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

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 = $$\frac{13}{21}$$

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