enVisionmath 2.0: Grade 6, Volume 1

Chapter 3 Numeric And Algebraic Expressions

Vocabulary Review

Page 167 Exercise 1 Answer

An exponent tells the number of times the base is used as a factor.

Result

exponent

Page 167 Exercise 2 Answer

A letter or symbol that represents an unknown quantity is a variable.

Result

variable

Page 167 Exercise 3 Answer

A diagram that shows the prime factors of a composite number is a factor tree.

Result

factor tree

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

Page 167 Exercise 4 Answer

To find the LCM of two numbers, start by finding their prime factorizations:

9 = 3 × 3

6 = 2 × 3

The LCM is the product of the greatest number of times each factor appears in either prime factorization. The greatest number of times the factor 2 appeared is once and the greatest number of times the factor 3 appeared is twice. The LCM is then:

2 × 3 × 3 = 6 × 3 = 18

Result

18

Page 167 Exercise 5 Answer

To find the LCM of two numbers, start by finding their prime factorizations:

9 = 3 × 3

12 = 2 × 6 = 2 × 2 × 3
​
The LCM is the product of the greatest number of times each factor appears in either prime factorization. The greatest number of times the factor 2 appeared is twice and the greatest number of times the factor 3 appeared is twice. The LCM is then:

2 × 2 × 3 × 3 = 4 × 9 = 36

Result

36

Page 167 Exercise 6 Answer

To find the LCM of two numbers, start by finding their prime factorizations:

8 = 2 × 4 = 2 × 2 × 2

7 = prime
​
The LCM is the product of the greatest number of times each factor appears in either prime factorization. The greatest number of times the factor 2 appeared is three times and the greatest number of times the factor 7 appeared is once. The LCM is then:

2 × 2 × 2 × 7 = 8 × 7 = 56

Result

56

Page 167 Exercise 7a Answer

Like terms must have the same variable part.

The terms of 3a + 3z have different variables since the first term has a as the variable and the second term has z as the variable.

We must then write N as the answer since they are NOT like terms.

Result

N

Page 167 Exercise 7b Answer

Like terms must have the same variable part.

The terms of \(\frac{x}{3}\) + \(\frac{x}{4}\) have the same variable parts since the first term has x as the variable and the second term has x as the variable.

We must then write Y as the answer since they are like terms.

Result

Y

Page 167 Exercise 7c Answer

Like terms must have the same variable part.

The terms of 4j − j + 3.8j have the same variable parts since all three terms have j as the variable.

We must then write Y as the answer since they are like terms.

Result

Y

Page 168 Exercise 1 Answer

To find the GCF of a pair of numbers, first find the prime factorization of each number.

30 = 2 x 15 = 2 x 3 x 5

100 = 2 x 50 = 2 x 2 x 25 = 2 x 2 x 5

The GCF of the pairs is the product of the common factors. The pair has common factors of 2 and 5 sot he GCf is 2 x 5 = 10.

Using the GCF and Distributive Property to find the sum of the pair gives:

30 + 100 = 10(3) + 10(10) Rewrite each number as a product.

= 10(3 + 10) Use the Distributive Property

= 10(13) Add

= 130 Add

Result

GCF: 10

Sum: 130

Page 168 Exercise 2 Answer

To find the GCF of a pair of numbers, first find the prime factorization of each number:

8 = 2 × 4 = 2 × 2 × 2

52 = 2 × 26 = 2 × 2 × 13

The GCF of the pair is the product of the common factors. The pair has common factors of 2 and 2 so the GCF is 2 × 2 = 4.

Using the GCF and Distributive Property to find the sum of the pair gives:

= 2(4) + 2(26) Rewrite each number as a product.

= 2(4 + 26) Use the Distributive Property.

= 2(30) Add.

= 60 Add.

Result

GCF: 4

Sum: 60

Page 168 Exercise 3 Answer

To find the GCF of a pair of numbers, first find the prime factorization of each number:

28 = 2 × 14 = 2 × 2 × 7

42 = 2 × 21 = 2 × 3 × 7
​
The GCF of the pair is the product of the common factors. The pair has common factors of 2 and 7 so the GCF is 2 × 7 = 14.

Using the GCF and Distributive Property to find the sum of the pair gives:

= 14(2) + 14(3) Rewrite each number as a product.

= 14(2+3) Use the Distributive Property.

= 14(5) Add.

= 70 Add.

Result

GCF: 14

Sum: 70

Page 168 Exercise 4 Answer

To find the GCF of a pair of numbers, first find the prime factorization of each number.

37 and 67 are prime numbers so they can’t be written as the product of more than one prime factor.

The GCF of the pair is the product of the common factors. The pair has no common prime factors so the GCF is 1 since all numbers have 1 as a factor.

Since 37 and 67 have a GCF of 1, we can’t rewrite the sum using the Distributive Property so we need to find the sum using normal addition:

37 + 67 = 104
​
Result

GCF: 1

Sum: 104

Page 168 Exercise 5 Answer

To find the GCF of a pair of numbers, first find the prime factorization of each number:

12 = 2 × 6 = 2 × 2 × 3

24 = 2 × 12 = 2 × 2 × 6 = 2 × 2 × 2 × 3
​
The GCF of the pair is the product of the common factors. The pair has common factors of 2, 2, and 3 so the GCF is 2 × 2 × 3 = 12.

Using the GCF and Distributive Property to find the sum of the pair gives:

= 12(1) + 12(2) Rewrite each number as a product.

= 12(1 + 2) Use the Distributive Property.

= 12(3) Add.

= 36 Add.
​
Result

GCF: 12

Sum: 36

Page 168 Exercise 6 Answer

To find the GCF of a pair of numbers, first find the prime factorization of each number:

8 = 2 × 4 = 2 × 2 × 2

12 = 2 × 6 = 2 × 2 × 3
​
The GCF of the pair is the product of the common factors. The pair has common factors of 2 and 2 so the GCF is 2 × 2 = 4.

Using the GCF and Distributive Property to find the sum of the pair gives:

= 4(2) + 4(3) Rewrite each number as a product.

= 4(2+3) Use the Distributive Property.

= 4(5) Add.

= 20 Add.
​
Result

GCF: 4

Sum: 20

Page 168 Exercise 7 Answer

To find the LCM of a pair of numbers, first find the prime factorization of each number:

4 = 2 × 2

9 = 3 × 3
​
The LCM of the pair is the product of the greatest number of times each factor appears in either prime factorization. The factor of 2 appeared twice and the factor of 3 appeared twice so the LCM is:

2 × 2 × 3 × 3 = 4 × 9 = 36

Result

36

Page 168 Exercise 8 Answer

For a given pair of numbers, when the smaller number is a factor of the larger number, then the LCM is always the larger number.

For the pair 3 and 6, we know that 3 is a factor of 6 so the LCM must be 6.

Result

6

Page 168 Exercise 9 Answer

To find the LCM of a pair of numbers, first find the prime factorization of each number:

8 = 2 × 4 = 2 × 2 × 2

10 = 2 × 5

The LCM of the pair is the product of the greatest number of times each factor appears in either prime factorization. The factor of 2 appeared three times and the factor of 5 appeared once so the LCM is:

2 × 2 × 2 × 5 = 4 × 10 = 40

Result

40

Page 168 Exercise 10 Answer

For a given pair of numbers where both numbers are prime, the LCM is always the product of the two prime numbers.

For the pair 3 and 5, both numbers are prime so the LCM is 3 × 5 = 15.

Result

15

Page 168 Exercise 11 Answer

To find the LCM of a pair of numbers, first find the prime factorization of each number:

12 = 2 × 6 = 2 × 2 × 3

5 = prime
​
The LCM of the pair is the product of the greatest number of times each factor appears in either prime factorization. The factor of 2 appeared twice, the factor of 3 appeared once, and the factor of 5 appeared once so the LCM is:

2 × 2 × 3 × 5 = 4 × 15 = 60
​
Result

60

Page 168 Exercise 12 Answer

To find the LCM of a pair of numbers, first find the prime factorization of each number:

4 = 2 × 2

11 = prime

The LCM of the pair is the product of the greatest number of times each factor appears in either prime factorization. The factor of 2 appeared twice and the factor of 11 appeared once so the LCM is:

2 × 2 × 11 = 4 × 11 = 44
​
Result

44

Page 168 Exercise 1 Answer

To write an expression with repeated multiplication as an expression using an exponent, we need to count how many times the number is used as a factor. The number is called the base and how many times it’s used as a factor is the exponent.

For the expression 8 × 8 × 8 × 8 × 8 × 8 × 8, the number being multiplied is 8 so the base is 8. The base is used as a factor 7 times so the exponent is 7.

Therefore, 8 × 8 × 8 × 8 × 8 × 8 × 8 = 8⁷.

Result

8⁷

Page 168 Exercise 2 Answer

To write an expression with repeated multiplication as an expression using an exponent, we need to count how many times the number is used as a factor. The number is called the base and how many times it’s used as a factor is the exponent.

For the expression 4, the number 4 is being multiplied 1 time so the base is 4 and the exponent is 1.

Therefore, 4 = 4¹.

Result

4¹

Page 168 Exercise 3 Answer

To write an expression with repeated multiplication as an expression using an exponent, we need to count how many times the number is used as a factor. The number is called the base and how many times it’s used as a factor is the exponent.

For the expression 10 × 10 × 10 × 10, the number being multiplied is 10 so the base is 10. The base is used as a factor 4 times so the exponent is 4.

Therefore, 10 × 10 × 10 × 10 = 10⁴.

Result

10⁴

Page 168 Exercise 4 Answer

9² = 9 × 9 = 81

Result

81

Page 168 Exercise 5 Answer

99¹ = 99

A number with an exponent of 1 is always equal to the number itself.

Result

99

Page 168 Exercise 6 Answer

3,105⁰ = 1

A nonzero number with an exponent of 0 is always equal to 1.

Result

1

Page 168 Exercise 7 Answer

222 = 22 × 22 = 484

Result

484

Page 168 Exercise 8 Answer

To evaluate a power, rewrite as repeated multiplication and then find the product of the factors:

27 = 2 × 2 × 2 × 2 × 2 × 2 × 2

= 4 × 2 × 2 × 2 × 2 × 2

= 8 × 2 × 2 × 2 × 2

= 16 × 2 × 2 × 2

= 32 × 2 × 2

= 64 × 2

= 128
​
Result

128

Page 168 Exercise 9 Answer

To evaluate a power, rewrite as repeated multiplication and then find the product of the factors:

3⁴ = 3 × 3 × 3 × 3

= 9 × 9

= 81

Result

81

Page 169 Exercise 1 Answer

We need to write an algebraic expression that represents:

22 less than 5 times a number f

The phrase “22 less than” means we need to subtract by 22.

The phrase “5 times a number f” is represented by 5f.

The algebraic expression is then 5f − 22.

Result

5f − 22

Page 169 Exercise 2 Answer

We need to write an algebraic expression that represents:

48 times a number of game markers, g

There is only one operation of multiplication and the two factors are 48 and g so the algebraic expression is 48g.

Result

48g

Page 169 Exercise 3 Answer

We need to write an algebraic expression that represents:

A number of eggs, e, divided by 12

There is only one operation of division and the two quantities being divided are e and 12 so the algebraic expression is e ÷ 12.

Result

e ÷ 12

Page 169 Exercise 4 Answer

We need to write an algebraic expression that represents:

3 times the sum of m and 7

The phrase “sum of” means we need to add so we get m + 7.

The phrase “3 times the sum” means we need to multiply the sum m + 7 by 3.

