Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 7 Lesson 7.2 Evaluating Algebraic Expressions to score better marks in the exam.

## Math in Focus Grade 6 Course 1 A Chapter 7 Lesson 7.2 Answer Key Evaluating Algebraic Expressions

### Math in Focus Grade 6 Chapter 7 Lesson 7.2 Guided Practice Answer Key

**Evaluate each algebraic expression for the given value of x.**

Question 1.

Answer:

Explanation:

An expression is a term in that describes a group of variables, numbers and operators.

Operators include division, multiplication, addition and subtraction.

Variables in expression are usually denoted as x and y, but it can be and other symbol.

The expressions can be written as verbal phrase or algebraic expression.

### Math in Focus Course 1A Practice 7.2 Answer Key

**Evaluate each expression for the given value of the x.**

Question 1.

x + x + 5 when x = 7

Answer:

19

Explanation:

x = 7

by substituting x value as 7

x + x + 5 = 7 + 7 + 5 = 19

Question 2.

3x + 5 when x = 5

Answer:

20

Explanation:

3x + 5, where as

x = 5

by substituting x value =5

3 x 5 + 5 = 15 + 5 = 20

Question 3.

5y – 8 when y = 3

Answer:

7

Explanation:

5 x y – 8

y = 3

by substituting y value = 3

5 x 3 – 8

15 – 8 = 7

Question 4.

40 – 9y when y = 2

Answer:

22

Explanation:

40 – 9y

y = 2

by substituting y value = 2

40 – 9 x 2 = 40 – 18 = 22

Question 5.

33 – 7w + 6 when w = 4

Answer:

19

Explanation:

33 – 7w + 6

w = 4

by substituting w value = 4

33 – 7 x 4 = 33 – 14 = 19

Question 6.

\(\frac{7w}{2}\) when w = 18

Answer:

63

Explanation:

= \(\frac{7w}{2}\)

w = 18

by substituting w value = 18

= \(\frac{7 x 18}{2}\)

= \(\frac{126}{2}\) = 63

Question 7.

4 + \(\frac{5z}{6}\) when z = 12

Answer:

14

Explanation:

4 + \(\frac{5z}{6}\)

z = 12

by substituting z value = 12

= 4 + \(\frac{5×12}{6}\)

= 4 + \(\frac{60}{6}\)

= 4+10 = 14

Question 8.

\(\frac{4+5z}{2}\) when z = 12

Answer:

32

Explanation:

\(\frac{4+5z}{2}\)

z = 12

by substituting z value =12

= \(\frac{4+5 x 12}{2}\)

= \(\frac{4+60}{2}\)

= \(\frac{64}{2}\) = 32

Question 9.

20 – \(\frac{4r}{5}\) when r = 10

Answer:

16

Explanation:

20 – \(\frac{4r}{5}\)

r = 10

by substituting r value =10

= 20 – \(\frac{4x 10}{5}\)

= 20 – \(\frac{40}{5}\)

= 20 – 8 = 16

Question 10.

\(\frac{8r}{9}\) – 15 when r = 27

Answer:

9

Explanation:

\(\frac{8r}{9}\) – 15

r = 27

by substituting r value =27

= \(\frac{8 x 27}{9}\) – 15

= \(\frac{216}{9}\) – 15

24 – 15 = 9

Question 11.

16 – \(\frac{2 z-4}{3}\) when z = 18

Answer:

\(\frac{16}{3}\)

Explanation:

16 – \(\frac{2 z-4}{3}\) when z = 18

by substituting z value = 18

16 – \(\frac{2 z-4}{3}\)

=16 – \(\frac{2 x 18 – 4}{3}\)

=16 – \(\frac{36-4}{3}\)

=16 – \(\frac{32}{3}\)

=\(\frac{48-32}{3}\)

\(\frac{16}{3}\)

Question 12.

16 – \(\frac{2 z}{3}\) – 4 when z = 18

Answer:

0

Explanation:

16 – \(\frac{2 z}{3}\) – 4 when z = 18

by substituting z value = 18

16 – \(\frac{2 x 18}{3}\) – 4

= 16 – \(\frac{36}{3}\) – 4

=16 – 12 – 4

=0

**Evaluate each expression when x = 3.**

Question 13.

