Chapter 3 Numeric And Algebraic Expressions
Section 3.6: Generate Equivalent Expressions
Page 155 Exercise 1a Answer
Evaluating 8n + 6 for n = 1 gives:
8n + 6 = 8(1) + 6 Substitute n = 1.
= 8 + 6 Multiply.
= 14 Add.
Evaluating 2(4n + 3) for n = 1 gives:
2(4n + 3) = 2(4.1 + 3) Substitute n = 1.
= 2(4 + 3) Multiply.
= 2 . 7 Add.
= 14 Multiply.
Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions
Evaluating 14n for n = 1 gives:
14n = 14(1) Substitute n = 1.
= 14 Multiply.
Result
Page 155 Exercise 1b Answer
Evaluating 8n + 6 for n = 2 gives:
= 8(2) + 6 Substitute n = 2.
= 16 + 6 Multiply.
= 22 Add.
Evaluating 2(4n+3) for n = 2 gives:
2(4n+3) = 2(4⋅2+3) Substitute n = 2.
= 2(8 + 3) Multiply.
= 2 . 11 Add.
= 22 Multiply.
Evaluating 14n for n = 2 gives:
= 14(2) Substitute n = 2.
= 28 Multiply.
Result
Page 155 Exercise 1c Answer
Expressions are equivalent if they give the same result when evaluated for different values of the variable.
In parts A and B, we made the following tables:
From the tables, all three expressions had the same result when n = 1 but they did not all have the same result when n = 2. This means the three expressions are not equivalent.
Notice that the first two expressions have the same result through for n = 1 and n = 2. Using the Distributive Property gives 2(4n + 3) = 2(4n) + 2(3) = 8n + 6 so 8n + 6 and 2(4n + 3) are equivalent.
Result
All three expressions are not equivalent. Only 8n+6 and 2(4n+3) are equivalent.
Page 155 Exercise 1 Answer
Two expressions are only equivalent if they have the same result for all values of the variable. Therefore, the expressions only need to have different results one time before you know they are not equivalent.
Page 156 Exercise 1 Answer
To write an expression which is equivalent to 3y − 9 use the Distributive Property in reverse order. To do so, look for a common factor of both terms that is greater than 1. Since y is an unknown, find a common factor of 3 and 9.
A common factor of 3 and 9 is 3, since 9 can be writen as 3 × 3.
3y − 9
= 3(y) − 3(3) (Rewrite both terms as products using a common factor.)
= 3(y−3) (Extract the common factor.)
3y − 9 is equivalent to 3(y−3).
Equivalent expressions have the same value even when the value of the variable in the expression changes. Using different properties of operations you can write equivalent expressions in more than one way. Either use the Distributive and Associative Properties or use the Distributive Propert in reverse order.
Result
3(y−3)
Page 157 Exercise 2 Answer
Use different properties of operations to write the expressions in more than one way and than compare them.
10y + 5
= 5(2y) + 5(1) (Look for a common factor of both terms.)
= 5(2y+1) (Use distributive property in reverse.)
Notice that the first expression 10y + 5 can be written differently as the third expression 5(2y+1), thus they are equivalent.
However, the second expression has only one term and can’t be written as either the first or as the third expression.
Result
The first, 10y + 5, and the second, 5(2y+1), expressions are equivalent.
Page 157 Exercise 3 Answer
2(x−3) + 1
= 2(x) − 2(3) + 1 (Use the Distributive Property.)
= 2x – 6 + 1 (Use the Associative Property of Multiplication.)
= 2x − 5 (Find the sum.)
Notice that 2(x−3) + 1 ≠ 2x + 6, so the expressions are not equivalent.
Substitute the variable x with any number to show that the expression are not equivalent. Suppose x = 3.
2(x−3) + 1
= 2x − 5 (The expression can be written differently.)
= 2(3) − 5 (Substitute x with 3.)
= 6 − 5 (Find the product.)
= 1 (Find the difference.)
2x + 6
= 2(3) + 6 (Substitute x with 3.)
= 6 + 6 (Find the product.)
= 12 (Find the sum.)