The algebraic expression is then 3(m+7).

Result

3(m+7)

Page 169 Exercise 5 Answer

12x – 7 = 12(4) – 7 Substitute x = 7.

= 48 – 7 Multiply.

= 41 Subtract.
​
Result

41

Page 169 Exercise 6 Answer

x² ÷ y = 4² ÷ 8 Substitute x = 4 and y = 8.

= 16 ÷ 8 Evaluate the power.

= 2 Divide.
​
Result

2

Page 169 Exercise 7 Answer

5z + 3n − z³ = 5(1) + 3(7) − 13 Substitute n = 7 and z = 1.

= 5(1) + 3(7) – 1 Evaluate the power.

= 5 + 21 – 1 Multiply.

= 26 – 1 Add.

= 25 Subtract.
​
Result

25

Page 169 Exercise 8 Answer

= \(\frac{8^2}{2(4)}+3(7)-1\) Substitute n = 7, x = 4, y = 8, and z = 1.

= \(\frac{64}{2(4)}+3(7)-1\) Evaluate the power

= \(\frac{64}{8}+21-1\) Multiply

= 8 + 21 – 1 Divide

= 29 – 1 Add

= 28 Subtract

Result

28

Page 169 Exercise 1 Answer

The Order of Operations is:

Evaluate parentheses and brackets from inside out.

Evaluate any powers.

Multiply and divide in order from left to right.

Add or subtract in order from left to right.

Using the order of operations to evaluate the given expression gives:

80 − 42 ÷ 8

= 80 − 16 ÷ 8 Evaluate the power.

= 80 – 2 Divide.

= 78 Subtract.

Result

78

Page 169 Exercise 2 Answer

The Order of Operations is:

Evaluate parentheses and brackets from inside out.

Evaluate any powers.

Multiply and divide in order from left to right.

Add or subtract in order from left to right.

Using the order of operations to evaluate the given expression gives:

​92.3 – (3.2 ÷ 0.4) x 2³

= 92.3 – 8 x 2³ Evaluate inside the parentheses by dividing.

= 92.3 – 8 x 8 Evaluate the power.

= 92.3 – 64 Multiply.

= 28.3 Subtract.
​
Result

28.3

Page 169 Exercise 3 Answer

The Order of Operations is:

Evaluate parentheses and brackets from inside out.

Evaluate any powers.

Multiply and divide in order from left to right.

Add or subtract in order from left to right.

Using the order of operations to evaluate the given expression gives:

= \(\left[(8 \times 2.5) \div \frac{1}{2}\right]+120\) Evaluate inside the parentheses by evaluating the power.

= \(\left[20 \div \frac{1}{2}\right]+120\) Evaluate inside the parentheses by multiplying

= [20 x 2] + 120 Rewrite the division as multiplying by the reciprocal

= 40 + 120 Evaluate inside the brackets by multiplying.

= 160 Add

Result

160

Page 169 Exercise 4 Answer

The Order of Operations is:

Evaluate parentheses and brackets from inside out.

Evaluate any powers.

Multiply and divide in order from left to right.

Add or subtract in order from left to right.

Using the order of operations to evaluate the given expression gives:

[20 + (2.5 x 3)] – 3³

= [20 + 7.5] – 3³ Evaluate inside the parentheses by multiplying.

= 27.5 – 3³ Evaluate inside the brackets by adding.

= 27.5 – 27 Evaluate the power.

= 0.5 Subtract

Result

0.5

Page 169 Exercise 5 Answer

The Order of Operations is:

Evaluate parentheses and brackets from inside out.

Evaluate any powers.

Multiply and divide in order from left to right.

Add or subtract in order from left to right.

Using the order of operations to evaluate the given expression gives:

= \(\left[(2 \times 1) \div \frac{1}{3}\right]+8\) Evaluate inside the parentheses by evaluating the power.

= \(\left[2 \div \frac{1}{3}\right]+8\) Evaluate inside the parentheses by multiplying

= [2 x 3] + 8 Rewrite the division as multiplying by the reciprocal

= 6 + 8 Multiply

= 14 Add

Result

14

Page 170 Exercise 1 Answer

9y + 4 − 6y

= 9y – 6y + 4 Use the Commutative Property of Addition.

= (9 – 6)y + 4 Use the Distributive Property.

= 3y + 4 Subtract.
​
Result

3y + 4

Page 170 Exercise 2 Answer

​3x + 5 + 7x

= 3x + 7x + 5 Use the Commutative Property of Addition.

= (3 + 7)x + 5 Use the Distributive Property.

= 10x + 5 Add.
​
Result

10x + 5

Page 170 Exercise 3 Answer

​= 8x + 13 – 3x + 9

= 8x – 3x + 13 + 9 Use the Commutative Property of Addition.

= (8 – 3)x + (13 + 9) Use the Distributive Property.

= 5x + 22 Add and subtract.
​
Result

5x + 22

Page 170 Exercise 4 Answer

y² + 3y²

= (1 + 3)y² Use the Distributive Property.

= 4y² Add.
​
Result

4y²

Page 170 Exercise 5 Answer

4x + 15 − 3x + 10

= 4x – 3x + 15 + 10 Use the Commutative Property of Addition.

= (4 – 3)x + (15 + 10) Use the Distributive Property.

= x + 25 Subtract.
​
Result

x + 25

Page 170 Exercise 6 Answer

10x + 2x − 12x

= (10 + 2 – 12)x Use the Distributive Property.

= (12 – 12)x Add.

= 0 Subtract.

Result

0

Page 170 Exercise 1 Answer

We need to evaluate the expressions 5(2.2y + 1) − 3, 11y + 5 − y, and 11y + 2 for y = 1, 2, and 3.

Evaluating the expressions for y = 1 gives:

5(2.2y + 1) − 3

= 5(2.2⋅1 + 1) − 3

= 5(2.2 + 1) − 3

= 5(3.2) − 3

= 16 − 3

= 13
​
11y + 5 − y

= 11⋅1 + 5 − 1

= 11 + 5 − 1

= 16 − 1

= 15
​
11y + 2

= 11⋅1 + 2

= 11 + 2

= 13

Evaluating the expressions for y = 2 gives:

5(2.2y + 1) − 3

= 5(2.2⋅2 + 1) − 3

= 5(4.4+1) − 3

= 5(5.4) − 3

= 27 − 3

= 24
​
11y + 5 − y

= 11⋅2 + 5 − 2

= 22 + 5 − 2

= 27 − 2

= 25
​
11y + 2

= 11 ⋅ 2 + 2

= 22 + 2

= 24
​
Evaluating the expressions for y = 3 gives:

5(2.2y+1) − 3

= 5(2.2⋅3+1) − 3

= 5(6.6+1) − 3

= 5(7.6) − 3

= 38 − 3

= 35
​
11y + 5 − y

= 11⋅3 + 5 − 3

= 33 + 5 − 3

= 38 − 3

= 35
​
11y + 2

= 11 ⋅ 3 + 2

= 33 + 2

= 35

The completed table is then:

​

From the table, we can see the equivalent expressions are 5(2.2y+1) − 3 and 11y + 2 since they have the same value for all three values of y.

Result

Equivalent expressions: 5(2.2y+1) − 3 and 11y + 2

Page 170 Exercise 2 Answer

We need to determine if 10x − 3 + 2x − 5 and 4(3x−2) are equivalent expressions.

Two expressions are equivalent if we can use the properties of operations to rewrite one or both expressions so they equal the same expression.

Using the Commutative Property and then combining like terms for the first expression gives:

10x − 3 + 2x − 5 = (10x+2x) + (−3−5) = 12x − 8

Using the Distributive Property to rewrite the second expression gives:

4(3x−2) = 4(3x) − 4(2) = 12x − 8

Both expressions equal 12x − 8 so Yes the expressions are equivalent.

Result

Yes

Page 170 Exercise 3 Answer

We need to determine if 3y + 3 and \(9\left(y+\frac{1}{3}\right)\) are equivalent expressions.

Two expressions are equivalent if we can use the properties of operations to rewrite one or both expressions so they equal the same expression.

Using the Distributive Property to rewrite the second expression gives:

Since 3y + 3 ≠ 9y + 3, then No, the expressions are not equivalent.

Result

No

Page 170 Exercise 4 Answer

We need to determine if 6(3x+1) and 9x + 6 + 9x are equivalent expressions.

Two expressions are equivalent if we can use the properties of operations to rewrite one or both expressions so they equal the same expression.

Using the Distributive Property to rewrite the first expression gives:

6(3x+1) = 6(3x) + 6(1) = 18x + 6

Using the Commutative Property and then combining like terms for the second expression gives:

9x + 6 + 9x = (9x+9x) + 6 = 18x + 6

Both expressions equal 18x + 6 so Yes the expressions are equivalent.

Result

Yes

Page 170 Exercise 5 Answer

The Distributive Property across Addition states a(b+c) = a(b) + a(c). Therefore:

2(x + 4) = 2(x) + 2(4) = 2x + 8

Result

2x + 8

Page 170 Exercise 6 Answer

The Distributive Property across Subtraction states a(b−c) = a(b) − a(c). Therefore:

5x − 45 = 5(x) − 5(9) = 5(x−9)

Result

5(x−9)

Page 170 Exercise 7 Answer

The Distributive Property across Addition states a(b+c) = a(b) + a(c). Therefore:

3(x+7) = 3(x) + 3(7) = 3x + 21

Result

3x + 21

Chapter 4 Represent And Solve Equations And Inequalities

Section 4.0: Review What You Know

Page 175 Exercise 1 Answer

In the expression 6x, x is a variable.

Variable is a symbol, usually a letter, which represents a number, called the value of the variable. Variable can be either arbitrary, not fully specified, or unknown.

Result

Variable

Page 175 Exercise 2 Answer

x + 5 is an algebraic expression.

An algebraic expression as a type of math expression that has at least one variable and at least one operation.

Result

Algebraic expression.

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

Page 175 Exercise 3 Answer

Evaluate an expression to find its value.

To evaluate an expression one can use substitution to replace a variable with its numerical value. Then use the order of operations to simplify.

Result

Evaluate.

Page 175 Exercise 4 Answer

The expression on each side of the equal sign in an equation are equal.

For example,

4 + 5 = 3 + 6,

both sides are equal to 9.

Result

Equation.

Page 175 Exercise 5 Answer

An equation is given:

6 + 2 = 2 + 6.

Both 2 + 6 and 6 + 2 is equal to 8, thus the equation is true.

Result

True.

Page 175 Exercise 6 Answer

An equation is given:

2.5 − 1 = 1 − 2.5.

The left side, 2.5 − 1 equals 1.5, and the right side, 1 − 2.5 equals − 1.5, which means that the left and the right side are not equal. The equation is false.

Result

False.

Page 175 Exercise 7 Answer

An equation is given:

Both sides are equal to \(\frac{3}{2}\). Thus, the equation is true.

Result

True.

Page 175 Exercise 8 Answer

An equation is given:

Since dividing by 5 is equal to multiplying by its reciprocal, which is \(\frac{1}{5}\), both sides are equal to \(\frac{3}{20}\).

Thus, the equation is true.

Result

True.

Page 175 Exercise 9 Answer

An equation is given:

5 ÷ \(\frac{1}{3}\) = \(\frac{5}{3}\).

Since dividing by a number is equal to multiplying by its reciprocal, the left side is equal to 5 × 3, that is 15. The left and the right side are not equal. Thus, the equation is false.

Result

False.

