Chapter 3 Numeric And Algebraic Expressions
Section 3.7: Simplify Algebraic Expressions
Page 161 Exercise 1 Answer
Using the properties of operations we get:
x + 5 + 2x + 2
= x + 2x + 5 + 2 Use the Commutative Property of Addition.
= (x+2x) + (5+2) Use the Associative Property of Addition.
= (1+2)x + (5+2) Use the Distributive Property.
= 3x + 7 Add
Result
3x + 7
Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions
Page 161 Exercise 1 Answer
Two expressions are equivalent if they have the same value each time a number is substituted for the variable.
Evaluating x + 5 + 2x + 2 and 3x + 7 for x = 1 gives:
x + 5 + 2x + 2 = 1 + 5 + 2(1) + 2
= 1 + 5 + 2 + 2
= 10
3x + 7 = 3(1) + 7
= 3 + 7
= 10
Evaluating x + 5 + 2x + 2 and 3x + 7 for x = 2 gives:
x + 5 + 2x + 2 = 2 + 5 + 2(2) + 2
= 2 + 5 + 4 + 2
= 13
3x + 7 = 3(2) + 7
= 6 + 7
= 13
Evaluating x + 5 + 2x + 2 and 3x + 7 for x = 3 gives:
x + 5 + 2x + 2 = 3 + 5 + 2(3) + 2
= 3 + 5 + 6 + 2
= 16
3x + 7 = 3(3) + 7
= 9 + 7
= 16
The expressions x + 5 + 2x + 2 and 3x + 7 have the same value for all values of x so they are equivalent.
Result
The expressions x + 5 + 2x + 2 and 3x + 7 have the same value for all values of x so they are equivalent.
Page 162 Exercise 1 Answer
Using the properties of operations we get:
4z − z + z − 2z
= 4z – 1z + 1z – 2z Use the Identity Property of Multiplication.
= (4 – 1 + 1 – 2)z Use the Distributive Property.
= 2z Add and subtract.
The simplified expression is then 2z.
Like terms are terms that have the same variable part. The terms in 2x + 4y have different variables so the terms are not like terms. This means the terms cannot be combined to get 6xy. We then know that 2x + 4y is not equivalent to 6xy.
We can verify this by substituting values for x and y. If x = 1 and y = 2, then:
2x + 4y = 2(1) + 4(2)
= 2 + 8
= 10
6xy = 6(1)(2)
= 6(2)
= 12
The two expressions do not have the same value so they are not equivalent.
Result
4z − z + z − 2z
= 4z − 1z + 1z − 2z
= (4−1+1−2)z
= 2z
The simplified expression is 2z.
We know 2x + 4y is not equivalent to 6xy since the terms 2x and 4y are not like terms so they can’t be combined. Verifying this using substitution for x = 1 and y = 2 gives 2x + 4y = 10 and 6xy = 12, which are not the same value so the expressions are not equivalent.
Page 163 Exercise 2 Answer
Using the properties of operations we get:
\(\frac{1}{2} n+n+\frac{1}{2}\)
= \(\frac{1}{2} n+1 n+\frac{1}{2}\) Use the Identity Property of Multiplication.
= \(\left(\frac{1}{2}+1\right) n+\frac{1}{2}\) Use the Distributive Property.
= \(\frac{3}{2} n+\frac{1}{2}\) Add.
= \(1 \frac{1}{2} n+\frac{1}{2}\) Rewrite as a mixed number.
Result
Page 163 Exercise 3 Answer
In Example 3, the expression 2.5(20.50x + 5.50x – 10) represented her total earnings where 2.5 was how many times the amount she charged and her expenses increased by.
If the amount she charges and her expenses increase by 3.5 times, then we need to replace the 2.5 in the expression with 3.5. The first expression is then 3.5(20.50x + 5.50x − 10).
Using the properties of operations, an equivalent expression is:
3.5(20.50x + 5.50x − 10)
= 3.5(26x−10) Combine like terms inside the parentheses.
= 3.5(26x) − 3.5(10) Use the Distributive Property.
= 91x − 35 Multiply.