Result
The expressions are not equivalent. When x = 3, 2(x−3) + 1 = 1 and 2x + 6 = 12 so they don’t have the same value.
Page 158 Exercise 1 Answer
To identify and write equivalent expressions, we can use the properties of operations, which are the Commutative Properties of Addition and Multiplication, Associative Properties of Addition and Multiplication, and the Distributive Properties across Addition and Subtraction.
Page 158 Exercise 2 Answer
Two possible answers are shown below:
For the expression y + \(\frac{1}{2}\), we could use the Commutative Property of Addition to write the equivalent expression \(\frac{1}{2}\) + y.
For the expression y + \(\frac{1}{2}\), we could use the Distributive Property to write the an equivalent expression that is a product:
\(y+\frac{1}{2}=\frac{1}{2}(2 y)+\frac{1}{2}(1)=\frac{1}{2}(2 y+1)\)Result
Possible answers:
We can use the Commutative Property of Addition to write the equivalent expression \(\frac{1}{2}\) + y.
We can use the Distributive Property to write the equivalent expression \(\frac{1}{2}\)(2y+1).
Page 158 Exercise 3 Answer
The expressions z3 and 3z are not equivalent expressions. The expression 3z means 3 × z while z3 means z × z × z.
You can verify the expressions are not equivalent using substitution. When z = 1, we get:
z3 = 13 = 1
3z = 3(1) = 3
Since 1 ≠ 3, then z3 and 3z are not equivalent.
Result
No, when z = 1 we get z3 = 1 and 3z = 3 so the expressions are not equivalent.
Page 158 Exercise 4 Answer
When y = 1, we get:
3(y+1) = 3(1+1) = 3(2) = 6
3y + 3 = 3(1) + 3 = 3 + 3 = 6
Both expressions equal 6 when y = 1 so the expressions are equivalent when y = 1.
When y = 2, we get:
3(y+1) = 3(2+1) = 3(3) = 9
3y + 3 = 3(2) + 3 = 6 + 3 = 9
Both expressions equal 9 when y = 2 so the expressions are equivalent when y = 2.
When y = 3, we get:
3(y+1) = 3(3+1) = 3(4) = 12
3y + 3 = 3(3) + 3 = 9 + 3 = 12
Both expressions equal 12 when y = 3 so the expressions are equivalent when y = 3.
Result
The expressions are equivalent for y = 1, y = 2, and y = 3.
Page 158 Exercise 5 Answer
Using the Distributive Property a(b+c) = a(b) + a(c) gives:
3(y+1) = 3(y) + 3(1) = 3y + 3
Therefore, 3(y+1) and 3y + 3 are equivalent for any value of y.
Result
By the Distributive Property 3(y+1) = 3y + 3 so they are equivalent for any value of y.
Page 158 Exercise 6 Answer
2(r+3)
= 2(r) + 2(3) (Use the Distributive Property.)
= 2r + 6 (Multiply.)
Result
2r + 6
Page 158 Exercise 7 Answer
6(4s−1)
= 6(4s) − 6(1) (Use the Distributive Property.)
= (6×4)s − 6(1) (Use the Associative Property of Multiplication.)
= 24s − 6 (Multiply.)
Result
24s − 6
Page 158 Exercise 8 Answer
8t + 2
= 2(4t) + 2(1) (Find a common factor of both terms.)
= 2(4t+1) (Use the Distributive Property in reverse.)
Result
2(4t+1)
Page 158 Exercise 9 Answer
To complete the table, we need to evaluate the expressions 12x − 6, 3x + 3, and 6(2x−1) for x = 1, 2, and 3.
Evaluating the expressions for x = 1 gives:
12x − 6 = 12(1) − 6 = 12 − 6 = 6
3x + 3 = 3(1) + 3 = 3 + 3 = 6
6(2x−1) = 6(2⋅1−1) = 6(2−1) = 6(1) = 6
Evaluating the expressions for x = 2 gives:
12x − 6 = 12(2) − 6 = 24 − 6 = 18
3x + 3 = 3(2) + 3 = 6 + 3 = 9
6(2x−1) = 6(2⋅2−1) = 6(4−1) = 6(3) = 18
Evaluating the expressions for x = 3 gives:
12x − 6 = 12(3) − 6 = 36 − 6 = 30
3x + 3 = 3(3) + 3 = 9 + 3 = 12
6(2x−1) = 6(2⋅3−1) = 6(6−1) = 6(5) = 30
The completed table is then:
Result
Page 158 Exercise 10 Answer
In Exercise 9, the first and the third expression are equivalent.