Page 175 Exercise 10 Answer

An equation is given:

\(\frac{2}{3}\) x 5 = \(\frac{10}{15}\)

When multiplying a fraction by a whole number the result is a fraction where the denominator is the same and the numerator is the product of the whole number and the numerator. So, the left side is equal to \(\frac{10}{3}\). The left and the right side are not equal. Thus, the equation is false.

Result

False.

Page 175 Exercise 11 Answer

To evaluate the expression x − 2 use substitution, that is substitute x with its value x = 8.

x − 2 = 8 − 2 (Substitute the variable with its value.)

= 6 (Find the difference.)
​
Result

6

Page 175 Exercise 12 Answer

To evaluate the expression 2b use substitution, that is substitute b with its value b = 9.

2b = 2(9) (Substitute the variable with its value.)

= 18 (Find the product.)
​
Result

18

Page 175 Exercise 13 Answer

To evaluate the expression \(3 \frac{3}{4}\) + y use substitution, that is substitute y with its value y = \(\frac{5}{6}\).

\(3 \frac{3}{4}\) + y = \(3 \frac{3}{4}\) + \(\frac{5}{6}\) (Substitute the variable with its value.)

= \(\frac{15}{4}\) + \(\frac{5}{6}\) (Rewrite the mixed number as a fraction.)

= \(\frac{45}{12}\) + \(\frac{10}{12}\) (Rewrite friactions with 12 as denominators.)

= \(\frac{55}{12}\) (Find the sum.)

= \(4 \frac{7}{12}\) (Rewrite the result as a mixed number.)
​
Result

Page 175 Exercise 14 Answer

To evaluate the expression \(\frac{15}{x}\) use substitution, that is substitute x with its value x = 3.

\(\frac{15}{x}\) = \(\frac{15}{3}\) (Substitute the variable with its value.)

= 5 (Find the quotient.)
​
Result

5

Page 175 Exercise 15 Answer

To evaluate the expression 5.6t use substitution, that is substitute t with its value t = 0.7.

5.6t = 5.6(0.7) (Substitute the variable with its value.)

= 3.92 (Find the product.)
​
Result

3.92

Page 175 Exercise 16 Answer

To evaluate the expression 4x use substitution, that is substitute x with its value x = \(\frac{1}{2}\).

4x = 4(\(\frac{1}{2}\)) (Substitute the variable with its value.)

= 2 (Find the product.)

Result

2

Page 175 Exercise 17 Answer

Follow the order of operations, first evaluate parentheses and brackets from inside out, then evaluate powers, finally calculate any products and quotients, sums and differences.

[(33 ÷ 3) + 1] − 22 = [11 + 1] − 22 (Find the quotient.)

= 12 – 22 (Find the sum.)

= 12 – 4 (Evaluate the power.)

= 8 (Find the difference.)

Result

8

Page 175 Exercise 18 Answer

To plot the point A(−6,2) find −6 on the x-axis, since it is its x-coordinate, and find 2 on the y-axis, since it is its y-coordinate. Follow the gridlines from these points to where they meet. Mark that spot A.

Result

To plot the point A(−6,2) find −6 on the x-axis and find 2 on the y-axis. Follow the gridlines from these points to where they meet. Mark that spot A.

Chapter 4 Represent And Solve Equations And Inequalities

Midpoint

Page 209 Exercise 1 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 209 Exercise 2 Answer

The equation is f × 4 = \(\frac{1}{2}\). To solve the equation either divide both sides by four or multiply both sides by the reciprocal of 4, that is by \(\frac{1}{4}\).

​f x 4 = \(\frac{1}{2}\)

\(f \times 4 \times \frac{1}{4}=\frac{1}{2} \times \frac{1}{4}\) (Multiply both sides by \(\frac{1}{4}\).)

f = \(\frac{1}{2} \times \frac{1}{4}\)

f = \(\frac{1 \times 1}{2 \times 4}\)

f = \(\frac{1}{8}\) (Multiply the fractions.)
​
Result

f = \(\frac{1}{8}\)

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

Page 209 Exercise 3 Answer

The equations is 832 ÷ n = 16. Start by multiplying by n.

832 ÷ n = 16

832 ÷ n x n = 16 x n (Multiply both sides by n.)

832 = 16 x n (Simplify.)

832 ÷ 16 = 16 × n ÷ 16 (Divide both sides by 16.)
​
52 = n

Result

n = 52

Page 209 Exercise 4 Answer

The equation is x − 10 = 6. To solve start by adding 10 to both sides.

x − 10 = 6

x – 10 + 10 = 6 + 10 (Add 10 to both sides.)

x = 16 (Evaluate.)

Result

x = 16

Page 209 Exercise 5 Answer

n – n – 9 = 12 – n (The Subtraction Property of Equality was used.)

n – 9 + 12 = 12 – 9 (The equations are not equivalent.)

n – 9 + 9 = 12 + 9 (The Addition Property of Equality was used.)

n – 9 – n = 12 – n (The Subtraction Property of Equality was used.)

n – 9 + 9 = 12 – 12 (The equations are not equivalent..)

The equations which are not equivalent to n − 9 = 12 are: n − 9 + 9 = 12 − 12 and n − 9 + 12 = 12 − 9.

Result

n − 9 + 9 = 12 − 12, n − 9 + 12 = 12 − 9.

Page 209 Exercise 6 Answer

To answer the question use substitution – substitute d for each of the values and check it the equation is true.

9 = 18 ÷ d

9 = 18 ÷ 0.5

9 = 180 ÷ 5 (Multiply both numbers by 10)

9 ≠ 36

9 = 18 ÷ \(\frac{10}{5}\)

9 = 18 ÷ 2 (Simplify)

9 = 9

9 = 18 ÷ 162

9 = \(\frac{18}{162}\)

9 ≠ \(\frac{1}{9}\)

9 = 18 ÷ \(\frac{1}{4}\)

9 = 18 x 4 (Multiply by the reciprocal of \(\frac{1}{4}\))

9 ≠ 72

The only value of d that made the equation 9 = 18 ÷ d true was \(\frac{10}{5}\).

Result

Page 209 Exercise 7 Answer

Substitute the variables with their values in the expression A = bh. Solve the equation to find the height.
​
A = bh

15.3 = bh

15.3 = 4.5 h (Substitute variables with their values.)

15.3 ÷ 4.5 = h (Divide both sides by 4.5.)
​
3.4 = h

The height of the parallelogram is 3.4 centimeters.

Result

3.4 centimeters.

Page 210 Exercise 1a Answer

To solve the equation \(\frac{3}{4}\)h = \(3 \frac{3}{8}\) using a reciprocal, multiply both sides of the equation by the reciprocal of \(\frac{3}{4}\) which is \(\frac{4}{3}\).

The equation than look like:

Result

Page 210 Exercise 1b Answer

From Part A, we know the equation \(\frac{3}{4} h=3 \frac{3}{8}\) can be used to find the height h of the finished totem. In Part A, we determined that the equation \(\frac{3}{4} h \times \frac{4}{3}=3 \frac{3}{8} \times \frac{4}{3}\) was an equivalent equation. Simplifying this equivalent equation gives:

​\(\frac{3}{4} h \times \frac{4}{3}=3 \frac{3}{8} \times \frac{4}{3}\)

h = \(3 \frac{3}{8} \times \frac{4}{3}\) Simplify the left side.

h = \(\frac{27}{8} \times \frac{4}{3}\) Rewrite the mixed number as an improper fraction.

h = \(\frac{108}{24}\) Multiply.

h = \(\frac{9}{2}\) Reduce the fraction.

h = \(4 \frac{1}{2}\) Rewrite as a mixed number.
​
The height of the finished totel pole is then \(4 \frac{1}{2}\) ft.

Next, we need to write and solve an equation to find the height, s, of the section that has not been carved. We know that \(3 \frac{3]{8}\) ft of the totem has been completed so the finished totem pole will have a height of s + \(3 \frac{3}{8}\) feet. Since the height of the finished totem pole is \(4 \frac{1}{2}\) ft, then the equation is \(s+3 \frac{3}{8}=4 \frac{1}{2}\).

Solving this equation for s we get:

​\(s+3 \frac{3}{8}=4 \frac{1}{2}\)

\(s+3 \frac{3}{8}-3 \frac{3}{8}=4 \frac{1}{2}-3 \frac{3}{8}\) Subtract \(3 \frac{3}{8}\) on both sides.

s = \(4 \frac{4}{8}-3 \frac{3}{8}\) Get a common denominator.

s = \(1 \frac{1}{8}\) Subtract.

The height of the section that has not been carved is then \(1 \frac{1}{8}\)ft.

Result

Page 210 Exercise 1c Answer

To answer the question solve the equation $10.50 + x = $19.35 Start by subtracting 10.50 from both sides.

10.50 + x − 10.50 = 19.35 − 10.50

x = 8.85
​
The cost of the wood Ronald used is $8.85.

Result

$8.85

Page 210 Exercise 1d Answer

To solve the equation $10.50 + y = $35.19 Ronald could use the Subtraction Property of Equality, that is he could subtract 10.50 from both sides. If he does so the unknown y stays on one side of the equation, since he used a property of equality the equation stayed balanced and the other side of the equation than shows the value of y.

$10.50 + y = $35.19

10.50 + y − 10.50 = 35.19 − 10.50

y = $24.69
​
Result

y = $24.69

Chapter 4 Represent And Solve Equations And Inequalities

Page 223 Exercise 1 Answer

In the video, we see someone trying to fit as many shoes as possible into a suitcase. Some possible questions that come to mind are then:

Why does he need to bring so many shoes?

How many shoes will fit inside the suitcase?

What will be the weight of the bag once his shoes are packed?

Result

Possible questions:

Why does he need to bring so many shoes?

How many shoes will fit inside the suitcase?

What will be the weight of the bag once his shoes are packed?

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

Page 223 Exercise 2 Answer

In the video, we saw someone trying to fit as many shoes as possible into a suitcase. We were also given in the video the weights of the suitcase as he added each pair of shoes and we were told the weight limit was 50 pounds. The Main Question for the video is then:

How many shoes can he fit in his suitcase if the weight limit is 50 pounds?

Result

How many shoes can he fit in his suitcase if the weight limit is 50 pounds?

Page 223 Exercise 3 Answer

We need to make a prediction about how many shoes will fit. To make a prediction, we need to use an estimate for the weight of each pair of shoes and the weight of the suitcase.

Assuming the suitcase weighs 6 pounds, then the shoes will weigh 50 − 6 = 44 pounds. Assuming each pair of shoes weighs 2 pounds, then a prediction could be 22 pairs of shoes since 44 ÷ 2 = 22.

Result

Possible answer:

I predict that 22 pairs of shoes will fit by using an estimate of 6 pounds for the weight of the suitcase and 2 pounds for each pair of shoes.

Page 223 Exercise 4 Answer

We know the weight limit is 50 pounds.

A number that is too small to be the answer could then be 2, since this would mean 2 pairs of shoes and the suitcase would have a combined weight of 50 pounds, which is unrealistic.

A number that is too large to be the answer could be 100 since this would mean each pair of shoes weighed less than 50 ÷ 100 = 0.5 pounds, which is unrealistic.

Writing these two numbers on the given number line gives:

Result

Possible answers:

Too small: 2

Too large: 100

Page 223 Exercise 5 Answer

The prediction I made in Exercise 3 was 22 pounds. Plotting this on the number line from Exercise 4 then gives:

Result

Plot the prediction you made in Exercise 3 on the number line from Exercise 4.

Page 224 Exercise 6 Answer

To determine how many pairs of shoes he can fit, there are three pieces of information we need to know:

How much the suitcase weighs.

How much each pair of shoes weighs.