\(\frac{x+1}{2}\) + \(\frac{5 x-3}{10}\)

Answer:

\(\frac{16}{5}\)

Explanation:

by substituting x value = 3 in the given algebraic equation

\(\frac{x+1}{2}\) + \(\frac{5 x-3}{10}\)

= \(\frac{3+1}{2}\) + \(\frac{5 x 3 – 3}{10}\)

= \(\frac{4}{2}\) + \(\frac{12}{10}\)

= 2 + \(\frac{6}{5}\)

= \(\frac{10 + 6}{5}\)

= \(\frac{16}{5}\)

Question 14.

\(\frac{11+x}{2}\) – \(\frac{9 x-3}{4}\)

Answer:

1

Explanation:

by substituting x value = 3 in the given algebraic equation

\(\frac{11+x}{2}\) – \(\frac{9 x-3}{4}\)

= \(\frac{11+3}{2}\) – \(\frac{9 x 3-3}{4}\)

= \(\frac{14}{2}\) – \(\frac{27-3}{4}\)

= \(\frac{7}{1}\) – \(\frac{24}{4}\)

7 – 6 = 1

Question 15.

\(\frac{7 x-6}{3}\) + 4(8 + 2x)

Answer:

61

Explanation:

by substituting x value = 3 in the given algebraic equation

\(\frac{7 x-6}{3}\) + 4(8 + 2x)

= \(\frac{7 x 3 -6}{3}\) + 4(8 + 2 x 3)

= \(\frac{21- 6}{3}\) + 4(8 + 6)

= \(\frac{15}{3}\) + 56

= 5 + 56

= 61

Question 16.

13(11 – 3x) – \(\frac{5(16-4 x)}{2}\)

Answer:

16

Explanation:

by substituting x value = 3 in the given algebraic equation

13(11 – 3x) – \(\frac{5(16-4 x)}{2}\)

13(11 – 3 x 3) – \(\frac{5(16-4 x 3)}{2}\)

13(11 – 9) – \(\frac{5(16 – 12)}{2}\)

13(2) – \(\frac{5(4)}{2}\)

26 – \(\frac{20}{2}\)

= 26 – \(\frac{20}{2}\)

= 26 – 10

= 16

Question 17.

5(x + 2) + 2(6 – x) + \(\frac{2 x+3}{3}\)

Answer:

34

Explanation:

by substituting x value = 3 in the given algebraic equation

5(x + 2) + 2(6 – x) + \(\frac{2 x+3}{3}\)

5(3 + 2) + 2(6 – 3) + \(\frac{2 x 3+3}{3}\)

5(5) + 2(3) + \(\frac{6+3}{3}\)

25 + 6 + 3 = 34

Question 18.

\(\frac{5 x-3}{4}\) + \(\frac{5(x+5)}{8}\) + 3(13 – 2x)

Answer:

29

Explanation:

by substituting x value = 3 in the given algebraic equation

\(\frac{5 x-3}{4}\) + \(\frac{5(x+5)}{8}\) + 3(13 – 2x)

= \(\frac{5 x 3 – 3}{4}\) + \(\frac{5(3+5)}{8}\) + 3(13 – 2 x 3)

= \(\frac{15-3}{4}\) + \(\frac{5(8)}{8}\) + 3(13 – 6)

= \(\frac{12}{4}\) + \(\frac{40}{8}\) + 3(7)

= 3 + 5 + 21

= 29

Question 19.

\(\frac{2 x+4}{5}\) – \(\frac{x+1}{4}\) + \(\frac{x}{6}\)

Answer:

\(\frac{3}{2}\)

Explanation:

by substituting x value = 3 in the given algebraic equation

\(\frac{2 x+4}{5}\) – \(\frac{x+1}{4}\) + \(\frac{x}{6}\)

= \(\frac{2 x 3+4}{5}\) – \(\frac{3+1}{4}\) + \(\frac{3}{6}\)

= \(\frac{10}{5}\) – \(\frac{4}{4}\) + \(\frac{1}{2}\)

= 2-1+\(\frac{1}{2}\)

=1+\(\frac{1}{2}\)

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

Question 20.

