 7 + 6 = 13

The perimeter of rectangle ABCD is

2 × 7 + 2 × 13 = 14 + 26 = 40.

The area of rectangle ABCD is

7 × 13 = 91.

Result

The perimeter of rectangle ABCD is 40, and the area is 91.

Page 105 Exercise 12 Answer

The perimeter of the patio is the sum of the lengths of all sides of the patio.

All the vertices are in the same quadrant, Quadrant I. If the points have the same x-coordinates, the distance between them is the difference of absolute values of their y-coordinates. If the points have the same y-coordinates, the distance between them is the difference of absolute values of their x-coordinates.

AB = ∣4∣ − ∣2∣ = 4 − 2 = 2

BC = ∣7∣ − ∣3∣ = 7 − 3 = 4

CD = ∣7∣ − ∣4∣ = 7 − 4 = 3

DE = ∣9∣ − ∣7∣ = 9 − 7 = 2

EF = ∣7∣ − ∣2∣ = 7 − 2 = 5

AF = ∣9∣ − ∣3∣ = 9 − 3 = 6

The perimeter of the patio is

AB + BC + CD + DE + EF + AF = 2 + 4 + 3 + 2 + 5 + 6 = 22.

Result

The perimeter of the patio is 22 yards.

Page 106 Exercise 13 Answer

The total distance that Jordan ran is the perimeter of polygon HBLCS.

HB = \(\left|-\frac{3}{4}\right|+|1|=\frac{3}{4}+1=1 \frac{3}{4}\)

BL = \(\left|\frac{1}{2}\right|+\left|-\frac{1}{4}\right|=\frac{1}{2}+\frac{1}{4}=\frac{2}{4}+\frac{1}{4}=\frac{2+1}{4}=\frac{3}{4}\)

LP = \(|1|-\left|\frac{1}{2}\right|=1-\frac{1}{2}=\frac{2}{2}-\frac{1}{2}=\frac{2-1}{2}=\frac{1}{2}\)

PC = \(\left|-\frac{3}{4}\right|-\left|-\frac{1}{4}\right|=\frac{3}{4}-\frac{1}{4}=\frac{3-1}{4}=\frac{2}{4}=\frac{1}{2}\)

SC = \(\left|-\frac{3}{4}\right|+\left|\frac{1}{2}\right|=\frac{3}{4}+\frac{1}{2}=\frac{3}{4}+\frac{2}{4}=\frac{3+2}{4}=\frac{5}{4}=\frac{4}{4}+\frac{1}{4}=1 \frac{1}{4}\)

HS = \(\left|\frac{1}{2}\right|+\left|-\frac{3}{4}\right|=\frac{1}{2}+\frac{3}{4}=\frac{2}{4}+\frac{3}{4}=\frac{2+3}{4}=\frac{5}{4}=\frac{4}{4}+\frac{1}{4}=1 \frac{1}{4}\)

Thee perimeter of polygon HBLCS is

Result

The total distance that Jordan ran is 6 miles.

Page 106 Exercise 14 Answer

Three vertices of a rectangle are known, (−1.2,−3.5), (−1.2,4.4), and (5.5,4.4).

Point (−1.2,−3.5) is in Quadrant III, point (−1.2,4.4) in Quadrant II, and point (5.5,4.4) in Quadrant I. We can now conclude that points (5.5,4.4) and (−1.2,−3.5) are adjecent points to the fourth vertex. We can also graph the known points to check this.

From the graph, we can also conclude that the fourth vertex is in Quadrant IV. Its x-coordinate must be the same as the x-coordinate of point (5.5,4.4) and its y-coordinate must be the same as the y-coordinate of point (−1.2,−3.5).

Result

The coordinates of the fourth vertex are (5.5,−3.5).

Page 106 Exercise 15 Answer

To calculate the area of the backyard needed for the pool we need to first find the lengths of sides of the rectangular pool.

Points A and B have the same y-coordinates and are in different quadrants, so the distance from A to B is the sum of absolute values of their x-coordinates.

∣−5∣ + ∣1∣ = 5 + 1 = 6

The lengths of one of the sides is 6 yards.

Points B and C have the same x-coordinates and are in different quadrants, so the distance from B to C is the sum of absolute values of their y-coordinates.

∣7∣ + ∣−1∣ = 7 + 1 = 8

The lengths of one of the sides is 8 yards.

The area of the rectangular pool is the product

AB × BC = 6 × 8 = 48.

Result

The area of the backyard needed for the pool is 48 square yards.

Page 106 Exercise 16 Answer

Each point in a coordinate plane has its unique coordinates, real numbers which describe its location. Coordinates can be negative, zero, or positive. The distance between points can be calculated using their coordinates, however, coordinates can be negative, but distance is always positive, so we use absolute values of coordinates to calculate the distances.

The absolute value of the x-coordinate of a point represents the distance between that point and the x-axis, however the sign of the x-coordinate tells us in which direction from the origin the point is located. The same is true for the y-axis as well.

Result

The distance between points can be calculated using their coordinates, however, coordinates can be negative, but distance is always positive, so we use absolute values of coordinates to calculate the distances.

Page 106 Exercise 17 Answer

Since it is a square, all of its sides are of equal length. Perimeter is 10 units, to find the length of one side we can divide the perimeter by four.

10 ÷ 4 = 2.5

The length of one side is 2.5.

One vertex is (−0.5,−2), one of the adjecent vertices will have the same x-coordinate, and the other adjecent vertex will have the same y-coordinates.

If the x-coordinate is the same as that of the given vertex, than the other is either the difference −2 + 2.5 or −2 − 2.5. Since the difference −2 + 2.5 is equal 0.5, which is positive, so that point wouldn’t be in Quadrant III. So the other coordinate must be −2 − 2.5, which is equal to −4.5.

Second vertex is (−0.5,−4.5).

If the y-coordinate is the same as that of the given vertex, than the other is either the difference −0.5 + 2.5 or −0.5 − 2.5. Since the difference −0.5 + 2.5 is equal 2, which is positive, so that point wouldn’t be in Quadrant III. So the other coordinate must be −0.5 − 2.5, which is equal to −3.

Third vertex is (−3,−2).

To determine the coordinates of the fourth vertex we can graph the three known vertices and simply read them from the graph.

From the graph we can see that the fourth vertex has the same x-coordinates as point (−3,−2) and the same y-coordinates as point (−0.5,−4.5). The fourth vertex is (−3,−4.5).

Result

The coordinates of the other vertices are (−3,−2), (−0.5,−4.5), and (−3,−4.5).

Page 106 Exercise 18a Answer

Points A and B have the same x-coordinates and are in the same quadrant, Quadrant III, which we know because both of their coordinates are negative. To find the distance between points A and B we need to calculate the difference between their y-coordinates.

Result

The distance between points A and B is \(2 \frac{1}{2}\) units.

Page 106 Exercise 18b Answer

From point C we can move 8 units parallel to the x-axis in two directions, and also parallel to the y-axis in two directions.

If we move parallel to the x-axis, the y-coordinate is the same as that of point C. If we move parallel to the y-axis, the x-coordinate is the same as that of point C.

−3 + 8 = 5

−3 − 8 = −11

4 − 8 = −4

4 + 8 = 12

Two points which are 8 units from point C are (4,5) and (-4,-3).

Result

The coordinates of two points which are 8 units from point C are (4,5) and (−4,−3).