What the weight limit is.

We can then use this information to write an equation or an inequality that represents the total weight of the suitcase.

Result

We need to know how the suitcase weighs, how much each pair of shoes weighs, and what the weight limit is. we can use this information to write an equation or inequality that represents the total weight of the suitcase.

Page 224 Exercise 7 Answer

To get the information we need, we can use the information given by the Act 2 images in the online interactivity:

Weight of each pair of shoes: 2.2 lb, 2.5 lb, 2.2 lb, and 2.3 lb

Weight limit: 50 lb

Weight of suitcase: 6.2 lb

Result

Weight of each pair of shoes: 2.2 lb, 2.5 lb, 2.2 lb, and 2.3 lb

Weight limit: 50 lb

Weight of suitcase: 6.2 lb

Page 224 Exercise 8 Answer

To represent the situation, we need to use the following relationship:

weight of shoes + weight of suitcase ≤ weight limit

Possible solution:

We know that four pairs of shoes have weights of 2.2 lb, 2.5 lb, 2.2 lb, and 2.3 lb. To write an expression for the total weight of the shoes, we can use an estimate for the average weight of each pair of shoes. To find the estimated average, add the four weights and then divide by 4:

The average weight of each pair is then 2.3 lb.

Let p be the number of pairs of shoes. Since the average weight of each pair is about 2.3 lb, then the total weight of p pairs of shoes is 2.3p pounds.

Since the suitcase weighs 6.2 lb, then the total weight of the suitcase and shoes is 2.3p + 6.2 pounds.

The weight limit is 50 lb so 2.3p + 6.2 must be less than or equal to 50. The inequality is then 2.3p + 6.2 ≤ 50.

Solving this inequality gives:

2.3p + 6.2 ≤ 50

2.3p + 6.2 − 6.2 ≤ 50 − 6.2 Subtract 6.2 on both sides.

2.3p ≤ 43.8 Simplify.

2.3p ÷ 2.3 ≤ 43.8 ÷ 2.3 Divide both sides by 2.3.

p ≤ 19.04 Simplify.

The number of pairs must be a whole number so if the number of pairs is at most 19.04, then the greatest number of pairs is 19 pairs.

Note: Since we only know the weights of 4 pairs of shoes, the actual average weight of the shoes may not be 2.3 lb. There could be pairs of shoes that weigh less than 2.2 lb or weigh more than 2.5 lb. This means that you can represent the situation using an average weight that is not 2.3.

For example, you could use an average of 2.4 lb or 2.5 lb since it is possible that the rest of the shoes have weights closer to 2.5 lb or you could use an average weight of 2.2 lb or 2.1 lb since it is possible that the rest of the shoes have weights closer to 2.2 lb.

Result

Possible answers: 2.3p + 6.2 ≤ 50 19 pairs

Page 224 Exercise 9 Answer

From Exercise 8, we got an answer of 19 pairs. In Exercise 3, I predicted 22 pairs. The answer of 19 pairs is then lower than my prediction. This is because I used an average weight of 2 lb for each pair of shoes in my prediction but an average of 2.3 lb was used in Exercise 8. A larger average means fewer pairs of shoes can fit.

Page 225 Exercise 10 Answer

From the Act 3 video, he had weights of 47.2 lb for 18 pairs of shoes, 49.3 for 19 pairs, and 51.9 lb for 20 pounds. Since the limit is 50 lb, then the answer to the Main Question is 19 pairs since this was the most pairs he could fit and still have a weight less than 50 lb.

Result

19 pairs

Page 225 Exercise 11 Answer

Possible solution: In Exercise 8, I used an average weight of 2.3 lb per pair of shoes to write the inequality 2.3p + 6.2 ≤ 50. This inequality gave me an answer of 19 pairs, which matches the answer in the video.

If you did not use the same inequality I did, your answer likely won’t match. If you got a higher answer, it’s because you used a lower average weight so the reason for the difference is that the shoes weighed more than you thought. Similarly, if you got a lower answer, it’s because you used a higher average weight so the reason for the difference is that the shoes weighed less than you thought.

Result

Possible answer: In Exercise 8, I used an average weight of 2.3 lb per pair of shoes to write the inequality 2.3p + 6.2 ≤ 50. This inequality gave me an answer of 19 pairs, which matches the answer in the video.

Page 225 Exercise 12 Answer

Possible answer: In Exercise 8, I used the inequality 2.3p + 6.2 ≤ 50, which gave me an answer of 19 pairs. Since my answer matches the video, I would not change my model.

If you did not use the same inequality I did, your answer likely didn’t match so you should change your model. If you got a higher answer, it’s because you used a lower average weight so you need to change your model to use a higher average weight. Similarly, if you got a lower answer, it’s because you used a higher average weight so you need to change your model to use a lower average weight.

Result

Possible answer: In Exercise 8, I used the inequality 2.3p + 6.2 ≤ 50, which gave me an answer of 19 pairs. Since my answer matches the video, I would not change my model.

Page 226 Exercise 13 Answer

Possible answer: I used a mathematical model to represent the situation by writing an inequality. I wrote the inequality using the following relationship:

weight of shoes + weight of suitcase ≤ weight limit

To write an expression for the weight of shoes, I let p be the number of pairs of shoes and multiplied p by an estimate for the average weight of the shoes. I then used this expression, the given weight of 6.2 lb for the suitcase, and the given weight limit of 50 lb to write the inequality. Solving this inequality for p and rounding to the nearest whole number then allowed me to answer the Main Question.

Result

I wrote the inequality weight of shoes + weight of suitcase ≤ weight limit to represent the situation. To write an expression for the weight of the shoes, I let p be the number of pairs of shoes and then multiplied p by an estimate for the average weight of the shoes. I then used this expression and the given weights for the suitcase and weight limit to write the inequality. Solving this inequality for p allowed me to answer the Main Question.

Page 226 Exercise 14 Answer

An inequality was more useful to answer the Main Question. This is because the combined weight of the shoes and suitcase did not have to equal 50 lb, it could also be less than 50 lb.

Page 226 Exercise 15 Answer

If the weight limit is changed to 40 pounds, then the number of shoes would decrease.

Changing the inequality 2.3p + 6.2 ≤ 50 from Exercise 8 to 2.3p + 6.2 ≤ 40 and solving for p gives:

2.3p + 6.2 − 6.2 ≤ 40 − 6.2 Subtract 6.2 on both sides.

2.3p ≤ 33.8 Simplify.

2.3p ÷ 2.3 ≤ 33.8 ÷ 2.3 Divide both sides by 2.3

p ≤ 14.70 Simplify.

The maximum number of shoes would then decrease to 14 pairs.

Result

If the weight limit is changed to 40 pounds, then the maximum number of shoes would decrease to 14 pairs.

Chapter 1 Algebra: Understand Numerical And Algebraic Expressions

Review What You Know

Page 2 Exercise 1 Answer

A formula is a rule that connects two or more quantities using symbols. Therefore, the word that goes on the blank is formula.

Result

formula

Page 2 Exercise 2 Answer

A(n) numerical expression is a mathematical phrase that includes numbers and at least one operations.

Result

NUMERICAL EXPRESSION

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

Page 2 Exercise 3 Answer

A quantity that is unknown may be represented by a(n) variable.

Result

VARIABLE

Page 2 Exercise 4 Answer

56 − 27 + (16 ÷ 4) Evaluate

= 56 − 27 + 4 Solve parentheses

= 29 + 4 Subtract

= 33 Add

Result

33

Page 2 Exercise 5 Answer

94 − (5 × 6) ÷ 6 × 11 Evaluate

= 94 − 30 ÷ 6 × 11 Solve Parentheses

= 94 − 5 × 11 Divide

= 94 − 55 Multiply

= 39 Subtract

Result

39

Page 2 Exercise 6 Answer

[21 ÷ 3] + (18 ÷ 6) Evaluate

= 7 + (18 ÷ 6) Solve square brackets

= 7 + 3 Solve parentheses

= 10 Add

Result

10

Page 2 Exercise 7 Answer

2 × 36 − (12 + 7) Evaluate

= 2 × 36 − 19 Solve parentheses

= 72 − 19 Multiply

= 53 Simplify

Result

53

Page 2 Exercise 8 Answer

12 ÷ (2 + 1) × 15 ÷ 3 Evaluate

= 12 ÷ 3 × 15 ÷ 3 Solve Parentheses

= 4 × 5 Divide

= 20 Multiply

Result

20

Page 2 Exercise 9 Answer

15 − (2 × 3) × 2 + 9 Evaluate

= 15 − 6 × 2 + 9 Solve Parentheses

= 15 − 12 + 9 Multiply

= 3 + 9 Subtract

= 12 Add

Result

12

Page 2 Exercise 10 Answer

Length = 13 cm ; Width = 13 cm

Perimeter

= 2l + 2w

= 2(13) + 2(13) Substitute the value

= 26 + 26 Multiply

= 52 cm

Area

= lw

= (13)⋅(13) Substitute the value

= 169 sq. cm

Result

P = 52 cm

A = 169 cm²

Page 2 Exercise 11 Answer

Length = 21 in ; Width = 5 in

Perimeter

= 2l + 2w

= 2(21) + 2(5) Substitute the value

= 42 + 10 Multiply

= 52 in

Area

= lw

= (21) ⋅ (5) Substitute the value

= 105 sq. in

Result

P = 52 in

A = 105 in²

Page 2 Exercise 12 Answer

Length = 15 m ; Width = 9 m

Perimeter

= 2l + 2w

= 2(15) + 2(9) Substitute the value

= 30 + 18 Multiply

= 48 m

Area = lw

= (15)⋅(9) Substitute the value

= 135 sq. m

Result

P = 48 m

A = 135 m²

Chapter 1 Algebra: Understand Numerical And Algebraic Expressions

Independent Practice

Page 9 Exercise 1 Answer

How many times 4 is use as a factor in the expression \(4^{5}\)

Since exponent = 5$

So, 4 is used 5 times as a factor in the expression \(4^{5}\)

Write the numerical expression as repeated multiplication.

= 4 × 4 × 4 × 4 × 4 = 1024

Result

5 times

4 × 4 × 4 × 4 × 4

Page 9 Exercise 2 Answer

Write any power then evaluate the power.

= 2 × 2 × 2 = 4 × 2 = 8

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

Result

\(2^{3}\) = 2 × 2 × 2 = 8

Page 9 Exercise 3 Answer

81 = 3 × 3 × 3 × 3 Write 81 as a repeated multiplication of 3s

81 = 3 x 3 x 3 x 3 = \(3^{4}\) Write as a power

Result

3 × 3 × 3 × 3

Page 9 Exercise 4 Answer

\(\left(\frac{1}{2}\right)^3\) Exponent = 3

= \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\) Write as repeated multiplication

Result

Page 9 Exercise 5 Answer

9 × 9 × 9 × 9

9 is used 4 times as a factor.

Exponent = 4

Result

Exponent = 4

Page 9 Exercise 6 Answer

1.2 is used 9 times as a factor.

Exponent = 9

Result

Exponent = 9

Page 9 Exercise 7 Answer

\(\frac{1}{6}\) is used 3 times as a factor.

