enVisionmath 2.0: Grade 6 Volume 1 Chapter 1 Algebra: Understand Numerical And Algebraic Expressions Homework And Practice 6

Chapter 1 Algebra: Understand Numerical And Algebraic Expressions

Homework And Practice 6

Page 41 Exercise 1 Answer

5(m − 2) Given

= 5(m) − 5(2) Using Distributive Property

= 5m − 10 Multiply

Result

5m − 10

Page 41 Exercise 2 Answer

24x + 18y Given

= 6(4x) + 6(3y) Using Distributive Property

= 6(4x + 3y) 6 is the common factor

Result

6(4x + 3y)

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Page 41 Exercise 3 Answer

\(2\left(9 p-\frac{1}{2}\right)\) Given

= \(2(9 p)-2\left(\frac{1}{2}\right)\) Using Distributive Property

= 18p − 1 Multiply

Result

18p − 1

Page 41 Exercise 4 Answer

8(2x − 3) Given

= 8(2x) − 8(3) using Distributive Property

= 16x − 24 Multiply

Result

8(2x − 3) and 16x − 24 are equivalent

Page 41 Exercise 5 Answer

5(3x − 9) Given

= 5(3x) − 5(9) Using Distributive Property

= 15x − 45 Multiply

Result

5(3x − 9) and 15x − 45 are equivalent

Page 41 Exercise 6 Answer

6(2x + 9) Given

= 6(2x) + 6(9) Using Distributive Property

= 12x + 54 Multiply

Result

6(2x + 9) and 12x + 54 are equivalent

Page 41 Exercise 7 Answer

3(6x − 7) Given

= 3(6x) − 3(7) Using Distributive Property

= 18x − 21 Multiply

Result

18x − 21

Page 41 Exercise 8 Answer

4(9x − 2) Given

= 4(9x) − 4(2) Using Distributive Property

= 36x − 8 Multiply

Result

36x − 8

Page 41 Exercise 9 Answer

6(8x + 1) Given

= 6(8x) + 6(1) Using Distributive Property

= 48x + 6 Multiply

Result

48x + 6

Page 41 Exercise 10 Answer

35x + 30 Given

= 5(7x) + 5(6) Using Distributive Property

= 5(7x + 6) Multiply

Result

5(7x + 6)

Page 41 Exercise 11 Answer

4(x + 7) Given

= 4(x) + 4(7) Using Distributive Property

= 4x + 28 Multiply

Result

4x + 28

Page 41 Exercise 12 Answer

5x − 15y Given

= 5(x) − 5(3y) Using Distributive Property

= 5(x − 3y) 5 is the common factor

Result

5(x − 3y)

Page 41 Exercise 13 Answer

\(6\left(3 y-\frac{1}{2}\right)\) Given

= \(6(3 y)-6\left(\frac{1}{2}\right)\) Using Distributive Property

= 18y − 3 Multiply

Result

18y − 3

Page 41 Exercise 14 Answer

1.6 + (2z + 0.4) Given

= (1.6 + 2z) + 0.4 Using Associative Property of Addition

Result

(1.6 + 2z) + 0.4

Page 41 Exercise 15 Answer

8w − 16 Given

= 8(w) − 8(2) Using Distributive Property

= 8(w − 2) 8 is the common factor

Result

8(w – 2)

Page 41 Exercise 16 Answer

2.2x + 2.2 Given

= 2.2(x) + 2.2(1) Using Distributive Property

= 2.2(x + 1) 2.2 is the common factor

Result

2.2(x + 1)

Page 41 Exercise 17 Answer

\(100\left(z^2-5.38\right)\) Given

= 100\((z)^2\) – 100(5.38) Using Distributive Property

100\(z^2\) − 538 Multiply

Result

100\(z^2\) − 538

Page 41 Exercise 18 Answer

\(8 \cdot\left(y^3 \cdot \frac{3}{4}\right)\) Evaluate

= \(8 \cdot \frac{3 y^3}{4}\) Evaluate inside parentheses

= 6\(y^3\)Multiply

Result

6\(y^3\)

