Linear Equations and Inequalities in One Variable — Exercise 12.1

Solve the following equation:

$5x - 3 = 3x - 5$

Answer

Solving,

$\Rightarrow 5x - 3 = 3x - 5 \\[1em] \Rightarrow 5x - 3x = - 5 + 3 \\[1em] \Rightarrow 2x = - 2 \\[1em] \Rightarrow x = - \dfrac{2}{2} \\[1em] \Rightarrow x = - 1$

Hence, $\bm{x = -1}$.

Solve the following equation:

$3x - 7 = 3(5 - x)$

Answer

Solving,

$\Rightarrow 3x - 7 = 3(5 - x) \\[1em] \Rightarrow 3x - 7 = 15 - 3x \\[1em] \Rightarrow 3x + 3x = 15 + 7 \\[1em] \Rightarrow 6x = 22 \\[1em] \Rightarrow x = \dfrac{22}{6} \\[1em] \Rightarrow x = \dfrac{11}{3}$

Hence, $\bm{x = \dfrac{11}{3}}$.

Solve the following equation:

$4(2x + 1) = 3(x - 1) + 7$

Answer

Solving,

$\Rightarrow 4(2x + 1) = 3(x - 1) + 7 \\[1em] \Rightarrow 8x + 4 = 3x - 3 + 7 \\[1em] \Rightarrow 8x + 4 = 3x + 4 \\[1em] \Rightarrow 8x - 3x = 4 - 4 \\[1em] \Rightarrow 5x = 0 \\[1em] \Rightarrow x = \dfrac{0}{5} \\[1em] \Rightarrow x = 0$

Hence, $\bm{x = 0}$.

Solve the following equation:

$3(2p - 1) = 5 - (3p - 2)$

Answer

Solving,

$\Rightarrow 3(2p - 1) = 5 - (3p - 2) \\[1em] \Rightarrow 6p - 3 = 5 - 3p + 2 \\[1em] \Rightarrow 6p - 3 = 7 - 3p \\[1em] \Rightarrow 6p + 3p = 7 + 3 \\[1em] \Rightarrow 9p = 10 \\[1em] \Rightarrow p = \dfrac{10}{9} \\[1em] \Rightarrow p = 1\dfrac{1}{9}$

Hence, $\bm{p = 1\dfrac{1}{9}}$.

Solve the following equation:

$5y - 2[y - 3(y - 5)] = 6$

Answer

Solving,

$\Rightarrow 5y - 2[y - 3(y - 5)] = 6 \\[1em] \Rightarrow 5y - 2[y - 3y + 15] = 6 \\[1em] \Rightarrow 5y - 2[- 2y + 15] = 6 \\[1em] \Rightarrow 5y + 4y - 30 = 6 \\[1em] \Rightarrow 9y = 6 + 30 \\[1em] \Rightarrow 9y = 36 \\[1em] \Rightarrow y = \dfrac{36}{9} \\[1em] \Rightarrow y = 4$

Hence, $\bm{y = 4}$.

Solve the following equation:

$0.3(6 - x) = 0.4(x + 8)$

Answer

Solving,

$\Rightarrow 0.3(6 - x) = 0.4(x + 8) \\[1em] \Rightarrow 1.8 - 0.3x = 0.4x + 3.2 \\[1em] \Rightarrow 1.8 - 3.2 = 0.4x + 0.3x \\[1em] \Rightarrow - 1.4 = 0.7x \\[1em] \Rightarrow x = - \dfrac{1.4}{0.7} \\[1em] \Rightarrow x = - 2$

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

Solve the following equation:

$\dfrac{x - 1}{3} = \dfrac{x + 2}{6} + 3$

Answer

Given,

$\dfrac{x - 1}{3} = \dfrac{x + 2}{6} + 3$

L.C.M. of 3 and 6 = 6

Multiplying both sides by 6, we get :

$\Rightarrow \dfrac{(x - 1)}{3} \times 6 = \Big[\dfrac{(x + 2)}{6} + 3\Big] \times 6 \\[1em] \Rightarrow 2(x - 1) = \dfrac{(x + 2)}{6} \times 6 + 3 \times 6 \\[1em] \Rightarrow 2(x - 1) = (x + 2) + 18 \\[1em] \Rightarrow 2x - 2 = x + 20 \\[1em] \Rightarrow 2x - x = 20 + 2 \\[1em] \Rightarrow x = 22$

Hence, $\bm{x = 22}$.