Exponent = 3

Result

Exponent = 3

Page 9 Exercise 8 Answer

\(0.6^{2}\) Evaluate

= 0.6 × 0.6 Write as repeated multiplication

= 0.36 Multiply

Result

0.36

Page 9 Exercise 9 Answer

\(\left(\frac{1}{4}\right)^2\) Evaluate

= \(\frac{1}{4} \times \frac{1}{4}\) Write as repeated multiplication

= \(\frac{1}{16}\) Multiply

Result

Page 9 Exercise 10 Answer

\(2^{7}\) Evaluate

= 2 × 2 × 2 × 2 × 2 × 2 × 2 Write as repeated multiplication

= 128 Multiply

Result

128

Page 10 Exercise 12 Answer

base = 2 ; exponent = 5

\(2^{5}\) = 2 × 2 × 2 × 2 × 2 = 32

base = 5 ; exponent = 2

\(5^{2}\) = 5 × 5 = 25

Result

\(2^{5}\) = 32 ; \(5^{2}\) = 25

Page 10 Exercise 13 Answer

Her response was incorrect because

80000 × 25

= \(8 \times 10^4 \times 5^2\)

= \(2^3 \times 10^4 \times 5^2\)

Result

Page 10 Exercise 14 Answer

\(0.3^{3}\) Evaluate

= 0.3 × 0.3 × 0.3 Write as repeated multiplication

= 0.027 Multiply

Result

n = 0.027

Page 10 Exercise 15 Answer

1,000,000 = \(10^{6}\)

10 is used as a base because 10 is multiplied repeatedly 6 times as a factor.

Result

Because 10 is repeatedly multiplied

Page 10 Exercise 16 Answer

Zach invested $50 and tripled his money in two years.

$50 × 3 = $150

Kayla invested $50 and after two years, the amount was equal to 50 to the third power.

\(50^{3}\) = 50 × 50 × 50 = $125000

After two years Kayla had more money.

Result

Kayla had more money.

Page 10 Exercise 17 Answer

A) \(2^{10}\)

= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

= 1024

B) 5 x 5 x 5 x 5

= 625

C) \(4^{5}\)

= 4 x 4 x 4 x 4 x 4

= 1024

D) 4 x 4 x 4 x 4 x 4

= 1024

The expression 5 x 5 x 5 x 5 is Not equal to 1024

Result

5 x 5 x 5 x 5

Page 10 Exercise 18 Answer

A) \(\frac{1}{3} \times \frac{1}{6}\)

= \(\frac{1}{18}\)

B) \(\frac{1}{4} \times\left(\frac{1}{3}\right)^3\)

= \(\frac{1}{4} \times \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3}\)

= \(\frac{1}{108}\)

C) \(\left(\frac{1}{2}\right)^2 \times\left(\frac{1}{3}\right)^2\)

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

= \(\frac{1}{4} \times \frac{1}{9}\)

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

D) \(\frac{1}{2} \times \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3}\)

= \(\frac{1}{2} \times \frac{1}{27}\)

= \(\frac{1}{54}\)

The expression \(\left(\frac{1}{2}\right)^2 \times\left(\frac{1}{3}\right)^2\) is equal to \(\frac{1}{36}\)

Result

Chapter 1 Algebra: Understand Numerical And Algebraic Expressions

Homework And Practice

Page 11 Exercise 1 Answer

\(5^{12}\) = 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5

5 is repeatedly multiplied.

Base = 5

Result

5

Page 11 Exercise 2 Answer

= 1.2 × 1.2

1.2 is repeatedly multiplied.

Base = 1.2

Result

1.2

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

Page 11 Exercise 3 Answer

\(\left(\frac{1}{3}\right)^4\)
​
= \(\frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3}\)

\(\frac{1}{3}\) is repeatedly multiplied.

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

Result

Page 11 Exercise 4 Answer

7 × 7 × 7 × 7

= \(7^{4}\)

7 is multiplied 4 times as a factor.

Exponent = 4

Result

4

Page 11 Exercise 5 Answer

\(\frac{2}{3}\) is multiplied 8 times as a factor.

Exponent = 8

Result

8

Page 11 Exercise 6 Answer

0.5 × 0.5 × 0.5

= \(0.5^{3}\)

0.5 is multiplied 3 times as a factor.

Exponent = 3

Result

3

Page 11 Exercise 7 Answer

\(9^{3}\) Evaluate

= 9 x 9 x 9 Write as repeated multiplication

= 729 Multiply

Result

729

Page 11 Exercise 8 Answer

\(\left(\frac{1}{4}\right)^3\) Evaluate

= \(\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4}\) Write as repeated multiplication

= \(\frac{1}{64}\) Multiply

Result

Page 11 Exercise 9 Answer

\(3^{5}\) Evaluate

= 3 x 3 x 3 x 3 x 3 Write as repeated multiplication

= 243 Multiply

Result

243

Page 11 Exercise 10 Answer

\(\left(\frac{1}{8}\right)^3\) Evaluate

= \(\frac{1}{8} \times \frac{1}{8} \times \frac{1}{8}\) Write as repeated multiplication

= \(\frac{1}{512}\) Multiply

Result

Page 11 Exercise 11 Answer

\(99^{9}\) Evaluate

= 1 Any non zero raised to the exponent of 0 has a value of 1

Result

1

Page 11 Exercise 12 Answer

\(1.5^{2}\) Evaluate

= 1.5 × 1.5 Write as repeated multiplication

= 2.25 Multiply

Result

2.25

Page 12 Exercise 13 Answer

\(0.3^{4}\) = 0.3 × 0.3 × 0.3 × 0.3 = 0.0081 Evaluate

\(0.9^{2}\) = 0.9 × 0.9 = 0.81 Evaluate

Result

No \(0.3^{4}\) is NOT equal to \(0.9^{2}\)

Page 12 Exercise 14 Answer

27 = 3 × 3 × 3 × 3 Write as repeated multiplication

27 = \(3^{4}\) Write as power

Result

27 can be written using number 3 as :

3 × 3 × 3 × 3 and \(3^{4}\)

Page 12 Exercise 15 Answer

\(1^{102}\) = 1

The value of any power of 1 is equal to 1.

Example → \(1^{32}=1 ; 1^6=1 ; 1^{100}=1\)

Result

1

Page 12 Exercise 19 Answer

A) \(8^{3}\)

= 8 × 8 × 8

= 512

B) 6 × 6 × 6

= 216

C) \(7^{3}\)

= 7 × 7 × 7

= 343

D) 7 × 7 × 7 × 7

= 2401

Result

\(7^{3}\) is equal to 343

Page 12 Exercise 20 Answer

A) \(0.5^{3}\)

= 0.5 × 0.5 × 0.5

= 0.125

B) 0.5 × 3

= 1.5

C) 0.5 × 0.5 × 0.5

= 0.125

D) \(0.5 \times 0.5^2\)

= 0.5 × 0.5 × 0.5

= 0.125

Result

0.5 × 3 is NOT equal to 0.125

Chapter 1 Algebra: Understand Numerical And Algebraic Expressions

Guided Practice

Page 15 Exercise 1 Answer

80 ÷ 8 × 5 + 4 = 90

We should insert parentheses around (80 ÷ 8) and (5 + 4).

(80 ÷ 8) × (5 + 4)

= 10 × 9 Solve within parentheses

= 90

Result

(80 ÷ 8) × (5 + 4) = 90

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

Page 15 Exercise 2 Answer

(21 − 3) × (7 + 2) ÷ (12 − 4)

Using PEMDAS (parentheses, exponents, multiplications, division, addition, subtraction) for the order of operations:

First perform the operations within the parentheses.

Next, perform the multiplication.

Finally, perform the division.

Result

Last, we will perform division.

Page 15 Exercise 3 Answer

\(5^{2}\) + (6.7 − 3.1) Evaluate

= \(5^{2}\) + 3.6 Evaluate inside the parentheses

= 25 + 3.6 Evaluate the power

= 28.6 Add

Result

28.6

Page 15 Exercise 4 Answer

(8.2 + 5.3) ÷ 5 Evaluate

= 13.5 ÷ 5 Evaluate inside the parentheses

= 2.7 Divide

Result

2.7

Page 15 Exercise 5 Answer

[(7.3 + 3.6) − 4.7] + 1.8 − \(2^{2}\) Evaluate

= [10.9 − 4.7] + 1.8 − \(2^{2}\) Evaluate inside the parentheses

= 6.2 + 1.8 − \(2^{2}\) Evaluate inside the bracket

= 6.2 + 1.8 − 4 Evaluate the power

= 8 − 4 Add

= 4 Subtract

Result

4

Page 15 Exercise 6 Answer

\(\left[(11.2+8.8) \times \frac{1}{4}\right]-1.8\) Evaluate

= \(\left[20 \times \frac{1}{4}\right]-1.8\) Evaluate inside the parentheses

= 5 − 1.8 Evaluate inside the bracket

= 3.2 Subtract

Result

3.2

Page 15 Exercise 7 Answer

\(4^{2}\) − (3.1 + 6.4) + 4.5 Evaluate

= \(4^{2}\) − 9.5 + 4.5 Evaluate inside the parentheses

= 16 − 9.5 + 4.5 Evaluate the power

= 6.5 + 4.5 Subtract

= 11 Add

Result

11

Page 15 Exercise 8 Answer

\((8.7+3.3) \times\left(\frac{1}{2}\right)^2\) Evaluate

= \(12 \times\left(\frac{1}{2}\right)^2\) Evaluate inside the parentheses

= 12 x \(\frac{1}{4}\) Evaluate inside the parentheses

= 3 Multiply

Result

3

Page 15 Exercise 9 Answer

157.8 − (\(3^{2}\) + 6) × 3 Evaluate

= 157.8 − (9 + 6) × 3 Evaluate the power

= 157.8 − 15 × 3 Evaluate inside the parentheses

= 157.8 − 45 Multiply

= 112.8 Subtract

Result

112.8

Page 15 Exercise 10 Answer

4.3 + (8.4 − 5.1) Evaluate

= 4.3 + 3.3 Evaluate inside the parentheses

= 7.6 Add

Result

7.6

Page 15 Exercise 11 Answer

\(4^3-\left[(9.9 \div 3.3) \times \frac{1}{3}\right]\) Evaluate

= \(4^3-\left[3 \times \frac{1}{3}\right]\) Evaluate inside the parentheses

= \(4^{3}\) – 1 Evaluate inside the bracket

= 64 − 1 Evaluate the power

= 63 Subtract

Result

63

Page 15 Exercise 12 Answer

[\(2^{3}\) × (152 ÷ 8)] − 52 Evaluate

= [\(2^{3}\) × 19] − 52 Evaluate inside the parentheses

= [8 × 19] − 52 Evaluate the power

= 152 − 52 Evaluate inside the bracket

= 100 Subtract

Result

100

Page 16 Exercise 13 Answer

We should follow Order of Operations rule to find which part of the numerical expression to evaluate first.

Order of Operations:

1. → Evaluate inside the parentheses and brackets.

2. → Evaluate the powers.

3. → Divide and Multiply from left to right.

4. → Add and Subtract from left to right.

Evaluate the Expression:

(26 + 2.5) − [(8.3 × 3) + (\(1^{3}\) − 0.25)]

= (26 + 2.5) − [(8.3 × 3) + (1 − 0.25)] Evaluate the powers

= 28.5 − [24.9 + 0.75] Evaluate inside the parentheses

= 28.5 − 25.65 Evaluate inside the brackets

= 2.85

Result

2.85

Page 16 Exercise 14 Answer

Amount of plants an elk can eat in a day = 20 pounds

Number of days in a week = 7

Total number of elk = 18

18 × (20 × 7)

= 18 × 140

= 2520 pounds

Result

18 elk can eat 2520 pounds of plants in a week.

