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
\(\frac{y^2}{2 x}+3 n-z\)= \(\frac{8^2}{2(4)}+3(7)-1\) 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:
\(\left[\left(2^3 \times 2.5\right) \div \frac{1}{2}\right]+120\)= \(\left[(8 \times 2.5) \div \frac{1}{2}\right]+120\) 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:
\(\left[\left(2 \times 10^0\right) \div \frac{1}{3}\right]+8\)= \(\left[(2 \times 1) \div \frac{1}{3}\right]+8\) 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:
\(9\left(y+\frac{1}{3}\right)=9 y+9\left(\frac{1}{3}\right)=9 y+3\)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