These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.6

Solve the following equations.

Question 1.

\(\frac{8 x-3}{3 x}=2\)

Solution:

\(\frac{8 x-3}{3 x}=2\)

By cross-multiplication we get,

8x – 3 = 2(3x)

8x – 3 = 6x

By transposing -3 to the R.H.S. and 6x to L.H.S.

8x – 6x = 3

2x = 3

Dividing both sides by 2

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

∴ x = \(\frac{3}{2}\)

Question 2.

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

Solution:

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

By cross-multiplication, we get

9x = 15(7 – 6x)

9x = 105 – 90x

By transposing -90x to L.H.S., we get

9x + 90x = 105

99x = 105

Dividing both sides by 99, we get

\(\frac{99 \mathrm{x}}{99}=\frac{105}{99}\)

∴ x = \(\frac{35}{33}\)

Question 3.

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

Solution:

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

By cross-multiplication, we get

9z = 4(z + 15)

9z = 4z + 60

Transposing 4z to L.H.S, we get

9z – 4z = 60

5z = 60

Dividing both sides by 5, we get

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

∴ z = 12

Question 4.

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

Solution:

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

By cross-multiplication, we get

5(3y + 4) = -2(2 – 6y)

15y + 20 = -4 + 12y

Transposing 12y to L.H.S. and 20 to R.H.S.

15y – 12y = -4 – 20

3y = -24

Dividing both sides by 3, we get

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

∴ y = -8

Question 5.

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

Solution:

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

By cross-multiplication, we get

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

21y + 12 = -4y – 8

Transposing +12 to R.H.S. and -4y to L.H.S.

21y + 4y = -8 – 12

25y = -20

Dividing both sides by 25, we get

\(\frac{25 y}{25}=\frac{-20}{25}\)

∴ y = \(-\frac{4}{5}\)

Question 6.

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 four years

Age of Hari = (5x + 4) years

Age of Harry = (7x + 4) years

According to the question

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

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

[Product of the extremes is equal to the product of the means]

20x + 16 = 21x + 12

Transposing 16 to R.H.S. and 21x to L.H.S., we get

20x – 21x = 12 – 16

-x = -4

∴ x = 4

Present age of Hari = 5 × 4 = 20 years

Present age of Harry = 7 × 4 = 28 years

Question 7.

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 be ‘x’

then the denominator is x + 8

∴ The fraction is \(\frac{\mathrm{x}}{\mathrm{x}+8}\)

According to the question, we get

\(\frac{x+17}{(x+8)-1}=\frac{3}{2}\)

\(\frac{x+17}{x+8-1}=\frac{3}{2}\)

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

By cross-multiplication, we get

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

3x + 21 = 2x + 34

By transposing 21 to R.H.S. and 2x to L.H.S., we get

3x – 2x = 34 – 21

x = 13

∴ The fraction is \(\frac{13}{13+8}\) = \(\frac{13}{21}\)

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