Page 16 Exercise 15 Answer

Cost of each hairbrush = $3.99

Number of hairbrushes Lillian bought = 4

She had a coupon of = $1 off

Amount paid by her mother = Half of the total bill

Expression : [(3.99 × 4) − 1] ÷ 2

[(3.99 × 4) − 1] ÷ 2

= [15.96 − 1] ÷ 2 Evaluate inside the parentheses

= 14.96 ÷ 2 Evaluate inside the bracket

= 7.48

Result

Lillian paid $7.48 towards the purchase of hairbrushes.

Page 16 Exercise 16 Answer

Frederick : In the expression solved by Frederick there is parentheses and bracket.

He follow Order of operation and evaluate the parentheses and bracket first.

So he is Correct.

Lana : In the expression solved by Lana there is no parentheses and bracket.

She also follow Order of operation and operate Add and Subtract from left to right. So she is also Correct.

Result

Frederick is correct as he evaluate parentheses and bracket first while Lana is correct as she solved the expression by Adding and Subtracting from left to right.

Page 16 Exercise 17 Answer

12.3 × [(2 × 1.7) + 6.6] − \(2^{3}\)

= 12.3 × [3.4 + 6.6] − \(2^{3}\) Evaluate inside the parentheses

= 12.3 × 10 − \(2^{3}\) Evaluate inside the bracket

= 12.3 × 10 − 8 Evaluate the power

= 123 − 8 Multiply

= 115

\(2^{4}\) ÷ [(3.2 × 0.8) + 1.44]

= \(2^{4}\) ÷ [2.56 + 1.44] Evaluate inside the parentheses

= \(2^{4}\) ÷ 4 Evaluate inside the bracket

= 16 ÷ 4 Evaluate the power

= 4

= 6.2 + (1 + 4.8) Evaluate inside the parentheses

= 6.2 + 5.8 Evaluate inside the parentheses

= 12

[4 × (9.6 ÷ 3)] + 8.2

= [4 × 3.2] + 8.2 Evaluate inside the parentheses

= 12.8 + 8.2 Evaluate inside the bracket

= 21

Result

12.3 × [(2 × 1.7) + 6.6] − \(2^{3}\) = 115

\(2^{4}\) ÷ [(3.2 × 0.8) + 1.44] = 4

[4 × (9.6 ÷ 3)] + 8.2 = 21

Chapter 4 Represent And Solve Equations And Inequalities

Section 4.1: Understand Equations And Solutions

Page 177 Exercise 1 Answer

Write an equation where the left side represents adding 10 cubes to the pan with 3 cubes and adding 4 cubes to the pan with 9 cube. It the sides are equal, that is if the equation is true, the pans will balance.

10 + 3 = 4 + 9

13 = 13
​
The pans would balance.

Result

The pans would balance and the equation is 10 + 3 = 4 + 9.

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

Page 177 Exercise 1 Answer

To make the pans balance write an equation, let’s say the left side has 3 cube and the right side has 9 cubes.

One way to make the equation true is to add 6 additional cubes to the left side.

3 + 6 = 9

There are 12 cubes in all on the balance so a second way to make the equation true is to remove 3 cubes from the right side and add them tot he right side since this would give 6 cubes on each side.

3 + 3 = 9 – 3

Result

One way is to add 6 additional cubes to the left side. A second way is to remove 3 cubes from the right side and add them to the left side.

Page 178 Exercise 1 Answer

To determine which app Tracy should buy substitute the variable x with its possible values, that is prices of apps and check for which value of x the equation

$21.00 = x + $9.00

is true.

For x = $9.50.

$21.00 = x + $9.00

= $9.50 + $9.00

= $18.50 ≠ $21.00

For x = $10.50.

$21.00 = x + $9.00

= $10.50 + $9.00

= $19.50 ≠ $21.00

For x = $12.00.

$21.00 = x + $9.00

= $12.00 + $9.00

= $21.00
​
The equation is true for x = $12.00. Thus, Tracy bought a remote desktop.

Three values are possible, however two of them, $9.50 and $10.50, are not whole numbers, but $12.00 is.

The sum of a whole number and a decimal number will be a decimal number. Since on the left side is a whole number, and to x we add $9.00 – a whole number, x must as well be a whole number .

Thus, we know that the solution is x = $12.00, without using substitution.

Result

x = $12.00, Remote Desktop.

Page 179 Exercise 2 Answer

Use substitution – that is substitute y for each of the values and evaluate to see if any of Anthony’s friends guessed correctly.

y = 48

y ÷ 4 = 48 ÷ 4

=12 ≠ 13

y = 60

y ÷ 4 = 60 ÷ 4

= 15 ≠ 13

y = 120

y ÷ 4 = 120 ÷ 4

= 30 ≠ 13
​
None of Anthony’s friends guessed correctly.

Step 2

To find how many marbles Anthony has, multiply both sides of the equation by four.

y ÷ 4 = 13

y = 13 × 4

y = 52
​
Anthony has 52 marbles in all.

Result

None of Anthony’s friends guessed correctly, he has 52 marbles in all.

Page 180 Exercise 1 Answer

To determine whether a given number makes an equation true use substitution. Replace the variable with its value and evaluate both sides of the equation to make sure they have the same value. If they do, the given number makes the equation true.

Page 180 Exercise 2 Answer

An equation is true when both the left and the right side of the equation have the same value.

That is, if the equation contains a variable or variables, it is true for the variable or variables for which both sides have equal values.

For example, the equation

x + 15 = 18,

is true for x = 3, since

3 + 15 = 18.

However it is not true for x = 4, since

4 + 15 = 19 ≠ 18.
​
Result

An equation is true when both the left and the right side of the equation have the same value.

Page 180 Exercise 3 Answer

To check whether Ben is correct substitute the variable n in the expression 7n = 45 vith its value, that is with 5, as Ben says.

7n = 45

7(5) = 45

35 ≠ 45
​
Ben is not correct, n = 5 is not the solution of the equation.

However, n = 9 is the solution, since

7n = 45

7(9) = 45

45 = 45.

Result

Ben is not correct and n = 5 is not the solution of the equation.

Page 180 Exercise 4 Answer

To find for which c the equation 3 + c = 11 is true, use substitution and check Lucy’s answers, that is substitute c with 7, 8, 9, and 10, or do the following:

3 + c = 11

c = 11 – 3 (Subtract 3 from both sides.)

c = 8

Lucy should add 8 cubes to make the pans balance.

Result

8

Page 180 Exercise 5 Answer

The equation is d + 9 = 35.

The given values of d are 16, 22, 26, and 36. We will substitute d with a value to check whether any of them are the solution.

16 + 9 = 25 ≠ 35

22 + 9 = 31 ≠ 35

26 + 9 = 35

36 + 9 = 45 ≠ 35

The solution of the equation d + 9 = 35 is d = 26.

Result

d = 26

Page 180 Exercise 6 Answer

The equation is 14n = 35.

The given values of n are 2, 3, 3.5, and 4. We will substitute n with a value to check whether any of them are the solution.

14 ⋅ 2 = 28 ≠ 35

14 ⋅ 3 = 42 ≠ 35

14 ⋅ 3.5 = 49 ≠ 35

14 ⋅ 4 = 56 ≠ 35

Result

None of the given values are the solution of the equation 14n = 35.

Page 180 Exercise 7 Answer

The equation is 13.4 − g = 8.1.

The given values of g are 4.3, 5.3, 5.5, and 6.5. We will substitute g with a value to check whether any of them are the solution.

13.4 – 4.3 = 9.1 ≠ 8.1

13.4 – 5.3 = 8.1

13.4 − 5.5 = 7.9 ≠ 8.1

13.4 − 6.5 = 6.9 ≠ 8.1

The solution of the equation 13.4 − g = 8.1 is g = 5.3.

Result

g = 5.3

Page 180 Exercise 8 Answer

The equation is 4 = 36 ÷ m.

The given values of m are 4, 6, 8, and 9. We will substitute m with a value to check whether any of them are the solution.

36 ÷ 4 = 9 ≠ 4

36 ÷ 6 = 6 ≠ 4

36 ÷ 8 = \(\frac{9}{4}\) = \(2 \frac{1}{4}\) ≠ 4

36 ÷ 9 = 4

The solution of the equation 4 = 36 ÷ m is m = 9.

Result

m = 9

Page 180 Exercise 9 Answer

The equation is n = 54 − 36.

We will substitute n with 8 to check whether the equation is true for n = 8.

54 − 36 = 18 ≠ 8

Result

The equation is not true for n = 8.

Page 180 Exercise 10 Answer

The equation is 5n = 40.

We will substitute n with 8 to check whether the equation is true for n = 8.

5 ⋅ 8 = 40

Result

The equation is true for n = 8.

Page 180 Exercise 11 Answer

The equation is 152 ÷ n = 21.

We will substitute n with 8 to check whether the equation is true for n = 8.

152 ÷ 8 = 19 ≠ 21

Result

The equation is not true for n = 8.

Page 180 Exercise 12 Answer

The equation is n + 46 = 54.

We will substitute n with 8 to check whether the equation is true for n = 8.

8 + 46 = 54

Result

The equation is true for n = 8.

Page 181 Exercise 13 Answer

The equation is t − 2.1 = 0.

The given values of t are 2.1, 2.4, 2.6, and 2.8. We will substitute t with a value to check whether any of them are the solution.

2.1 – 2.1 = 0

2.4 – 2.1 = 0.3 ≠ 0

2.6 – 2.1 = 0.5 ≠ 0

2.8 − 2.1 = 0.7 ≠ 0

The solution of the equation t − 2.1 = 0 is t = 2.1.

Result

t = 2.1

Page 181 Exercise 14 Answer

We are given the equation 49 = 7r and need to determine if r = 3, 6, 7, or 9 is a solution.

Substitute each value of r into the right side and then simplify to see if it equals 49:

r = 3 : 7r = 7(3) = 21 ≠ 49

r = 6 : 7r = 7(6) = 42 ≠ 49
​
r = 7 : 7r = 7(7) = 49 ✓
​
r = 9 : 7r = 7(9) = 63 ≠ 49

The solution of the equation is then r = 7.

Result

r = 7

Page 181 Exercise 15 Answer

The equation $4.10 = $6.25 – y is given. Four possible solution are offered, to tell which is true use substitution. Substitute the variable y with each value and evaluate the expression to check whether it is true or false.

For y = $2.15

$4.10 = $6.25 – y

= $6.25 – $2.15

= $4.10

For y = $2.95

$4.10 = $6.25 – y

= $6.25 – $2.95

≠ $3.30

For y = $3.05.

$4.10 = $6.25 – y

= $6.25 – $3.15

≠ $3.20

For y = $3.15

$4.10 = $6.25

= $6.25 – $3.15

≠ $3.10

Result

y = $2.15

Page 181 Exercise 16 Answer

The equation 24 ÷ h = 6 is given. Four possible solution are offered, to tell which is true use substitution. Substitute the variable h with each value and evaluate the expression to check whether it is true or false.

For h = 1.

24 ÷ h = 6

24 ÷ 1 = 6

24 ≠ 6

For h = 3.

24 ÷ h = 6

24 ÷ 3 = 6

8 ≠ 6

For h = 6.

24 ÷ h = 6

24 ÷ 6 = 6

4 ≠ 6

For h = 8.

24 ÷ h = 6

24 ÷ 8 = 6

3 ≠ 6

Result

None of the offered solutions are correct.

Page 181 Exercise 17 Answer

Miles per hour is a measurement for speed, as it states clearly, it tells how many miles one covers in an hour.

Marcie’s father rode his bike 108 miles in 7.5 hours, let’s calculate how fast he was going. Divide 108 by 7.5 since that is how much miles he covered in the time he spent cycling.

