# Linear Equations in One Variable : Exercise 2.3 (Mathematics NCERT Class 8th)

Solve the following equations and check your results.

Q.1 3x = 2x + 18
Sol. Given, 3x = 2x + 18
Transposing 2x from RHS to LHS, we get,
3x â€“ 2x = 18
x = 18
Verification:
LHS = 3x = 3 x 18 = 54
RHS = 2x + 18 = 2 x 18 + 18 = 36 + 18 = 54
Here, LHS = RHS
Thus, the result obtained is correct.

Q.2 5t â€“ 3 = 3t â€“ 5
Sol. Given, 5t â€“ 3 = 3t â€“ 5
Transposing 3t from RHS to LHS and â€“ 3 from LHS to RHS, we get,
5t â€“ 3t = -5 â€“ (- 3)
2t = - 2
Dividing both the sides by 2, we get,
t = -1
Verification:
LHS = 5t - 3 = 5 x (-1) â€“ 3 = - 8
RHS = 3t â€“ 5 = 3 x (-1) â€“ 5 = -3 - 18 = -8
Here, LHS = RHS
Thus, the result obtained is correct.

Q.3 5x + 9 = 5 + 3x
Sol. Given, 5x + 9 = 5 + 3x
Transposing 3x from RHS to LHS and 9 from LHS to RHS
5x â€“ 3x = 5 â€“ 9
2x = -4
Dividing both the sides by 2, we get,
x = -2
Verification:
LHS = 5x + 9 = 5 x (-2) + 9 = - 1
RHS = 5 + 3x = 5 + 3 x (-2) = 5 - 6 = -1
Here, LHS = RHS
Thus, the result obtained is correct.

Q.4 4z + 3 = 6 + 2z
Sol. Given, 4z + 3 = 6 + 2z
Transposing 2z from RHS to LHS and 3 from LHS to RHS, we get,
4z â€“ 2z = 6 â€“ 3
2z = 3
Dividing by 2 on both the sides, we get,
${\rm{z}} = {3 \over 2}$
Verification:
LHS = 4z + 3 = 4 x $\left( {{3 \over 2}} \right)$+ 3 = 6 + 3 = 9
RHS = 6 + 2z = 6 + 2 x $\left( {{3 \over 2}} \right)$= 6 + 3 = 9
Here, LHS = RHS
Thus, the result obtained is correct.

Q.5 2x â€“ 1 = 14 â€“ x
Sol. Given, 2x â€“ 1 = 14 â€“ x
Transposing x from RHs to LHS and 1 from LHS to RHS, we get,
2x + x = 14 + 1
3x = 15
Dividing by 3 on both the sides, we get,
x = 5
Verification:
LHS = 2x - 1 = 2 x (5) - 1 = 10 - 1= 9
RHS = 14 - x = 14 - 5 = 9
Here, LHS = RHS
Thus, the result obtained is correct.

Q.6 8x + 4 = 3 (x â€“ 1) + 7
Sol. Given, 8x + 4 = 3 (x â€“ 1) + 7
8x + 4 = 3x â€“ 3 + 7
Transposing 3x from RHS to LHS and 4 from LHS to RHS, we get,
8x â€“ 3x = - 3 + 7 â€“ 4
5x = - 7 + 7
x = 0
Verification:
LHS = 8x + 4 = 8 x (0) + 4 = 4
RHS = 3(x â€“ 1) + 7 = 3(0 â€“ 1) + 7 = - 3 + 7 = 4
Here, LHS = RHS
Thus, the result obtained is correct.

Q.7 $x = {4 \over 5}(x + 10)$
Sol. Given, $x = {4 \over 5}(x + 10)$
Multiplying by 5 on both the sides, we get,
5x = 4(x + 10)
5x = 4x + 40
Transposing 4x from RHS to LHS, we get,
5x â€“ 4x = 40
x = 40
Verification:
LHS = x = 40
RHS =${4 \over 5}(x + 10)$= ${4 \over 5}(40 + 10)$= ${4 \over 5} \times 50$= 40
Here, LHS = RHS
Thus, the result obtained is correct.

Q.8 ${{2x} \over 3} + 1 = {{7x} \over {15}} + 3$
Sol. Given, ${{2x} \over 3} + 1 = {{7x} \over {15}} + 3$
Transposing ${{7x} \over {15}}$ from RHS to LHS and 1 from LHS to RHS, we get,
${{2x} \over 3} - {{7x} \over {15}} = 3 - 1$
${{5 \times 2x - 7x} \over {15}} = 2$
${{3x} \over {15}} = 2$
${x \over 5} = 2$
Multiplying by 5 on both the sides, we get,
x = 10
Verification:
LHS = ${{2x} \over 3} + 1$= ${{2 \times 10} \over 3} + 1$=${{2 \times 10 + 1 \times 3} \over 3} = {{22} \over 3}$
RHS = ${{7x} \over {15}} + 3$= ${{7 \times 2} \over {15}} + 3$= ${{14} \over {15}} + 3$=
${{14 + 3 \times 15} \over {15}} = {{23} \over 3}$=
Here, LHS = RHS
Thus, the result obtained is correct.

Q.9 $2y + {5 \over 3} = {{26} \over 3} - y$
Sol. Given, $2y + {5 \over 3} = {{26} \over 3} - y$
$2y + y = {{26} \over 3} - {5 \over 3}$
$3y = {{21} \over 3} = 7$
Dividing by 3 on both the sides, we get,
$y = {7 \over 3}$
Verification:
LHS = $2y + {5 \over 3}$= $2 \times {7 \over 3} + {5 \over 3}$ = ${{14} \over 3} + {5 \over 3}$ = ${{19} \over 3}$
RHS = ${{26} \over 3} - y$= ${{26} \over 3} - {7 \over 3}$= ${{19} \over 3}$
Here, LHS = RHS
Thus, the result obtained is correct.

Q.10 $3m = 5m - {8 \over 5}$
Sol. Given, $3m = 5m - {8 \over 5}$
Transposing 5m from RHS to LHS, we get,
$3m - 5m = - {8 \over 5}$
$- 2m = - {8 \over 5}$
Dividing by -2 on both the sides, we get,
$m = {4 \over 5}$
Verification:
LHS = $3m = 3 \times {4 \over 5} = {{12} \over 5}$
RHS = $5m - {8 \over 5} = 5 \times {4 \over 5} - {8 \over 5} = {{12} \over 5}$
Here, LHS = RHS
Thus, the result obtained is correct.