enVisionmath 2.0: Grade 6, Volume 1 Chapter 4 Represent And Solve Equations And Inequalities Section 4.4

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:

\(\text { number of days }=\frac{\text { total number of pages }}{\text { number of pages per day }}\)

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

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