108 ÷ 7.5 = 14.4

Macie’s father rode his bike 14.4 mph.

y marks the number of hours she spent cycling. Use the equation 108 ÷ y = 18 to find how long it took her to cycle 108 miles if she rode her bike at the speed of 18 mph.

108 ÷ y = 18

108 = 18y (Multiply both sides b y.)

108 ÷ 18 = y (Divide both sides by 18.)
​
y = 6

It took Marcie 6 hours to cycle 108 miles. She didn’t match either of her parents time.

Result

y = 6

Page 181 Exercise 18a Answer

To check if b = 6 is the solution use substitution. Substitute the variable b for its value in the given equation and evaluate.

8b = 48

8(6) = 48

48 = 48

Result

b = 6 is the solution.

Page 181 Exercise 18b Answer

To check if b = 6 is the solution use substitution. Substitute the variable b for its value in the given equation and evaluate.

11 – b = 6

11 – 6 = 6

5 ≠ 6

Result

b = 6 is not a solution.

Page 181 Exercise 18c Answer

To check if b = 6 is the solution use substitution. Substitute the variable b for its value in the given equation and evaluate.

b + 3 = 9

6 + 3 = 9

9 = 9

Result

b = 6 is the solution.

Page 181 Exercise 18d Answer

To check if b = 6 is the solution use substitution. Substitute the variable b for its value in the given equation and evaluate.

54 ÷ b = 9

54 ÷ 6 = 9

9 = 9

Result

b = 6 is the solution.

Page 181 Exercise 19 Answer

To find the answer, that is which activity should the friends choose subtitute n with 4, since it is a group of 4 friends, in each of the equations. The equation which is true for n = 4 is the one tied to the activity they should choose.

6n + 5 = 29

6(4) + 5 = 29

24 + 5 = 29 (Substitute n with 4.)

29 = 29 (Find the product.)
​
14n = 29

14(4) = 29 (Substitute n with 4.)

56 ≠ 29 (Find the sum.)
​
30n – 40 = 29

30(4) – 40 = 29 (Substitute n with 4.)

120 – 40 = 29 (Find the product.)

80 ≠ 29 (Find the difference.)

The friends should choose the activity tied to the first equation, that is a Raft Trip.

Result

The friends should choose the Raft Trip.

Page 182 Exercise 20 Answer

To find which one of them is right use substitution – substitute p for each of the values they give as solutions.

27 = 18 + p

27 = 18 + 5 (Substitute p with 5.)

27 ≠ 23 (Evaluate and compare.)
​
27 = 18 + p

27 = 18 + 8 (Substitute p with 5.)

27 ≠ 26 (Evaluate and compare.)
​
27 = 18 + p

27 = 18 + 9 (Substitute p with 5.)

27 = 27 (Evaluate and compare.)
​
Rachel is correct – if 9 pennies are added the pans will balance.

Result

Rachel is correct.

Page 182 Exercise 21 Answer

Since Gerard spend $5.12 in total, that is for a sandwich and a drink, that is one side of the equation. Since s stands for the price of the sandwich and the drink cost $1.30, their sum is the other side of the equation.

s + $1.30 = $5.12

Substitute for the price of each sandwich to find for which one the equation is true. That is the sandwich Gerard bought.

s + $1.30 = $5.12

$3.54 + $1.30 = $5.12 (substitute s for 3.54)

$4.84 ≠ $5.12

s + $1.30 = $5.12

$3.82 + $1.30 = $5.12 (substitute s for 3.82)

s + $1.30 = $5.12

$3.92 + $1.30 = $5.12 (substitute s for 3.92)

$4.22 ≠ $5.12

Gerard bought a tuna sandwich for 3.82

Result

Gerard bought a tuna sandwich for 3.82.

Page 182 Exercise 22 Answer

x will be the variable in the equation. Since 12 is the solution the equation must be true for x = 12.

83 = 83 (choose any number and write it on both sides)

12 + 71 = 83 (Write 83 as a sum. Don’t change the other side)

12 + 1 + 70 = 83 (Write 71 as a sum)

12 + 1 + 7 x 10 = 83 (Write 70 as a product)

x + 1 + 7 x 10 = 83 (substitute 12 with x)

We used substitution but the other way round. Instead of subsituing a variable by its value, we decided that the value will be 12 and than wrote an expression which is true and put x instead of 12.

Result

x + 1 + 7 x 10 = 83

Page 182 Exercise 23 Answer

Substitute m for each of the values and evaluate to find for which value the equation is true.

255 – m = 124

255 – 111 = 124 (Substitute m with 111.)

144 ≠ 124 (Evaluate and compare.)

255 – m = 124

255 – 121 = 124 (Substitute m with 121.)

134 ≠ 124 (Evaluate and compare.)

255 – m = 124

255 – 131 = 124 (Substitute m with 131.)

124 = 124 (Evaluate and compare.)

255 – m = 124

255 – 141 = 124(Substitute m with 141.)

114 ≠ 124 (Evaluate and compare.)
​
Gina’s family has driven 131 miles.

Result

Gina’s family has driven 131 miles.

Page 182 Exercise 24 Answer

The perimeter is the sum of the lengths of all sides, that is

Perimeter = 3 + 5 + 3 + 3 + m.

Since the length of the perimeter is 19 cm, the expression is

19 = 3 + 5 + 3 + 3 + m.

Solve for m.

19 = 3 + 5 + 3 + 3 + m

19 = 14 + m (Find the sum.)

19 – 14 = m (Subtract 14 from both sides.)

5 = m (Find the difference.)
​
The length of the missing side is 5.

Result

m = 5

Page 182 Exercise 25 Answer

To find out if either of Alisa’s guesses is correct use the given equation and substitution. Substitute n for both values she gives as possible solutions.

7n = 147

7(11) = 147 (Substitute n with 11.)

77 ≠ 147

7n = 147

7(31) = 147 (Substitute n with 31.)
​
217 ≠ 147

Since neither 11 nor 31 are correct solutions for the given equation, both of Alisa’s guesses are wrong.

7n = 147

n = 147 ÷ 7 (Divide both sides by 7.)
​
n = 21

The park has 21 times as many trees as her yard.

Result

Neither of Alisa’s guesses is correct. The park has 21 times as many trees as her yard.

Page 182 Exercise 26 Answer

To find the answer use the given equation and substitution. Substitute c for the price of each item she wants to buy.

$18.62 + c = $26.00

$18.62 + $7.32 = $26.00 (Substitute c for $7.32.)

$25.94 < $26.00 (Evaluate and compare.)

$18.62 + c = $26.00

$18.62 + $7.38 = $26.00 (Substitute c for $7.38.)

$26.00 = $26.00 (Evaluate and compare.)

$18.62 + c = $26.00

$18.62 + $8.48 = $26.00 (Substitute c for $8.48.)

$27.10 > $26.00 (Evaluate and compare.)

Trish can’t buy oil paints because the total cost will then be greater than $26.00, she can buy knitting needles and she will spend less than $26.00. However, if she buys silk flowers she will spend exactly $26.00, which means the silk flowers are the most expensive item she can buy.

Result

Silk flowers.

Chapter 4 Represent And Solve Equations And Inequalities

Section 4.2: Apply Property Of Equalities

Page 183 Exercise 1 Answer

The computation to do – add 5 to both sides of the equation.

4 + 8 = 12

4 + 8 + 5 = 12 + 5

17 = 17

The equation is true.

The computation to do – add 3 to the left side of the equation and add 5 to the right side.

4 + 8 = 12

4 + 8 + 3 = 12 + 5

15 ≠ 17

The equation is false.

The computation to do – divide the left side of the equation by 2 and multiply the right side by 2.

4 + 8 = 12

(4 + 8) ÷ 2 = 12 x 2

6 ≠ 24

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

The equation is false.

The computation to do – subtract 4 from both sides of the equation.

4 + 8 = 12

4 + 8 – 4 = 12 – 4

8 = 8

The equation is true.

The computations which keep the equation true are the kind that does the same thing to both sides of the equation.

To determine if an equation is true evaluate both sides of the equation. If the result is the same on both sides, the equation is true.

Result

4 + 8 + 5 = 12 + 5 is true.

4 + 8 + 3 = 12 + 6 is false.

(4+8) ÷ 2 = 12 × 2 is false.

4 + 8 − 4 = 12 − 4 is true.

The computations which keep the equation true are the kind that does the same thing to both sides of the equation.

Page 183 Exercise 1 Answer

An equation is given but two numbers are missing. To make sure the equation is true, fill with number such that the result on both sides is the same.

For example, let’s choose numbers so that the result on both sides is 8. To do so, add 1 to the left side, since 7 + 1 = 8, and subtract 2 from the right side, since10 – 2 = 8.

7 + 1 = 10 – 2

8 = 8

Add 10 to both sides of the equation.

7 + 1 + 10 = 10 − 2 + 10

8 + 10 = 8 + 10

18 = 18
​
Divide both sides by 4.

(7+1) ÷ 4 = (10−2) ÷ 4

8 ÷ 4 = 8 ÷ 4

2 = 2
​
Notice, when dividing (or multiplying) first put the entire left side in brackets and than divide, the same goes for the right side. Do this because the order of operations is not the same for addition, subtraction, multiplication, and division.

Result

Possible answer: 7 + 1 = 10 − 2

Two other operations that could be done to the completed equation to keep it true are add 10 to both sides of the equation and divide both sides by 4.

Page 184 Exercise 1 Answer

5y = 25

5y – 7 = 25 – 7

Since 7 was subtracted from both sides of the equation the property used is the Subtraction Property of Equality.

Other properties of equality one could apply are the Addition Property of Equality, the Multiplication Property of Equality, and the Division Property of Equality.

For example, the Addition Property of Equality – add 5 to both sides.

5y = 25

5y + 5 = 25 + 5

For example, the Multiplication Property of Equality – multiply both sides by 2.

5y = 25

5y x 2 = 25 x 2

For example, the Division Property of Equality – divide both sides by 5.

5y = 25

5y ÷ 5 = 25 ÷ 5

Result

Subtraction Property of Equality

Page 185 Exercise 2a Answer

4 ⋅ x = 36

(4⋅x) ÷ 4 = 36 ÷ 4

x = 9

The solution to 4 ⋅ x = 36 is given as x = 9. Check this by substituting x by 9 in 4 ⋅ x = 36. The equation is balanced. Thus, both sides can be divided by the same number, that is the scale can be balanced with only one blue x-block.

Notice, if both sides are divided by 4, from the first line we get the third line.

Result

4 . x = 36

(4 . x) ÷4 = 36 ÷ 4

x = 9

Page 185 Exercise 2b Answer

If 25 + d = 36 is true, than 25 + d − 25 = 36 − 20 is false, since from the left side 25 is subtracted, and from the right side only 20 is subtracted. The equation stays balanced if and only if the same number is subtracted from both sides, that is the Subtraction Property of Equality.

Result

If 25 + d = 36 is true, than 25 + d − 25 = 36 − 20 is false.

Page 186 Exercise 1 Answer

If we have an equation, we can use any of the properites of equality, that is we can either add to, subtract from, multiply, or divide both sides of the equation by the same number, and the result will be a new equation that is equivalent to the original one.

For example, use th Subtraction Property of Equality to write a new equation if x + 5 = 34 is given.

x + 5 – 4 = 34 – 4

To check that these two equation are really equivalent, find the value of x which is the solution and compare them. if and only if they are the same, the equations are equivalent.

x + 5 = 34

x + 5 – 5 = 34 – 5 (Subtract 5 from both sides)

x = 29

x + 5 – 4 = 34 – 4

x + 1 = 20 (Evaluate)

x + 1 – 1 = 30 – 1 (Subtract 1 from both sides)

x = 29

Result

If we have an equation, we can use any of the properties of equality, that is we can either add to, subtract from, multiply, or divide both sides of the equation by the same number, and the result will be a new equation that is equivalent to the original one.