Result
3.5(20.50x + 5.50x − 10) and 91x − 35
Page 164 Exercise 1 Answer
To simplify algebraic expressions, we can use properties of operations to combine like terms to get an equivalent expression with no like terms and no parentheses.
Page 164 Exercise 2 Answer
To combine terms in an algebraic expression, the terms must be like terms, which are terms that have the same variable part.
Page 164 Exercise 3 Answer
Using the Identity Property of Multiplication, 2y − y = 2y − 1y. Using the Distributive Property gives 2y−1y = (2−1)y. Subtracting then gives (2−1)y = y. Therefore, 2y − y can be written as y.
Page 164 Exercise 4 Answer
Using the Distributive Property gives \(\frac{1}{2}\)x + \(\frac{1}{2}\)x = \(\left(\frac{1}{2}+\frac{1}{2}\right) x\). Adding gives \(\left(\frac{1}{2}+\frac{1}{2}\right) x\) = 1x. Using the Identity Property of Multiplication then gives 1x = x so \(\frac{1}{2}\)x + \(\frac{1}{2}\)x and x are equivalent.
Page 164 Exercise 5 Answer
4z2 − z2 and 4 are not equivalent expressions since 4z2 − z2 does not equal 4 for all values of z.
For example, when z = 2 we get:
4z2 − z2 = 4(2)2 − 22
= 4(4) − 4
= 16 − 4
= 12 ≠ 4
Result
4z2 − z2 and 4 are not equivalent expressions since 4z2 − z2 does not equal 4 for all values of z.
Page 164 Exercise 6 Answer
Using the properties of operations we get:
x + x + x + x
= 1x + 1x + 1x + 1x Use the Identity Property of Multiplication.
= (1+1+1+1)x Use the Distributive Property.
= 4x Add.
Result
4x
Page 164 Exercise 7 Answer
Using the properties of operations we get:
4y − y
= 4y − 1y Use the Identity Property of Multiplication.
= (4−1)y Use the Distributive Property.
= 3y Subtract.
Result
3y
Page 164 Exercise 8 Answer
7y − 4.5 − 6y = 7y − 6y − 4.5
= y − 4.5
Result
y − 4.5
Page 164 Exercise 9 Answer
Using the properties of operations we get:
= \(4 x+2-\frac{1}{2} x\) Use the Commutative Property.
= \(\left(4-\frac{1}{2}\right) x+2\) Use the Distributive Property.
= \(\left(3 \frac{2}{2}-\frac{1}{2}\right)+2\) Regroup 4 as a mixed number.
= \(3 \frac{1}{2} x+2\) Subtract.
Result
\(3 \frac{1}{2} x+2\)Page 164 Exercise 10 Answer
Using the properties of operations we get:
3 + 3y − 1 + y
= 3 + 3y − 1 + 1y Use the Identity Property of Multiplication.
= 3y + y + 3 − 1 Use the Commutative Property of Addition.
= (3y+y) + (3−1) Use the Associative Property of Addition.
= (3+1)y + (3−1) Use the Distributive Property.
= 4y + 2 Add and subtract.
Result
4y + 2
Page 164 Exercise 11 Answer
Using the properties of operations we get:
x + 6x
= 1x + 6x Use the Identity Property of Multiplication.
= (1+6)x Use the Distributive Property.
= 7x Add.
Result
7x
Page 164 Exercise 12 Answer
Using the properties of operations we get:
0.5w + 1.7w − 0.5
= (0.5+1.7)w − 0.5 Use the Distributive Property.
= 2.2w − 0.5 Add.
Result
2.2w − 0.5
Page 164 Exercise 13 Answer
Using the properties of operations we get:
\(12 \frac{1}{3} b+6 \frac{2}{3}-10 \frac{2}{3} b\)
= \(12 \frac{1}{3} b-10 \frac{2}{3} b+6 \frac{2}{3}\) Use the Commutative Property of Addition.
= \(\left(12 \frac{1}{3}-10 \frac{2}{3}\right) b+6 \frac{2}{3}\) Use the Distributive Property.
= \(\left(11 \frac{4}{3}-10 \frac{2}{3}\right) b+6 \frac{2}{3}\) Regroup.
= \(1 \frac{2}{3} b+6 \frac{2}{3}\) Subtract.
