Simple Linear Equations — Exercise 14(C)

Solve :

$\dfrac{x}{2} + x$ = 9

Answer

Solving,

$\Rightarrow \dfrac{x}{2} + x = 9 \\[1em] \Rightarrow \dfrac{x + 2x}{2} = 9 \\[1em] \Rightarrow \dfrac{3x}{2} = 9 \\[1em] \Rightarrow 3x = 18 \\[1em] \Rightarrow x = 6.$

Hence, $\bm{x}$ = 6.

Solve :

$\dfrac{x}{5} + 2x$ = 33

Answer

Solving,

$\Rightarrow \dfrac{x}{5} + 2x = 33 \\[1em] \Rightarrow \dfrac{x + 10x}{5} = 33 \\[1em] \Rightarrow \dfrac{11x}{5} = 33 \\[1em] \Rightarrow 11x = 165 \\[1em] \Rightarrow x = 15.$

Hence, $\bm{x}$ = 15.

Solve :

$\dfrac{3x}{4} + 4x$ = 38

Answer

Solving,

$\Rightarrow \dfrac{3x}{4} + 4x = 38 \\[1em] \Rightarrow \dfrac{3x + 16x}{4} = 38 \\[1em] \Rightarrow \dfrac{19x}{4} = 38 \\[1em] \Rightarrow 19x = 152 \\[1em] \Rightarrow x = 8.$

Hence, $\bm{x}$ = 8.

Solve :

$\dfrac{x}{2} + \dfrac{x}{5}$ = 14

Answer

Solving,

$\Rightarrow \dfrac{x}{2} + \dfrac{x}{5} = 14 \\[1em] \Rightarrow \dfrac{5x + 2x}{10} = 14 \\[1em] \Rightarrow \dfrac{7x}{10} = 14 \\[1em] \Rightarrow 7x = 140 \\[1em] \Rightarrow x = 20.$

Hence, $\bm{x}$ = 20.

Solve :

$\dfrac{x}{3} - \dfrac{x}{4}$ = 2

Answer

Solving,

$\Rightarrow \dfrac{x}{3} - \dfrac{x}{4} = 2 \\[1em] \Rightarrow \dfrac{4x - 3x}{12} = 2 \\[1em] \Rightarrow \dfrac{x}{12} = 2 \\[1em] \Rightarrow x = 24.$

Hence, $\bm{x}$ = 24.

Solve :

$y + \dfrac{y}{2} = \dfrac{7}{4} - \dfrac{y}{4}$

Answer

Solving,

$\Rightarrow y + \dfrac{y}{2} = \dfrac{7}{4} - \dfrac{y}{4} \\[1em] \Rightarrow y + \dfrac{y}{2} + \dfrac{y}{4} = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{4y + 2y + y}{4} = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{7y}{4} = \dfrac{7}{4} \\[1em] \Rightarrow 7y = 7 \\[1em] \Rightarrow y = 1.$

Hence, $\bm{y}$ = 1.

Solve :

$\dfrac{4x}{3} - \dfrac{7x}{3}$ = 1

Answer

Solving,

$\Rightarrow \dfrac{4x}{3} - \dfrac{7x}{3} = 1 \\[1em] \Rightarrow \dfrac{4x - 7x}{3} = 1 \\[1em] \Rightarrow \dfrac{-3x}{3} = 1 \\[1em] \Rightarrow -x = 1 \\[1em] \Rightarrow x = -1.$

Hence, $\bm{x}$ = −1.

Solve :

$\dfrac{1}{2}m + \dfrac{3}{4}m - m$ = 2.5

Answer

Solving,

$\Rightarrow \dfrac{1}{2}m + \dfrac{3}{4}m - m = 2.5 \\[1em] \Rightarrow \dfrac{2m + 3m - 4m}{4} = 2.5 \\[1em] \Rightarrow \dfrac{m}{4} = 2.5 \\[1em] \Rightarrow m = 10.$

Hence, $\bm{m}$ = 10.

Solve :

$\dfrac{2x}{3} + \dfrac{x}{2} - \dfrac{3x}{4}$ = 1

Answer

Solving,

$\Rightarrow \dfrac{2x}{3} + \dfrac{x}{2} - \dfrac{3x}{4} = 1 \\[1em] \Rightarrow \dfrac{8x + 6x - 9x}{12} = 1 \\[1em] \Rightarrow \dfrac{5x}{12} = 1 \\[1em] \Rightarrow 5x = 12 \\[1em] \Rightarrow x = \dfrac{12}{5} \\[1em] \Rightarrow x = 2\dfrac{2}{5}.$

Hence, $\bm{x} = 2\dfrac{2}{5}$.