12x – 6
= 6(2x) – 6(1) (Find a common factor of both terms)
= 6(2x – 1) (Use the Distributive Property in reverse)
12x – 6 = 6(2x – 1)
Result
In Exercise 9, the first and the third expression are equivalent.
Page 159 Exercise 11 Answer
3(m+3)
= 3m + 3(3) (Use the Distributive Property.)
= 3m + 9 (Multiply.)
Result
3m + 9
Page 159 Exercise 12 Answer
20n − 4m
= 4(5n) − 4(m) (Find a common factor of both terms.)
= 4(5n – m) (Use the Distributive Property in reverse.)
Result
4(5n−m)
Page 159 Exercise 13 Answer
3(x−6)
= 3(x) − 3(6) (Use the Distributive Property.)
= 3x − 18 (Multiply.)
Result
3x − 18
Page 159 Exercise 14 Answer
2x + 10
= 2(x) + 2(5) (Find a common factor of both terms.)
= 2(x+5) (Use the Distributive Property in reverse.)
Result
2(x+5)
Page 159 Exercise 15 Answer
8(2y + \(\frac{1}{4}\)
= 8(2y) + 8(\(\frac{1}{4}\) (Use the Distributive Property.)
= (8 x 2)y + (8 x \(\frac{1}{4}\))(Use the Associative Property of Multiplication.)
= 16y + 2 (Multiply.)
Result
16y + 2
Page 159 Exercise 16 Answer
5.7 + (3z+0.3)
= 5.7 + (0.3+3z) (Use the Commutative Property of Addition.)
= (5.7+0.3) + 3z (Use the Associative Property of Addition.)
= 6 + 3z (Find the sum inside the parentheses.)
= 3(2) + 3(z) (Find a commmon factor of both terms.)
= 3(2+z) (Use the Distributive Property in reverse.)
Result
3(2+z)
Page 159 Exercise 17 Answer
5w − 15
= 5(w) – 5(3) (Find a commmon factor of both terms.)
= 5(w-3) (Use the Distributive Property in reverse.)
Result
5(w−3)
Page 159 Exercise 18 Answer
2x + 4y
= 2(x) + 2(2y) (Find a commmon factor of both terms.)
= 2(x + 2y) (Use the Distributive Property in reverse.)
Result
2(x + 2y)
Page 159 Exercise 19 Answer
10(y2 + 2.45)
= 10(y2) + 10(24.5) (Use the Distributive Property.)
= 10y2 + 24.5 (Multiply.)
Result
10y2 + 24.5
Page 159 Exercise 20 Answer
\(\frac{3}{4} \times\left(z^3 \times 4\right)\)
= \(\frac{3}{4}\) x 4 x z3) (Use the Commutative Property of Multiplication.)
= (\(\frac{3}{4}\) x 4) x z3 (Use the Associative Property of Multiplication.)
= 3 x z3 (Find the product inside the parentheses.)
= 3z3 (Multiply.)
Result
3z3
Page 159 Exercise 21a Answer
To check if the given expression, 5(2x+3), is equivalent to 10x + 15, rewrite the given expression.
5(2x+3) = 5(2x) + 5(3)
= 10x + 15
They are equivalent.
Result
They are equivalent.
Page 159 Exercise 21b Answer
To check if the given expression, 5(2x+3), is equivalent to 5x + 15 + 5x, rewrite both expressions.
5(2x+3) = 5(2x) + 5(3)
= 10x + 15
5x + 15 + 5x = 10x + 15
Result
They are equivalent.
Page 159 Exercise 21c Answer
To check if the given expression, 5(2x+3), is equivalent to 10x + 8, rewrite the given expression.
5(2x+3) = 5(2x) + 5(3)
= 10x + 15 ≠ 10x + 8
Result
They are not equivalent.
