Oops! It appears that you have disabled your Javascript. In order for you to see this page as it is meant to appear, we ask that you please re-enable your Javascript!

Linear Equations in One Variable : Exercise 2.5 (Mathematics NCERT Class 8th)


Solve the following linear equations.

Q.1 {x \over 2} - {1 \over 5} = {x \over 3} + {1 \over 4}
Sol. Given, {x \over 2} - {1 \over 5} = {x \over 3} + {1 \over 4}
Transposing {x \over 3} from RHS to LHS and {1 \over 5} from LHS to RHS, we get,
{x \over 2} - {x \over 3} = {1 \over 4} + {1 \over 5}
{{3x - 2x} \over 6} = {{5 + 4} \over {20}}
{x \over 6} = {9 \over {20}}
x = {{9 \times 6} \over {20}}
x = {{27} \over {10}}

Q.2 {n \over 2} - {{3n} \over 4} + {{5n} \over 6} = 21
Sol. Given, {n \over 2} - {{3n} \over 4} + {{5n} \over 6} = 21
{{6n - 9n + 10n} \over {12}} = 21
{{7n} \over {12}} = 21
n = {{21 \times 12} \over 7}
n = 36

Q.3x + 7 - {{8x} \over 3} = {{17} \over 6} - {{5x} \over 2}
Sol. Given, x + 7 - {{8x} \over 3} = {{17} \over 6} - {{5x} \over 2}
Transposing {{5x} \over 2} from RHS to LHS and 7 from LHS to RHS, we get,
x + {{5x} \over 2} - {{8x} \over 3} = {{17} \over 6} - 7
{{6x - 16x + 15x} \over 6} = {{17 - 42} \over 6}
{{5x} \over 6} = {{ - 25} \over 6}
x = {{ - 25 \times 6} \over {6 \times 5}}
x = - 5

Q.4 {{x - 5} \over 3} = {{x - 3} \over 5}
Sol. Given, {{x - 5} \over 3} = {{x - 3} \over 5}
5 (x - 5) = 3 (x - 3)
5x - 25 = 3x - 9
Transposing 3x from RHS to LHS and 25 from LHS to RHS, we get,
5x - 3x = -9 + 25
2x = 16
x = 8

Q.5 {{3t - 2} \over 4} - {{2t + 3} \over 3} = {2 \over 3} - t
Sol. Given, {{3t - 2} \over 4} - {{2t + 3} \over 3} = {2 \over 3} - t
Transposing from RHS to LHS, we get,
{{3t - 2} \over 4} - {{2t + 3} \over 3} + t = {2 \over 3}
{{3(3t - 2) - 4(2t + 3) + 12t} \over 4} = {2 \over 3}
{{9t - 6 - 8t - 12 + 12t} \over {12}} = {2 \over 3}
{{3t - 18} \over {12}} = {2 \over 3}
3 \times (13t - 18) = 2 \times 12
39t - 54 = 24
39t = 24 + 54
39t = 78
t = {{78} \over {39}}
t = 2

Q.6 m - {{m - 1} \over 2} + {{m - 2} \over 3} = 1
Sol. Given, m - {{m - 1} \over 2} + {{m - 2} \over 3} = 1
Transposing {{m - 2} \over 3}from RHS to LHS, we get,
m - {{m - 1} \over 2} + {{m - 2} \over 3} = 1
{{6m - 3(m - 1) + 2(m - 2)} \over 2} = 1
{{6m - 3m + 2 + 2m - 4} \over 6} = 1
{{5m - 1} \over 6} = 1
5m - 1 = 6
5m = 7
m = {7 \over 5}

Simplify and solve the following linear equations.

Q.7 3(t-3) = 5(2t+1)
Sol. Given, 3(t-3) = 5(2t+1)
3t - 9 = 10t + 5
3t-10t = 5 + 9
-7t = 14
t = {{14} \over { - 7}}
t = -2

Q.8 15 (y-4) -2 (y-9) + 5 (y+6) = 0
Sol. Given,  15 (y-4) -2 (y-9) + 5 (y+6) = 0
15y - 60 - 2y + 18 + 5y + 30 = 0
18y - 12 = 0
18y = 12
y = {{12} \over {18}}
y = {2 \over 3}

Q.9 3(5z - 7) - 2 (9z - 11) = 4 (8z  - 13) -17
Sol. Given, 3(5z - 7) - 2 (9z - 11) = 4 (8z  - 13) -17
15z - 21 -18z + 22 = 32z - 52 - 17
-3z + 1 = 32z - 69
-3z -32 z = - 69 - 1
-35 z = -70
z = {{ - 70} \over { - 35}}
z = 2

Q.10 0.25 (4f - 3) = 0.05 (10f - 9)
Sol. Given, 0.25 (4f - 3) = 0.05 (10f - 9)
1.00 f -0.75 = 0.50 f - 0.45
1.00 f -0.50 f = -0.45 +0.75
0.50 f = 0.3
f = {{0.3} \over {0.50}}
f = 0.6



3 Comments

Leave a Reply

Contact Us

Call us: 8287971571,0261-4890014

Or, Fill out the form & get a call back.!