Solve :

$\dfrac{3a}{4} + \dfrac{a}{6}$ = 66

Answer

Solving,

$\Rightarrow \dfrac{3a}{4} + \dfrac{a}{6} = 66 \\[1em] \Rightarrow \dfrac{9a + 2a}{12} = 66 \\[1em] \Rightarrow \dfrac{11a}{12} = 66 \\[1em] \Rightarrow 11a = 792 \\[1em] \Rightarrow a = 72.$

Hence, $\bm{a}$ = 72.

Solve :

$\dfrac{2p}{3} - \dfrac{p}{5}$ = 35

Answer

Solving,

$\Rightarrow \dfrac{2p}{3} - \dfrac{p}{5} = 35 \\[1em] \Rightarrow \dfrac{10p - 3p}{15} = 35 \\[1em] \Rightarrow \dfrac{7p}{15} = 35 \\[1em] \Rightarrow 7p = 525 \\[1em] \Rightarrow p = 75.$

Hence, $\bm{p}$ = 75.

Solve :

$0·6a + 0·2a = 0·4a + 8$

Answer

Solving,

$\Rightarrow 0.6a + 0.2a = 0.4a + 8 \\[1em] \Rightarrow 0.8a = 0.4a + 8 \\[1em] \Rightarrow 0.8a - 0.4a = 8 \\[1em] \Rightarrow 0.4a = 8 \\[1em] \Rightarrow a = \dfrac{8}{0.4} \\[1em] \Rightarrow a = \dfrac{80}{4} \\[1em] \Rightarrow a = 20.$

Hence, $\bm{a}$ = 20.

Solve :

$p + 1·4p$ = 48

Answer

Solving,

$\Rightarrow p + 1.4p = 48 \\[1em] \Rightarrow 2.4p = 48 \\[1em] \Rightarrow p = \dfrac{48}{2.4} \\[1em] \Rightarrow p = \dfrac{480}{24} \\[1em] \Rightarrow p = 20.$

Hence, $\bm{p}$ = 20.

Solve :

$10$% of $x$ = 20

Answer

Solving,

$\Rightarrow 10$% of $x$ = 20

$\Rightarrow \dfrac{10}{100} \times x = 20$

$\Rightarrow x = \dfrac{20 \times 100}{10}$ = 200.

Hence, $\bm{x}$ = 200.

Solve :

$y + 20$% of $y$ = 18

Answer

Solving,

$\Rightarrow y + 20$% of $y$ = 18

$\Rightarrow y + \dfrac{20}{100} \times y = 18$

$\Rightarrow y + \dfrac{y}{5} = 18$

$\Rightarrow \dfrac{5y + y}{5} = 18$

$\Rightarrow \dfrac{6y}{5} = 18$

$\Rightarrow 6y = 90$

$\Rightarrow y = \dfrac{90}{6}$ = 15.

Hence, $\bm{y}$ = 15.

Solve :

$x - 30$% of $x$ = 35

Answer

Solving,

$\Rightarrow x - 30$% of $x$ = 35

$\Rightarrow x - \dfrac{30}{100} \times x = 35$

$\Rightarrow x - \dfrac{3x}{10} = 35$

$\Rightarrow \dfrac{10x - 3x}{10} = 35$

$\Rightarrow \dfrac{7x}{10} = 35$

$\Rightarrow 7x = 350$

$\Rightarrow x = \dfrac{350}{7}$ = 50.

Hence, $\bm{x}$ = 50.

Solve :

$\dfrac{x + 4}{2} + \dfrac{x}{3}$ = 7

Answer

Solving,

$\Rightarrow \dfrac{x + 4}{2} + \dfrac{x}{3} = 7 \\[1em] \Rightarrow \dfrac{3(x + 4) + 2x}{6} = 7 \\[1em] \Rightarrow \dfrac{3x + 12 + 2x}{6} = 7 \\[1em] \Rightarrow \dfrac{5x + 12}{6} = 7 \\[1em] \Rightarrow 5x + 12 = 42 \\[1em] \Rightarrow 5x = 30 \\[1em] \Rightarrow x = 6.$

Hence, $\bm{x}$ = 6.

Solve :

$\dfrac{y+2}{3} + \dfrac{y+5}{4}$ = 6

Answer

Solving,

$\Rightarrow \dfrac{y + 2}{3} + \dfrac{y + 5}{4} = 6 \\[1em] \Rightarrow \dfrac{4(y + 2) + 3(y + 5)}{12} = 6 \\[1em] \Rightarrow \dfrac{4y + 8 + 3y + 15}{12} = 6 \\[1em] \Rightarrow \dfrac{7y + 23}{12} = 6 \\[1em] \Rightarrow 7y + 23 = 72 \\[1em] \Rightarrow 7y = 49 \\[1em] \Rightarrow y = 7.$

Hence, $\bm{y}$ = 7.