Page 159 Exercise 22a Answer
To check if the given expression, 4x − 8, is equivalent to 2(2x−6), rewrite one of the expressions.
2(2x−6) = 2(2x) − 2(6)
= 4x − 12 ≠ 4x − 8
Result
They are not equivalent.
Page 159 Exercise 22b Answer
To check if the given expression, 4x − 8, is equivalent to 2(2x−4), rewrite one of the expressions.
2(2x−4) = 2(2x) − 2(4)
= 4x − 8
Result
They are equivalent.
Page 159 Exercise 22c Answer
To check if the given expression, 4x − 8, is equivalent to x − 8 + 3x, rewrite one of the expressions.
x − 8 + 3x = 4x − 8
Result
They are equivalent.
Page 159 Exercise 23a Answer
To check if the given expression, 12x − 16, is equivalent to 9.6x − 16 + 2.4x, rewrite one of the expressions.
When 9.6x and 2.4x are added they give 12x, thus the expressions are equivalent.
9.6x − 16 + 2.4x = 12x − 16
Result
They are equivalent.
Page 159 Exercise 23b Answer
To check if the given expression, 12x − 16, is equivalent to 3(3x−5), rewrite one of the expressions.
3(3x−5) = 3(3x) − 3(5)
= 9x − 15 ≠ 12x − 16
Result
They are not equivalent.
Page 159 Exercise 23c Answer
To check if the given expression, 12x − 16, is equivalent to 4(3x−4), rewrite one of the expressions.
4(3x−4) = 4(3x) − 4(4)
= 12x − 16
Result
They are equivalent.
Page 159 Exercise 24a Answer
To check if the given expression, 2(6x + \(\frac{1}{2}\)), is equivalent to 12x + 2, rewrite one of the expressions.
= \(2\left(6 x+\frac{1}{2}\right)=2(6 x)+2\left(\frac{1}{2}\right)\)
= 12x + 1 ≠ 12x + 2
Result
They are not equivalent.
Page 159 Exercise 24b Answer
To check if the given expression, 2(6x + \(\frac{1}{2}\)), is equivalent to 12x + 1, rewrite one of the expressions.
= \(2\left(6 x+\frac{1}{2}\right)=2(6 x)+2\left(\frac{1}{2}\right)\)
= 12x + 1
Result
They are equivalent.
Page 159 Exercise 24c Answer
To check if the given expression, 2(6x + \(\frac{1}{2}\)), is equivalent to 6x + \(\frac{1}{2}\) + 6x + \(\frac{1}{2}\), rewrite both expressions.
= \(2\left(6 x+\frac{1}{2}\right)=2(6 x)+2\left(\frac{1}{2}\right)\)
= 12x + 1
= \(6 x+\frac{1}{2}+6 x+\frac{1}{2}=12 x+\frac{1}{2}+\frac{1}{2}\)
= 12x + 1
Result
They are equivalent.
Page 159 Exercise 25a Answer
Since Mr. Tonkery bought x soccer balls which cost $15 each, the product x × $15 represents what he paid for them.
Also, he bought 3 baseballs which cost $6 each, so the product3 × $6 represents how much he paid for them.
The sum of this two products represents the total amount Mr. Tonkery paid.
An algebraic expression which represents the situation is:
x × $15 + 3 × $6.
Result
x × $15 + 3 × $6
Page 159 Exercise 25b Answer
First, make an expression which describes what each of them bought. Since they all bought the same and there are three of them, multiply the first expression by three.
Each of them bought a baseball, which costs $6, and x pairs of sweat socks, which cost $5 per pair, that is the productx × $5 describes the second part.
Each of them paid:
$6 + x × $5.
All three of them paid:
3($6 + x × $5).
Result
3($6 + x × $5)
Page 159 Exercise 26 Answer
To check if the expression x × $15 + 3 × $6 and 3($6 + x × $5), if the value of x is the same, rewrite the expressions.
x × $15 + 3 × $6 = 15x + 18
3($6 + x × $5) = 18 + 3(x)(5)
= 15x + 18
Result
They are equivalent.