7x – \(\frac{x}{5}\) + \(\frac{7-x}{9}\)

Answer:

21 – \(\frac{3}{5}\) + \(\frac{4}{9}\)

Explanation:

by substituting x value = 3 in the given algebraic equation

7x – \(\frac{x}{5}\) + \(\frac{7-x}{9}\)

= 7 x 3 – \(\frac{3}{5}\) + \(\frac{7-3}{9}\)

= 21 – \(\frac{3}{5}\) + \(\frac{4}{9}\)

= 21 – \(\frac{3}{5}\) + \(\frac{4}{9}\)

**Evaluate each of the following when y = 7.**

Question 21.

(5y + 2) minus (2y + 5).

Answer:

18

Explanation:

by substituting y value = 7 in the given algebraic equation

(5y + 2) minus (2y + 5)

= (5 x 7 + 2) – (2 x 7 + 5).

= (35 + 2) – (14 + 5).

= 37 – 19

= 18

Question 22.

The sum of \(\frac{y}{3}\) and \(\frac{4y}{9}\)

Answer:

\(\frac{49}{9}\)

Explanation:

by substituting y value = 7 in the given algebraic equation

\(\frac{y}{3}\) + \(\frac{4y}{9}\)

= \(\frac{7}{3}\) + \(\frac{4 x 7}{9}\)

= \(\frac{7}{3}\) + \(\frac{28}{9}\)

= \(\frac{21+28}{9}\)

= \(\frac{49}{9}\)

Question 23.

The product of (y + 1)and(y – 1)

Answer:

\(\frac{4}{3}\)

Explanation:

by substituting y value = 7 in the given algebraic equation

(y + 1)and(y – 1)

= (7 + 1)and(7 – 1)

= (8)and(6)

= \(\frac{8}{6}\)

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

Question 24.

8(2y – 1) minus \(\frac{14 y+37}{5}\)

Answer:

8(2y – 1) minus \(\frac{14 y+37}{5}\)

Explanation:

by substituting y value = 7 in the given algebraic equation

8(2y – 1) minus \(\frac{14 y+37}{5}\)

= 8(2 X 7 – 1) minus \(\frac{14 X 7 + 37}{5}\)

= 8(2y – 1) minus \(\frac{14 y+37}{5}\)

Question 25.

The quotient of 9(7y – 15) and \(\frac{110-6 y}{4}\)

Answer:

18

Explanation:

by substituting y value = 7 in the given algebraic equation

9(7y – 15) and \(\frac{110-6 y}{4}\)

= 9(7 x 7 – 15) and \(\frac{110 – 6 x 7}{4}\)

= 9(49 – 15) and \(\frac{110-42}{4}\)

= 9(34) and \(\frac{68}{4}\)

= 306/17 = 18

Question 26.

The sum of \(\frac{5 y}{6}\) and 4(\(\frac{3 y}{7}\) + 2y)

Answer:

\(\frac{443}{6}\)

Explanation:

by substituting y value = 7 in the given algebraic equation

\(\frac{5 y}{6}\) and 4(\(\frac{3 y}{7}\) + 2y)

= \(\frac{5 x 7}{6}\) and 4(\(\frac{3 x 7}{7}\) + 2 x 7)

= \(\frac{35}{6}\) and 4(\(\frac{21}{7}\) + 14)

= \(\frac{35}{6}\) and 4(3 + 14)

= \(\frac{35}{6}\) +68

= \(\frac{35 + 408 }{6}\)

= \(\frac{443}{6}\)

Question 27.

The quotient of (\(\frac{y}{2}+\frac{2 y}{3}\)) and (\(\frac{5 y}{6}-\frac{y}{3}\))

Answer:

\(\frac{7}{3}\)

Explanation:

by substituting y value = 7 in the given algebraic equation

(\(\frac{y}{2}+\frac{2 y}{3}\)) and (\(\frac{5 y}{6}-\frac{y}{3}\))

= (\(\frac{7}{2}+\frac{2 x 7}{3}\)) and (\(\frac{5 x 7}{6}-\frac{7}{3}\))

= (\(\frac{7}{2}+\frac{14}{3}\)) and (\(\frac{35}{6}-\frac{7}{3}\))

= (\(\frac{21 + 28}{6}\)) and (\(\frac{35 – 14}{6}\))

= (\(\frac{49}{6}\)) and (\(\frac{21}{6}\))

= (\(\frac{49}{6}\)) X (\(\frac{6}{21}\))

= \(\frac{49}{21}\)

= \(\frac{7}{3}\)