These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 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 2.5

Solve the following linear equations.

Question 1.

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

Solution:

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

Transposing \(-\frac{1}{5}\) to R.H.S. and \(\frac{\mathrm{x}}{3}\) to L.H.S

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

\(\frac{3 x-2 x}{6}=\frac{5+4}{20}\)

\(\frac{x}{6}=\frac{9}{20}\)

By cross-multiplication, we get

20(x) = 9 × 6

20x = 54

Dividing both sides by 20

\(\frac{20 \mathrm{x}}{20}=\frac{54}{20}\)

x = \(\frac{27}{10}\)

Question 2.

\(\frac{\mathrm{n}}{2}-\frac{3 n}{4}+\frac{5 \mathrm{n}}{6}=21\)

Solution:

\(\frac{\mathrm{n}}{2}-\frac{3 n}{4}+\frac{5 \mathrm{n}}{6}=21\)

L.C.M of 2, 4 and 6 = 12

Multiplying each term by 12, we get

\(12 \times \frac{\mathrm{n}}{2}-12 \times \frac{3 \mathrm{n}}{4}+12 \times \frac{5 \mathrm{n}}{6}=21 \times 12\)

6n – 9n + 10n = 252

7n = 252

Dividing both sides by 7

\(\frac{7 \mathrm{n}}{7}=\frac{252}{7}\)

n = 36

Question 3.

\(x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}\)

Solution:

\(x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}\)

Transposing 7 to R.H.S. and \(\frac{-5 x}{2}\) to L.H.S

\(x-\frac{8 x}{3}+\frac{5 x}{2}=\frac{17}{6}-7\)

L.C.M of 3, 2 and 6 = 6

Multiplying each term by 6, we get

\(6 \times x-6 \times \frac{8 x}{3}+6 \times \frac{5 x}{2}=6 \times \frac{17}{6}-6 \times 7\)

6x – 16x + 15x = 17 – 42

5x = -25

Dividing both sides by 5

\(\frac{5 x}{5}=-\frac{25}{5}\)

x = -5

Question 4.

\(\frac{x-5}{3}=\frac{x-3}{5}\)

Solution:

\(\frac{x-5}{3}=\frac{x-3}{5}\)

By cross-multiplication we get

5(x – 5) = 3(x – 3)

5x – 25 = 3x – 9

Transposing -25 to R.H.S and 3x to L.H.S.

5x – 3x = 25 – 9

2x = 16

Dividing both sides by 2

\(\frac{2 \mathrm{x}}{2}\) = \(\frac{16}{2}\)

x = 8

Question 5.

\(\frac{3 \mathrm{t}-2}{4}-\frac{2 \mathrm{t}+3}{3}=\frac{2}{3}-\mathrm{t}\)

Solution:

\(\frac{3 \mathrm{t}-2}{4}-\frac{2 \mathrm{t}+3}{3}=\frac{2}{3}-\mathrm{t}\)

Transposing -t to L.H.S., we get

\(\frac{3 \mathrm{t}-2}{4}-\frac{2 \mathrm{t}+3}{3}+\mathrm{t}=\frac{2}{3}\)

L.C.M of 4 and 3 is 12

Multiplying each term by 12

\(12 \frac{(3 \mathrm{t}-2)}{4}-12 \frac{(2 \mathrm{t}+3)}{3}+12 \mathrm{t}=12 \times \frac{2}{3}\)

3(3t – 2) – 4(2t + 3) + 12t = 8

9t – 6 – 8t – 12 + 12t = 8

13t – 18 = 8

Transposing -18 to R.H.S., we get

13t = 8 + 18

13t = 26

Dividing both sides by 13

\(\frac{13 t}{13}=\frac{26}{13}\)

t = 2

Question 6.

\(m-\frac{m-1}{2}=1-\frac{m-2}{3}\)

Solution:

\(m-\frac{m-1}{2}=1-\frac{m-2}{3}\)

Transposing \(\left(\frac{m-2}{3}\right)\) to L.H.S. we get

\(\mathrm{m}-\frac{\mathrm{m}-1}{2}+\frac{\mathrm{m}-2}{3}=1\)

L.C.M of 2 and 3 is 6

Multiplying each term by 6, we get

6m – \(\frac{6(m-1)}{2}+\frac{6(m-2)}{3}\) = 1 × 6

6m – 3(m – 1) + 2(m – 2) = 6

6m – 3m + 3 + 2m – 4 = 6

5m – 1 = 6

Transposing -1 to the R.H.S., we get

5m = 6 + 1

5m = 7

Dividing both sides by 5, we get

\(\frac{5 m}{5}=\frac{7}{5}\)

m = \(\frac{7}{5}\)

Simplify and solve the following linear equations.

Question 7.

3(t – 3) = 5 (2t + 1)

Solution:

3(t – 3) = 5 (2t + 1)

Opening the brackets, we get

3t – 9 = 10t + 5

Transposing -9 to R.H.S. and 10t to L.H.S.

3t – 10t = 5 + 9

-7t = 14

Dividing both sides by -7, we get

\(\frac{-7 t}{-7}=\frac{14}{-7}\)

t = -2

Question 8.

15(y – 4) – 2(y – 9) + 5 (y + 6) = 0

Solution:

15(y – 4) – 2(y – 9) + 5(y + 6) = 0

Opening the brackets, we get

15y – 60 – 2y + 18 + 5y + 30 = 0

15y – 2y + 5y – 60 + 18 + 30 = 0

18y – 12 = 0

Transposing -12 to R.H.S.

18y = 12

Dividing both sides by 18, we get.

\(\frac{18 y}{18}=\frac{12}{18}\)

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

Question 9.

3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

Solution:

3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

Opening the brackets, we get

15z – 21 – 18z + 22 = 32z – 52 – 17

15z – 18z + 22 – 21 = 32z – 69

-3z + 1 = 32z – 69

Transposing 1 to R.H.S. and 32z to L.H.S.

-3z – 32z = – 69 – 1

-35z = -70

Dividing both sides by -35, we get

\(\frac{-35 z}{-35}=\frac{-70}{-35}\)

z = 2

Question 10.

0.25(4f – 3) = 0.05(10f – 9)

Solution:

0.25(4f – 3) = 0.05(10f – 9)

Opening the brackets, we get

0.25(4f) – 0.25 × 3 = 0.05 × 10f – 0.05 × 9

f – 0.75 = 0.5f – 0.45

f – 0.5f = 0.75 – 0.45

0.5f = 0.3

Dividing both sides by 0.5, we get

\(\frac{0.5 \mathrm{f}}{0.5}=\frac{0.3}{0.5}\)

f = \(\frac{3}{5}\) = 0.6

f = 0.6