Page 42 Exercise 19 Answer

Cost of 1 Pencil Pack = $1.50

Cost of 1 Notebook = $2

Cost of 1 Marker = $2.50

Number of Pencil Packs ordered = 5

Number of Notebooks ordered = n

Number of Markers ordered = 5 sets

Algebraic Expression for the total cost of Ms. Thomas′s ordered:

Number of Pencil pack order × Cost of each pencil pack + Number of Notebooks order × Cost of each Notebook + Number of Markers order × Cost of each Marker

= 5 × 1.50 + n × 2 + 5 × 2.50

Result

5 × 1.50 + n × 2 + 5 × 2.50

Page 42 Exercise 20 Answer

5 × 1.50 + n × 2 + 5 × 2.50 Algebraic Expression

= 7.50 + 2n + 12.50 Multiply

= 7.50 + 12.50 + 2n Commutative Property of Addition

= 20 + 2n Add

= 2(10) + 2(n) Distributive Property

= 2(10 + n) 2 is the common factor

Result

2(10 + n)

Page 42 Exercise 21 Answer

2(10 + n) Evaluate

= 2(10 + 20) Substitute n = 20

= 2(30) Add

= $60 Multiply

Result

The total cost of Ms. Thomas′s Order if she ordered 20 Notebooks is $60

Page 42 Exercise 22 Answer

2l + 2w Evaluate

= 2(l) + 2(w) Using Distributive Property

=2(l + w) 2 is the common factor

Result

2l + 2w and 2(l + w) are equivalent expressions.

Page 42 Exercise 23 Answer

It is easier to use 2(l + w) than to use 2l + 2w because →

2l + 2w: We need to multiply length with 2 and then width with 2 and then add to get the solution.

2(l + w): We need to add length and width and then Multiply it with 2 to get the solution.

Example: Let length = 5 units and Width = 3 units

2l + 2w = 2 ⋅ 5 + 2 ⋅ 3 = 10 + 2 ⋅ 3 = 10 + 6 = 16 units

2(l + w) = 2(5 + 3) = 2(8) = 16 units

Result

Because it is less time consuming than the expression 2l+2w

Page 42 Exercise 24 Answer

\(4 \frac{1}{2}+\left(3 t+1 \frac{1}{2}\right)\)

= \(\left(4 \frac{1}{2}+3 t\right)+1 \frac{1}{2}\) Using Associative Property of Addition

\(4 \frac{1}{2}+\left(3 t+1 \frac{1}{2}\right)\)

= \(\left(4 \frac{1}{2}+1 \frac{1}{2}\right)+3 t\) Using Associative Property of Addition

\(4 \frac{1}{2}+\left(3 t+1 \frac{1}{2}\right)\)

= \(\left(4 \frac{1}{2}+1 \frac{1}{2}\right)+3 t\) Using Associative Property of Addition

= 6 + 3t Adding

\(4 \frac{1}{2}+\left(3 t+1 \frac{1}{2}\right)\)

= \(\left(4 \frac{1}{2}+1 \frac{1}{2}\right)+3 t\) Using Associative Property of Addition

= 6 + 3t Adding

= 3(2) + 3(t) Using Distributive Property

= 3(2 + t) 2 is the common factor

Result

\((4+3 t)+1 \frac{1}{2}\) \(\left(4+1 \frac{1}{2}\right)+3 t\)

6 + 3t

3(2 + t)

Page 42 Exercise 25 Answer

8(x − 3)

= 8(x) − 8(3) Using Distributive Property

= 8x − 24

8(x − 24)

= 8(x) − 8(24) Using Distributive Property

= 8x − 192

9(x − 3) − (x – 3)

= 9(x) − 9(3) − x + 3 Using Distributive Property

= 9x − 27 − x + 3 Multiply

= 9x − x − 27 + 3 Using Commutative Property of Addition

= 8x − 24

(5 + 3)x − 24

= x(5) + x(3) − 24 Using Distributive Property

= 5x + 3x − 24 Multiply

= 8x − 24

Result

8x − 24 is equivalent to the following:

8(x − 3)

9(x − 3) − (x − 3)

(5 + 3)x − 24

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