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.3$x + 7 - {{8x} \over 3} = {{17} \over 6} - {{5x} \over 2}$
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