Result
\(1 \frac{2}{3} b+6 \frac{2}{3}\)
Page 164 Exercise 14 Answer
Using the properties of operations we get:
\(\frac{3}{4} x+2+3 x-\frac{1}{2}\)
= \(\frac{3}{4} x+3 x+2-\frac{1}{2}\) Use the Commutative Property of Addition.
= \(\left(\frac{3}{4} x+3 x\right)+\left(2-\frac{1}{2}\right)\) Use the Associative Property of Addition.
= \(\left(\frac{3}{4}+3\right) x+\left(2-\frac{1}{2}\right)\) Use the Distributive Property.
= \(\left(\frac{3}{4}+3\right) x+\left(1 \frac{2}{2}-\frac{1}{2}\right)\) Rewrite 2 as a mixed number.
\(3 \frac{3}{4} x+1 \frac{1}{2}\) Add and subtract.
Result
\(3 \frac{3}{4} x+1 \frac{1}{2}\)
Page 164 Exercise 15 Answer
Using the properties of operations we get:
3.2x + 6.5 − 2.4x − 4.4
= 3.2x − 2.4x + 6.5 − 4.4 Use the Commutative Property of Addition.
= (3.2x−2.4x) + (6.5−4.4) Use the Associative Property of Addition.
= (3.2−2.4)x + (6.5−4.4) Use the Distributive Property.
= 0.8x + 2.1 Subtract.
Result
0.8x + 2.1
Page 165 Exercise 16 Answer
2.1x2 + 3 − 0.5x2 − 1 = (2.1x2 − 0.5x2) + (3−1)
= 1.6x2 + 2
Result
1.6x2 + 2
Page 165 Exercise 17 Answer
\(\frac{2}{3} n+6+3 n-\frac{2}{3}=\frac{2}{3} n+3 n+6-\frac{2}{3}\)= \(3 \frac{2}{3} n+\frac{18}{3}-\frac{2}{3}\)
= \(3 \frac{2}{3} n+\frac{16}{3}\)
= \(3 \frac{2}{3} n+5 \frac{1}{3}\)
Result
\(3 \frac{2}{3} n+5 \frac{1}{3}\)Page 165 Exercise 18 Answer
5 + 3w + 3 − w = 3w − w + 5 + 3
= 2w + 8
Result
2w + 8
Page 165 Exercise 19 Answer
5w – 5w = 0
Result
0
Page 165 Exercise 20 Answer
2x + 5 + 3x + 6 = 2x + 3x + 5 + 6
= 5x + 11
Result
5x + 11
Page 165 Exercise 21 Answer
\(\frac{3}{4} z^3+4-\frac{1}{4} z^3=\frac{3}{4} z^3-\frac{1}{4} z^3+4\)= \(\frac{2}{4} z^3+4\)
= \(\frac{1}{2} z^3+4\)
Result
\(\frac{1}{2} z^3+4\)Page 165 Exercise 22 Answer
3.4m + 2.4m = 5.8m
Result
5.8m
Page 165 Exercise 23 Answer
4.2n + 5 – 3.2n = 4.2n – 3.2n + 5
= n + 5
Result
n + 5
Page 165 Exercise 24 Answer
q5 + q5 + q5 = 3q5
Result
3q5
Page 165 Exercise 25 Answer
\(3 x+\frac{1}{4}+2 y+\frac{1}{4}+7 x-y=3 x+7 x+2 y-y+\frac{1}{4}+\frac{1}{4}\)= 10x + y + \(\frac{2}{4}\)
= 10x + y + \(\frac{1}{2}\)
Result
10x + y + \(\frac{1}{2}\)
Page 165 Exercise 26 Answer
1.5z2 + 4.5 + 6z − 0.3 − 3z + z2 = 1.5z2 + z2 + 6z − 3z + 4.5 − 0.3
= 2.5z2 + 3z + 4.2
Result
2.5z2 + 3z + 4.2
Page 165 Exercise 27a Answer
The rooms have hourly rental fees, in order, of $25, $20, and $50.
Since Yolanda rents all the rooms for t hours, multiply each hourly rental fee by t to find the cost for each room. The total cost of the rooms is then $25t + $20t + $50t.
