Gujarat BoardĀ GSEB Textbook Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6

Question 1.

Solve the following equations.

(i) \(\frac { 8x – 3 }{ 3x }\) = 2

(ii) \(\frac { 9x }{ 7 – 6x }\) = 15

(iii) \(\frac { z }{ z + 15 }\) = \(\frac { 4 }{ 9 }\)

(iv) \(\frac { 3y + 4 }{ 2 – 6y }\) = \(\frac { -2 }{ 5 }\)

(v) \(\frac { 7y + 4 }{ y + 2 }\) = \(\frac { -4 }{ 3 }\)

Solution:

(i) \(\frac { 8x – 3 }{ 3x }\) = 2

Multiplying both sides by 3x, we have

\(\frac { 8x – 3 }{ 3x }\) = 2 Ć 3x = 2 Ć 3x

or 8x – 3 = 6x

Transposing (-3) to RHS and 6x to LHS, we have

8x – 6x = 3 or 2x = 3

Dividing both sides by 2, we have x = \(\frac { 3 }{ 2 }\)

(ii) \(\frac { 9x }{ 7 – 6x }\) = 15

Multiplying both sides by 7 – 6x, we have

\(\frac { 9x }{ 7 – 6x }\) Ć (7 – 6x) = 15 Ć (7 – 6x)

or 9x = 105 – 90x

Transposing (-90x) to LHS, we have

9x + 90x = 105 or 99x = 105

or x = \(\frac { 105 }{ 99 }\) (Dividing both sides by 99)

or x = \(\frac { 35 }{ 33 }\)

(iii) \(\frac { z }{ z + 15 }\) = \(\frac { 4 }{ 9 }\)

by cross-multiplication, we have

9z = 4(z +15) ā 9z = 4z + 60

Transposing 4z to LHS, we have

5z = 60 ā z = \(\frac { 60 }{ 5 }\) = 12

ā“ z = 12

(iv) \(\frac { 3y + 4 }{ 2 – 6y }\) = \(\frac { -2 }{ 5 }\)

By cross-multiplication, we have

5(3y + 4) = -2(2 – 6y) .

or 15y + 20 = – 4 + 12y

Transposing 20 to RHS and 12y to LHS, we

have

15y – 12y = – 4 – 20 or 3y = -24

or y = \(\frac { -24 }{ 3 }\) = -8

(Dividing both sides by 3) or y = -8

(v) \(\frac { 7y + 4 }{ y + 2 }\) = \(\frac { -4 }{ 3 }\)

By cross-multiplication, we have

3 x (7y + 4) = -4 x (y + 2)

or 21y + 12 = -4y – 8

Transposing 12 to RHS and (-4y) to LHS,

we have

21y + 4y = -8 – 12

or 25 y = -20 or y = \(\frac { -20 }{ 25 }\) = \(\frac { -4 }{ 5 }\)

(Dividing both sides by 25)

ā“ y = \(\frac { -4 }{ 5 }\)

Question 2.

The ages of Hari and Harry are in the ratio 5 : 7. Four years from now the ratio of their ages will be 3 : 4. Find their present ages.

Solution:

Let the present age of Hari = 5x years and the present age of Harry = 7x years After 4 years, age of Hari = (5x + 4) yearn Age of Harry = (7x + 4)’years According to the condition,

(5x + 4) : (7x + 4) = 3 : 4 or \(\frac { 5x + 4 }{ 7x + 4 }\) = \(\frac { 3 }{ 4 }\)

by cross multiplication, we have:

4(5x + 4) = 3(7x + 4)

or 20x + 16 = 21x + 12

Transposing 1+6 to RHS and 21x to LHS, we have

20x – 21x = 12 – 16 -x = -4 ā x = 4

ā“ Present age of Hari = 5 x 4 = 20 years

Present age of Harry = 7 x 4 = 28 years

Question 3.

The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is \(\frac { 3 }{ 2 }\). Find the rational number.

Solution:

Let the numerator = x

ā“Denominator = x + 8

New numerator = x + 17

New denominator = (x + 8) – 1 = x + 1

ā“ The new number = \(\frac { x + 17 }{ x + 7 }\)

According to the condition, we have

\(\frac { x + 17 }{ x + 7 }\) = \(\frac { 3 }{ 2 }\)

By cross multiplication, we have

2(x + 17) = 3(x + 7)

2x + 34 = 3x + 21

Transposing 34 to RHS and 3x to LHS, we have

2x – 3x = 21 – 34 ā -x = -13 x = 13

ā“ x = 13 ā Numerator =13

x + 8 = 13 + 8 = 21āDenominator = 21

ā“ The rational number = \(\frac { 13 }{ 21 }\)