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