Solve :

$\dfrac{3a-2}{7} - \dfrac{a-2}{4}$ = 2

Answer

Solving,

$\Rightarrow \dfrac{3a - 2}{7} - \dfrac{a - 2}{4} = 2 \\[1em] \Rightarrow \dfrac{4(3a - 2) - 7(a - 2)}{28} = 2 \\[1em] \Rightarrow \dfrac{12a - 8 - 7a + 14}{28} = 2 \\[1em] \Rightarrow \dfrac{5a + 6}{28} = 2 \\[1em] \Rightarrow 5a + 6 = 56 \\[1em] \Rightarrow 5a = 50 \\[1em] \Rightarrow a = 10.$

Hence, $\bm{a}$ = 10.

Solve :

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

Answer

Solving,

$\Rightarrow \dfrac{1}{2}(x + 5) - \dfrac{1}{3}(x - 2) = 4 \\[1em] \Rightarrow \dfrac{3(x + 5) - 2(x - 2)}{6} = 4 \\[1em] \Rightarrow \dfrac{3x + 15 - 2x + 4}{6} = 4 \\[1em] \Rightarrow \dfrac{x + 19}{6} = 4 \\[1em] \Rightarrow x + 19 = 24 \\[1em] \Rightarrow x = 5.$

Hence, $\bm{x}$ = 5.

Solve :

$\dfrac{x-1}{2} - \dfrac{x-2}{3} - \dfrac{x-3}{4}$ = 0

Answer

Solving,

$\Rightarrow \dfrac{x - 1}{2} - \dfrac{x - 2}{3} - \dfrac{x - 3}{4} = 0 \\[1em] \Rightarrow \dfrac{6(x - 1) - 4(x - 2) - 3(x - 3)}{12} = 0 \\[1em] \Rightarrow \dfrac{6x - 6 - 4x + 8 - 3x + 9}{12} = 0 \\[1em] \Rightarrow \dfrac{-x + 11}{12} = 0 \\[1em] \Rightarrow -x + 11 = 0 \\[1em] \Rightarrow x = 11.$

Hence, $\bm{x}$ = 11.

Solve :

$\dfrac{x+1}{3} + \dfrac{x+4}{5} = \dfrac{x-4}{7}$

Answer

Given,

⇒ $\dfrac{x + 1}{3} + \dfrac{x + 4}{5} = \dfrac{x - 4}{7}$

L.C.M. of 3, 5 and 7 = 105.

Multiplying both sides by 105,

$\Rightarrow \left(\dfrac{x + 1}{3} + \dfrac{x + 4}{5}\right) \times 105 = \left(\dfrac{x - 4}{7}\right) \times 105 \\[1em] \Rightarrow 35(x + 1) + 21(x + 4) = 15(x - 4) \\[1em] \Rightarrow 35x + 35 + 21x + 84 = 15x - 60 \\[1em] \Rightarrow 56x + 119 = 15x - 60 \\[1em] \Rightarrow 56x - 15x = -60 - 119 \\[1em] \Rightarrow 41x = -179 \\[1em] \Rightarrow x = -\dfrac{179}{41} \\[1em] \Rightarrow x = -4\dfrac{15}{41}.$

Hence, $\bm{x}$ = $-4\dfrac{15}{41}$.

Solve :

$15 - 2(5 - 3x) = 4(x - 3) + 13$

Answer

Solving,

$\Rightarrow 15 - 2(5 - 3x) = 4(x - 3) + 13 \\[1em] \Rightarrow 15 - 10 + 6x = 4x - 12 + 13 \\[1em] \Rightarrow 5 + 6x = 4x + 1 \\[1em] \Rightarrow 6x - 4x = 1 - 5 \\[1em] \Rightarrow 2x = -4 \\[1em] \Rightarrow \dfrac{2x}{2} = \dfrac{-4}{2} \\[1em] \Rightarrow x = -2.$

Hence, $\bm{x}$ = −2.

Solve :

$\dfrac{2x+1}{3x-2} = 1\dfrac{1}{4}$

Answer

Solving,

$\Rightarrow \dfrac{2x + 1}{3x - 2} = 1\dfrac{1}{4} \\[1em] \Rightarrow \dfrac{2x + 1}{3x - 2} = \dfrac{5}{4} \\[1em] \Rightarrow 4(2x + 1) = 5(3x - 2) \\[1em] \Rightarrow 8x + 4 = 15x - 10 \\[1em] \Rightarrow 4 + 10 = 15x - 8x \\[1em] \Rightarrow 14 = 7x \\[1em] \Rightarrow x = 2.$

Hence, $\bm{x}$ = 2.