Page 186 Exercise 2 Answer

Since the equation 7 + 5 = 12 is balanced, if 4 units are removed from one side the same must be done to the other side.

7 + 5 = 12

7 + 5 – 4 = 12 − 4

Result

Subtract 4 from the other side as well.

Page 186 Exercise 3 Answer

Since the equation 32 + 43 = 66 is balanced, if one side is multiplied by 3 than the other side must be multiplied by 3 as well to keep the sides equal.

23 + 43 = 66

66 = 66 (The equation is balanced.)
​
66 x 3 = 66 x 3

(23 + 43) x 3 = 66 x 3

Result

The other side must be multiplied by 3 as well.

Page 186 Exercise 4 Answer

If x + 5 = 8 is balanced, than adding 9 to one side and (4+5) will keep the equation balanced, since 4 + 5 = 9, s 9 was actually added to both sides.

x + 5 = 8

x + 5 + 9 = 8 + (4+5)

x + 5 + 9 = 8 + 9
​
Result

The equation will stay equal.

Page 186 Exercise 5 Answer

If 23 + 37 = 60 is true, than 23 + 37 + 9 = 60 + 9 is true as well. 9 is added to both sides so the Addition Property of Equality is used. The equation stays balanced.

23 + 37 = 60

23 + 37 + 9 = 60 + 9

69 = 69
​
Result

If 23 + 37 = 60 is true, than 23 + 37 + 9 = 60 + 9 is true as well.

Page 186 Exercise 6 Answer

16 + 1 = 16

(16+1) − 1 = 17 − 2

16 ≠ 15

1 is subtracted from the left side and 2 is subtracted from the right side. Since the first equation is balanced, the second is not. The equation stays balanced if and only if the same number is subtracted from both sides, which is not the case here.

Result

If 16 + 1 = 17 is true, than (16+1) − 1 = 17 − 2 is false.

Page 186 Exercise 7 Answer

To apply the Multiplication Property of Equality multiply both sides of the equation by 5.

7n = 28

7n x 5 = 28 × 5

Result

7n × 5 = 28 × 5

Page 186 Exercise 8 Answer

Solve 6m = 9 for m.

6m = 9

m = \(\frac{9}{6}\) (Divide both sides by 6.)
​
m = \(\frac{3}{2}\)

Tomas changed the equation – he divided the left side by 2 and the right side by 3.

6m ÷ 2 = 9 ÷ 3

​3m = 3

m = \(\frac{3}{3}\) (Divide both sides by 3 to find m.)
​
m = 1 ≠ \(\frac{3}{2}\)

Tomas is not correct. If one side of the equation is divided by 2 and the other by 3, the equation won’t stay balanced. The result is not the same, thus the equations are not equivalent.

Result

Tomas is not correct.

Page 187 Exercise 9 Answer

5m + 4 = 19

5m + 4 – 3 = 19 − 3
​
The same number, 3, was subtracted from both sides. That is the equation is balanced.

The property of equality used is Subtraction Property of Equality.

Result

Subtraction Property of Equality.

Page 187 Exercise 10 Answer

3t = 20

3t ÷ 2 = 20 ÷ 2
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Both sides are divided by the same number, 2. That is the equation is balanced.

The property of equality used is Division Property of Equality.

Result

Division Property of Equality.

Page 187 Exercise 11 Answer

\(\frac{n}{6}\) = 9

(\(\frac{n}{6}\)) x 5 = 9 × 5
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Both sides are multiplied by the same number, 5. That is the equation is balanced.

The property of equality used is Multiplication Property of Equality.

Result

Multiplication Property of Equality.

Page 187 Exercise 12 Answer

5b – 6 = 14

5b – 6 + 2 = 14 + 2

The same number, 2, is added to both sides. That is the equation is balanced.

The property of equality used is Addition Property of Equality.

Result

Addition Property of Equality.

Page 187 Exercise 13 Answer

r + 9 = 42

r + 9 – 9 = 42 + 9

r + 9 – 9 ≠ 42 + 9

If equality r + 9 = 42 is true, than equality r + 9 − 9 = 42 + 9 is false since the equality stays true if and only if the same number is either added or subtracted from both sides. In this case, to one side 9 was added and from the other side 9 was subtracted. The equation is not balanced.

Result

If r + 9 = 42, than r + 9 − 9 ≠ 42 + 9.

Page 187 Exercise 14 Answer

6s = 24

6s ÷ 6 = 24 ÷ 6

Since both sides were divided by the same number, that is by 6, the equation stays balanced and is true.

Result

If 6s = 24, than 6s ÷ 6 = 24 ÷ 6.

Page 187 Exercise 15 Answer

On the left side of the scale is 12 and on the right 2 + 7 + 3 + 16, since the scale is not even, it is an inequality:

12 ≠ 2 + 7 + 3 + 16.

To make this two sides equal, evaluate the right side.

2 + 7 + 3 + 16 = 28
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The right side is equal to 28. To make the scale balance we need to make the left side equal to 28 as well.

Since 28 − 12 = 16, that is 28 can be written as 12 + 16, add 16 to the left side to make the scale balance.

12 + 16 = 2 + 7 + 3 + 16

Result

12 + 16 = 2 + 7 + 3 + 16

Page 187 Exercise 16 Answer

It is given that the scale balanced with 3 green blocks on one side and 1 blue x-block on the other side so we know that:

3 = x

The scale now has 15 green blocks on the left side. Since 3 ⋅ 5 = 15, then the left side of the equation was multiplied by 5. To make the completed equation true, the right side must also be multiplied by 5:

15 = 5 ⋅ x

Result

15 = 5 ⋅ x

Page 187 Exercise 17 Answer

Since 8x = 24 is true, and your friend divides the right side by 4, to make the equation true again, divide the left side by 4 as well.

8x = 24

8x ≠ 24 ÷ 4

8x ÷ 4 = 24 ÷ 4

Result

Divide the left side by 4.

Page 187 Exercise 18 Answer

The scale originally had 1 blue x-block and 20 green blocks on the left side and 40 green blocks on the right side. The balance was then represented by the equation: x + 20 = 40

If the scale now has 1 blue x-block and 3 green blocks on the left side, then 17 green blocks were removed from the left side since 20 − 17 = 3.

The scale and equation are only balanced if the same thing is done to both sides so you must then remove 17 green blocks from the right side to make the scale balanced:

x + 20 − 17 = 40 − 17

Result

17 green blocks

Page 188 Exercise 19 Answer

(y+6) ÷ 3 = 15
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The second equation is not equivalent to the first because the left side was divided by 3 and the right side was not. That is, the Division Property of Equality is not used.

To make these two equations equivalent Bobbie must divide the right side by 3 as well.

Result

Bobbie must divide the right side of the equation by 3 as well.

Page 188 Exercise 20 Answer

5 + 5 = 10

5 + 5 + n = 11 + n
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Since the same change was made to both sides of the equation, the equations are equivalent.

John used the Addition Property of Equality – he added the same number, that is n, to both sides of the equation.

Result

The equations are equivalent.

Page 188 Exercise 21 Answer

4 + 3 − 1 = 7 − 1

Since the scientist took one unit of mass from each side of a pan, and in the equation this is represented by subtracting one from both sides of the equation, the pan stays balanced. The equation stays balanced since the same change was made to both sides, that is he used the Subtraction Property of Equality.

Result

Subtraction Property of Equality.

Page 188 Exercise 22 Answer

Lexi used a property of equality to write an equation equivalent to n − 3 = 4. She could have used either of these four properites: Addition Property of Equality, Subtraction, Multiplication, or Division.

For example, if she used Multiplication Property of Equality, she multiplied both sides of the equation by the same number.

n − 3 = 4

(n-3) x 5 = 4 × 5
​
Notice, the left side must be in brackets.

The equations are equivalent because the same change was made to both sides of the first equation, and the result is the second equation. Also, one could check by finding n for both equation. The equation are equivalent if and only if the n is the same for both of them.

Result

(n−3) × 5 = 4 × 5

Page 188 Exercise 23 Answer

Since a quarter can be thought of, either as \(\frac{25}{100}\), that is as 25 cents, or more literally as \(\frac{1}{4}\), Emil’s money can be represented as a mixed number \(1 \frac{1}{4}\). Jade’s money, her 5 quarters, can similarly be represented as 5 x \(\frac{1}{4}\).

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

= \(\frac{5 \times 1}{1 \times 4}\)
​
= \(\frac{5}{4}\)

If Emil gives Jade $1 and she gives him four quarters, which is equal to one, they each still have the same amount of money.

\(1 \frac{1}{4}-1=\frac{1}{4}\)(Emil gives Jade $1.)

\(\frac{1}{4}+4 \times \frac{1}{4}=\frac{1}{4}+\frac{4}{4}\) (Jade gives Emil four quarters.)

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

\(5 \times \frac{1}{4}+1=\frac{5}{4}+\frac{4}{4}\) (Emil give Jade $1.)

= \(\frac{9}{4}\)

\(\frac{9}{4}-4 \times \frac{1}{4}=\frac{9}{4}-\frac{4}{4}\) (Jade gives Emil four quarters.)
​
= \(\frac{5}{4}\)

Result

After the exchange, they each still have the same amount of money.

Page 188 Exercise 24 Answer

To change 7w = 49 into 7w ÷ 7 = 49 ÷ 7 both side were divided by 7.

The property used to find these equivalent equations is Division Property of Equality.

Result

Division Property of Equality.

Page 188 Exercise 25 Answer

The equation 12b = 24 is given. The find the quantity that equals 4b divide both sides by three.

12b = 24

12b ÷ 3 = 24 ÷ 3

4b = 8
​
Result

Divide both sides by 3.

Page 188 Exercise 26 Answer

For the equation to be equivalent to n + 4 = 11 whatever is done to one side of the equation, must be done to the other side of the equation as well.

(n+4) × 2 = 11
​
Since only one side is multiplied by 2, this equation is not equivalent to n + 4 = 11.

(n+4) × 2 = 11 ÷ 2
​
Since one side is multiplied by 2 and the other is divided by 2, this equation is not equivalent to n + 4 = 11.

(n+4) × 2 = 11 × 4
​
Since one side is multiplied by 2 and the other by 4, this equation is not equivalent to n + 4 = 11.

(n+4) × 2 = 11 × 2
​
Since both sides are multiplied by 2, the equation marked D, is equivalent to n + 4 = 11.

Result

D

Page 188 Exercise 27 Answer

8p ÷ 8 = 12 ÷ 8

To get 8p ÷ 8 = 12 ÷ 8 from 8p = 12 divide both sides by 8. Since the same thing is done to both sides, the equation stays balanced. Thus, these two are equivalent.

8p ÷ 8 = 12 ÷ 12

Since one side is divided by 8 and the other by 12 the equation doesn’t stay balanced, that is the solution will not be the same, thus these two equations are not equivalent.

8p + 4 = 12 + 4
​
To get 8p + 4 = 12 + 4 from 8p = 12 add 4 to both sides. Since the same thing is done to both sides, the equation stays balanced. Thus, these two are equivalent.

8p − 2 = 12 − 2

To get 8p − 2 = 12 − 2 from 8p = 12 subtracted 2 from both sides. Since the same thing is done to both sides, the equation stays balanced. Thus, these two are equivalent.

Result

B