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

NCERT Solutions Class 8 Mathematics 
Chapter – 2 (Linear Equations in One Variable) 

The NCERT Solutions in English Language for Class 8 Mathematics Chapter – 2 Linear Equations in One Variable Exercise 2.5 has been provided here to help the students in solving the questions from this exercise. 

Chapter 2: Linear Equations in One Variable

Exercise – 2.5 

Solve the following linear equations.

1. \mathbf{\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}

⇒ \frac{x}{2} – \frac{x}{3} = \frac{1}{4} +\frac{1}{5}

⇒ \frac{3x-2x}{6} = \frac{5+4}{20}

⇒ 3x – 2x = \frac{9}{20}\times 6

⇒ x = \frac{9}{10}\times 3

⇒ x = \frac{27}{10}

2. \mathbf{\frac{n}{2} - \frac{3n}{4}+\frac{5n}{6}} = 21

Solution –

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

⇒ \frac{6n - 9n +10n}{12} = 21

⇒ \frac{7n}{12} = 21

⇒ 7n = 21 × 12

⇒ n = \frac{21\times 12}{7}
⇒ n = 3 \times 12
⇒ n = 36

3. x + 7 – \mathbf{\frac{8x}{3}} = \mathbf{\frac{17}{6}-\frac{5x}{2}}

Solution –

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

⇒ x – \frac{8x}{3} + \frac{5x}{2} = \frac{17}{6} – 7

⇒ \frac{6x-16x+15x}{6} = \frac{17-42}{6}

⇒ \frac{5x}{6} = – \frac{25}{6}

⇒ 5x = – 25

⇒ x = -\frac{25}{5}

⇒ x = – 5

4. \mathbf{\frac{(x-5)}{3} = \frac{(x-3)}{5}}

Solution –

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

⇒ 5 (x – 5) = 3 (x – 3)
⇒ 5x – 25 = 3x – 9
⇒ 5x – 3x = -9 + 25
⇒ 2x = 16
⇒ x = 8

5. \mathbf{\frac{(3t-2)}{4}-\frac{(2t+3)}{3} = \frac{2}{3} -t}

Solution –

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

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

⇒ (3t-2)\times3 -(2t+3)\times 4 = \frac{(2-3t)}{3} \times 12

⇒ 9t – 6 – 8t – 12 = (2-3t)\times 4
⇒ t – 18 = 8 – 12t
⇒ t + 12t = 8 + 18
⇒ 13t = 26
⇒ t = 2

6. m – \mathbf{m-\frac{m-1}{2} = 1-\frac{(m-2)}{3}}

Solution –

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

⇒ m – \frac{m-1}{2} +\frac{m-2}{3} = 1

⇒ \frac{6m -3\times (m-1)+2\times (m-2)}{6} = 1

⇒ 6m – 3(m – 1) + 2(m – 2) = 6
⇒ 6m – 3m +3 + 2m – 4 = 6
⇒ 5m – 1 = 6
⇒ 5m = 7
⇒ m = \frac{7}{5}

Simplify and solve the following linear equations.

7. 3 (t – 3) = 5(2t + 1)

Solution –
3(t – 3) = 5(2t + 1)
⇒ 3t – 9 = 10t + 5
⇒ 3t – 10t = 5 + 9
⇒ -7t = 14
⇒ t = \frac{14}{-7}

⇒ t = -2

8. 15(y – 4) –2(y – 9) + 5(y + 6) = 0

Solution –
15(y – 4) –2(y – 9) + 5(y + 6) = 0
⇒ 15y – 60 -2y + 18 + 5y + 30 = 0
⇒ 15y – 2y + 5y = 60 – 18 – 30
⇒ 18y = 12

⇒ y = \frac{12}{18}

⇒ y = \frac{2}{3}

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

Solution –
3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
⇒ 15z – 21 – 18z + 22 = 32z – 52 – 17
⇒ 15z – 18z – 32z = -52 – 17 + 21 – 22
⇒ -35z = -70

⇒ z = \frac{-70}{-35}
⇒ z = 2

10. 0.25(4f – 3) = 0.05(10f – 9)

Solution –
0.25(4f – 3) = 0.05(10f – 9)
⇒ f – 0.75 = 0.5f – 0.45
⇒ f – 0.5f = -0.45 + 0.75
⇒ 0.5f = 0.30

⇒ f = \frac{0.30}{0.5}

⇒ f = \frac{3}{5}

⇒ f = 0.6

 

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