For the sound system in each room Yolanda needs to pay, in order, $15 and $10. There is no charge for the sound system in the third room. The total cost of the sound systems is then $15 + $10.
The total cost is the sum of the total cost of the rooms and the total cost of the sound systems.
The expression which describes the total amount Yolanda needs to pay is:
$25t + $20t + $50t + $15 + $10.
Result
$25t + $20t + $50t + $15 + $10
Page 165 Exercise 27b Answer
The expression is $25t + $20t + $50t + $15 + $10. Using the Distributive Property across Addition.
$25t + $20t + $50t + $15 + $10 = ($25+$20+$50)t + $15 + $10
= $95t + $25
Result
$95t + $25
Page 166 Exercise 28 Answer
The perimeter of a rectangle is the sum of its four side lengths. From the figure, the rectangle has side lengths of y, 2y + 1, y, and 2y + 1. The perimeter is then:
y + 2y + 1 + y + 2y + 1
= (y+2y+y+2y) + (1+1) Group like terms.
= 6y + 2 Combine like terms.
Result
6y + 2
Page 166 Exercise 29 Answer
The expression from Exercise 28 is 6y + 2.
6 and 2 have a greatest common factor of 2 so we can use the Distributive Property to rewrite 6y + 2 as a product of two factors:
6y + 2
= 2(3y) + 2(1) Rewrite each term as a product.
= 2(3y + 1)Use the Distributive Property.
Result
2(3y + 1)
Page 166 Exercise 30 Answer
The expressions from Exercise 28 and 29 are 6y + 2 and 2(3y + 1).
In Exercise 29, we used the Distributive Property to rewrite 6y + 2 as 2(3y + 1). Since we used one of the properties of operations to rewrite the expression, then the two expressions must be equivalent.
Result
6y + 2 and 2(3y + 1) are equivalent by the Distributive Property.
Page 166 Exercise 31 Answer
Rodney rewrote \(\frac{1}{2}\)(2x+7) as 3 + \(3 \frac{1}{2}\) so he rewrote a product of two factors as a sum.
To rewrite an expression that is a product of two factors as a sum, you need to use the Distributive Property. We can verify this by using the Distributive Property on \(\frac{1}{2}\)(2x+7), which gives:
\(\frac{1}{2}(2 x+7)=\frac{1}{2}(2 x)+\frac{1}{2}(7)=1 x+\frac{7}{2}=x+3 \frac{1}{2}\)
Result
Distributive Property
Page 166 Exercise 32 Answer
I don’t agree with Annie. She rewrote the expression using the Distributive Property so the two expressions must be equivalent but she forgot to combine like terms:
6.5(x+0.5x+1) = 6.5(x) + 6.5(0.5x) + 6.5(1) = 6.5x + 3.25x + 6.5
Simplified expressions must have no like terms so to simplify the expression we need to combine the like terms of 6.5x and 3.25x to get:
6.5x + 3.25x + 6.5 = 9.75x + 6.5
Result
I don’t agree. She used the Distributive Property to write an equivalent expression but she didn’t combine the like terms. The simplified expression should be 9.5x + 6.5.
Page 166 Exercise 33 Answer
Thea is correct. Combining like terms gives 4x − 3x + 2 = x + 2 since 4x − 3x = (4−3)x = 1x = x.
Page 166 Exercise 34 Answer
Using the properties of operations gives:
= \(\frac{a}{3}+\frac{a}{3}+\frac{a}{3}\)
= \(a\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)\) Use the Distributive Property.
= a(\(\frac{3}{3}\)) Add.
= a(1) Simplify.
= a Multiply.
Result
\(\frac{a}{3}+\frac{a}{3}+\frac{a}{3}=a\)Page 166 Exercise 35 Answer
Simplifying the given expression gives:
2x + 7 + 6x − x
= (2x+6x−x) + 7 Group like terms.
= 7x + 7 Combine like terms.
= 7 + 7x Use the Commutative Property of Addition.
The two expressions that are equivalent to 2x + 7 + 6x − x are then 7x + 7 and 7 + 7x. The two expressions that are NOT equivalent are 2x + 13 and 14x. The table is then completed as:
Result