Solve :

$21 - 3(x - 7) = x + 20$

Answer

Solving,

$\Rightarrow 21 - 3(x - 7) = x + 20 \\[1em] \Rightarrow 21 - 3x + 21 = x + 20 \\[1em] \Rightarrow 42 - 3x = x + 20 \\[1em] \Rightarrow -3x - x = 20 - 42 \\[1em] \Rightarrow -4x = -22 \\[1em] \Rightarrow x = \dfrac{-22}{-4} \\[1em] \Rightarrow x = \dfrac{11}{2} \\[1em] \Rightarrow x = 5\dfrac{1}{2}.$

Hence, $\bm{x}$ = $5\dfrac{1}{2}$.

Solve :

$\dfrac{2(x-3)}{7} - \dfrac{x-4}{3}$ = 0

Answer

Solving,

$\Rightarrow \dfrac{2(x - 3)}{7} - \dfrac{x - 4}{3} = 0 \\[1em] \Rightarrow \dfrac{6(x - 3) - 7(x - 4)}{21} = 0 \\[1em] \Rightarrow \dfrac{6x - 18 - 7x + 28}{21} = 0 \\[1em] \Rightarrow \dfrac{-x + 10}{21} = 0 \\[1em] \Rightarrow -x + 10 = 0 \\[1em] \Rightarrow x = 10.$

Hence, $\bm{x}$ = 10.

Solve :

$\dfrac{2x-3}{3} - (x - 5) = \dfrac{x}{3}$

Answer

Solving,

$\Rightarrow \dfrac{2x - 3}{3} - (x - 5) = \dfrac{x}{3} \\[1em] \Rightarrow (2x - 3) - 3(x - 5) = x \\[1em] \Rightarrow 2x - 3 - 3x + 15 = x \\[1em] \Rightarrow -x + 12 = x \\[1em] \Rightarrow 12 = 2x \\[1em] \Rightarrow x = 6.$

Hence, $\bm{x}$ = 6.

Solve :

$\dfrac{x-4}{7} = \dfrac{x+3}{7} + \dfrac{x+4}{5}$

Answer

Given,

⇒ $\dfrac{x - 4}{7} = \dfrac{x + 3}{7} + \dfrac{x + 4}{5}$

L.C.M. of 7 and 5 = 35.

Multiplying both sides by 35,

$\Rightarrow \left(\dfrac{x - 4}{7}\right) \times 35 = \left(\dfrac{x + 3}{7} + \dfrac{x + 4}{5}\right) \times 35 \\[1em] \Rightarrow 5(x - 4) = 5(x + 3) + 7(x + 4) \\[1em] \Rightarrow 5x - 20 = 5x + 15 + 7x + 28 \\[1em] \Rightarrow 5x - 20 = 12x + 43 \\[1em] \Rightarrow 5x - 12x = 43 + 20 \\[1em] \Rightarrow -7x = 63 \\[1em] \Rightarrow x = -9.$

Hence, $\bm{x}$ = −9.

Solve :

$\dfrac{x-1}{5} - \dfrac{x}{3} = 1 - \dfrac{x-2}{2}$

Answer

Given,

⇒ $\dfrac{x - 1}{5} - \dfrac{x}{3} = 1 - \dfrac{x - 2}{2}$

L.C.M. of 5, 3 and 2 = 30.

Multiplying both sides by 30,

$\Rightarrow \left(\dfrac{x - 1}{5} - \dfrac{x}{3}\right) \times 30 = \left(1 - \dfrac{x - 2}{2}\right) \times 30 \\[1em] \Rightarrow 6(x - 1) - 10x = 30 - 15(x - 2) \\[1em] \Rightarrow 6x - 6 - 10x = 30 - 15x + 30 \\[1em] \Rightarrow -4x - 6 = 60 - 15x \\[1em] \Rightarrow -4x + 15x = 60 + 6 \\[1em] \Rightarrow 11x = 66 \\[1em] \Rightarrow x = 6.$

Hence, $\bm{x}$ = 6.

Solve :

$2x + 20$% of $x$ = 12·1

Answer

Solving,

$\Rightarrow 2x + 20$% of $x$ = 12.1

$\Rightarrow 2x + \dfrac{20}{100} \times x = 12.1$

$\Rightarrow 2x + \dfrac{x}{5} = 12.1$

$\Rightarrow \dfrac{10x + x}{5} = 12.1$

$\Rightarrow \dfrac{11x}{5} = 12.1$

$\Rightarrow 11x = 60.5$

$\Rightarrow x = \dfrac{60.5}{11} = \dfrac{605}{110}$

$\Rightarrow x = \dfrac{11}{2}$ = $5\dfrac{1}{2}$.

Hence, $\bm{x} = 5\dfrac{1}{2}$.