Page 159 Exercise 27 Answer
If soccer balls cost \(2 \frac{1}{2}\) times as much as baseballs, than the following equality must be true:
\(2 \frac{1}{2}\) x $6 = $15
\(\frac{5}{2} \times \frac{6}{1}=15\) = 15
\(\frac{5 \times 6}{2 \times 1}\) = 15
\(\frac{30}{2}\) = 15
15 = 15
The equality is balanced, that is it is true.
Wendy is right.
Result
Yes, \(2 \frac{1}{2}\) × $6 = $15.
Page 160 Exercise 28 Answer
Since the length of the rectangular rug is given as 2(x – 1), its width is 5, and the area is calculated as the product of the width and length, the area is:
A = 5 x 2(x – 1)
= 10(2x – 1)
= 10(2x) – 10(1)
= 10x – 10
Result
10x – 10
Page 160 Exercise 29 Answer
Jamie may not be right, the expressions may be equivalent despite one having a term which is added and the other having a term which is subtracted.
To check, rewrite the expressions using properties of operations.
6x − 2x + 4 = 4x + 4
4(x+1) = 4(x) + 4(1)
= 4x + 4
When the expressions are rewritten we can see that they are equivalent.
Result
No, the expression are equivalent.
Page 160 Exercise 30 Answer
Use properties of operations to check if the expressions are equivalent.
4(n+3) − (3+n)
= 4(n) + 4(3) − (3) − (n) (Use the Distributive Property.)
= 4n + 12 – 3 – n (Multiply.)
= 4n – n + 12 – 3 (Use the Commutative Property.)
= (4n – n) + (12 – 3) (Use the Associative Property.)
= 3n + 9 (Subtract.)
Result
The expressions are equivalent.
Page 160 Exercise 31 Answer
Use properties of operations to rewrite 4n − 2 as 2(2n−1).
4n − 2 = 2(2n) − 2(1)
= 2(2n−1)
Chris is right.
Result
Chris is right.
Page 160 Exercise 32 Answer
Using the properties of operations we get:
(f .g2) + 5 – (g2 . f)
= fg2 + 5 – g2f Multiply.
= fg2 – g2f + 5 Use the Commutative Property of Addition.
= fg2 – fg2 + 5 Use the Commutative Property of Multiplication.
= 5 Subtract.
Result
5
Page 160 Exercise 33 Answer
5(g+14) = 5(g) + 5(14)
= 5g + 70 ≠ 5g + 14
When the expression 5(g+14) is rewritten we can see that it is not equivalent to the expression 5g + 14 because of the Distributive Property across Addition.
Result
No, the expressions are not equivalent.
Page 160 Exercise 34 Answer
(8.5+2s) + 0.5 = 8.5 + (2s+0.5) (Use Associative Prop. of Addition.)
(8.5+0.5) + 2s = 8.5 + (0.5+2s) (Use Associative Prop. of Addition.)
= 8.5 + (2s+0.5) (Use Comm. Prop. of Addition.)
Using the properties of Addition the given expression can be rewritten as:
8.5 + 2s + 0.5 = 8.5 + 0.5 + 2s
= 9 + 2s.
However, even now it is not equivalent to 9 + 2.
2(4.5+s) = 2(4.5) + 2(s)
= 9 + 2s
8.5(2s+0.5) = 8.5(2s) + 8.5(0.5)
= 17s + 4.25 ≠ 9 + 2s
The expressions equivalent to the given expression are: (8.5+2s) + 0.5, (8.5+0.5) + 2s, and 2(4.5+s).
Result
(8.5+2s) + 0.5, (8.5+0.5) + 2s, and 2(4.5+s).
Page 160 Exercise 35 Answer
Using the Distributive Property, the given expression can also be written as
5(n+4) = 5(n) + 5(4) = 5n + 20.
5n + 4 ≠ 5n + 20
5n + 20 = 5(n+4)
15n + 5n + = 5n + 15 + 5 (Use the Commutative Property of Addition.)
= 5n + 20
5(n + 3) + 5 = 5n + 15 + 5(Use the Distributive Property.)
= 5n + 20
5n + 54 ≠ 5n + 20
Result
5n + 20, 15 + 5n + 5, and 5(n+3) + 5.