Solve the following equation:

$\dfrac{x + 7}{3} = 1 + \dfrac{3x - 2}{5}$

Answer

Given,

$\dfrac{x + 7}{3} = 1 + \dfrac{3x - 2}{5}$

L.C.M. of 3 and 5 = 15

Multiplying both sides by 15, we get:

$\Rightarrow \dfrac{(x + 7)}{3} \times 15 = \Big[1 + \dfrac{(3x - 2)}{5}\Big] \times 15 \\[1em] \Rightarrow 5(x + 7) = 1 \times 15 + \dfrac{(3x - 2)}{5} \times 15 \\[1em] \Rightarrow 5(x + 7) = 15 + 3(3x - 2) \\[1em] \Rightarrow 5x + 35 = 15 + 9x - 6 \\[1em] \Rightarrow 5x + 35 = 9x + 9 \\[1em] \Rightarrow 35 - 9 = 9x - 5x \\[1em] \Rightarrow 26 = 4x \\[1em] \Rightarrow x = \dfrac{26}{4} \\[1em] \Rightarrow x = \dfrac{13}{2} = 6\dfrac{1}{2}$

Hence, $\bm{x = 6\dfrac{1}{2}}$.

Solve the following equation:

$\dfrac{y + 1}{3} - \dfrac{y - 1}{2} = \dfrac{1 + 2y}{3}$

Answer

Given,

$\dfrac{y + 1}{3} - \dfrac{y - 1}{2} = \dfrac{1 + 2y}{3}$

L.C.M. of 3 and 2 = 6

Multiplying both sides by 6, we get:

$\Rightarrow \Big[\dfrac{(y + 1)}{3} - \dfrac{(y - 1)}{2}\Big] \times 6 = \dfrac{(1 + 2y)}{3} \times 6 \\[1em] \Rightarrow \dfrac{(y + 1)}{3} \times 6 - \dfrac{(y - 1)}{2} \times 6 = 2(1 + 2y) \\[1em] \Rightarrow 2(y + 1) - 3(y - 1) = 2(1 + 2y) \\[1em] \Rightarrow 2y + 2 - 3y + 3 = 2 + 4y \\[1em] \Rightarrow - y + 5 = 2 + 4y \\[1em] \Rightarrow 5 - 2 = 4y + y \\[1em] \Rightarrow 3 = 5y \\[1em] \Rightarrow y = \dfrac{3}{5}$

Hence, $\bm{y = \dfrac{3}{5}}$.

Solve the following equation:

$\dfrac{p}{3} + \dfrac{p}{4} = 55 - \dfrac{p + 40}{5}$

Answer

Given,

$\dfrac{p}{3} + \dfrac{p}{4} = 55 - \dfrac{p + 40}{5}$

L.C.M. of 3, 4 and 5 = 60

Multiplying both sides by 60, we get:

$\Rightarrow \Big[\dfrac{p}{3} + \dfrac{p}{4}\Big] \times 60 = \Big[55 - \dfrac{(p + 40)}{5}\Big] \times 60 \\[1em] \Rightarrow \dfrac{p}{3} \times 60 + \dfrac{p}{4} \times 60 = 55 \times 60 - \dfrac{(p + 40)}{5} \times 60 \\[1em] \Rightarrow 20p + 15p = 3300 - 12(p + 40) \\[1em] \Rightarrow 35p = 3300 - 12p - 480 \\[1em] \Rightarrow 35p = 2820 - 12p \\[1em] \Rightarrow 35p + 12p = 2820 \\[1em] \Rightarrow 47p = 2820 \\[1em] \Rightarrow p = \dfrac{2820}{47} \\[1em] \Rightarrow p = 60$

Hence, $\bm{p = 60}$.

Solve the following equation:

$n - \dfrac{n - 1}{2} = 1 - \dfrac{n - 2}{3}$

Answer

Given,

$n - \dfrac{n - 1}{2} = 1 - \dfrac{n - 2}{3}$

L.C.M. of 2 and 3 = 6

Multiplying both sides by 6, we get:

$\Rightarrow \Big[n - \dfrac{(n - 1)}{2}\Big] \times 6 = \Big[1 - \dfrac{(n - 2)}{3}\Big] \times 6 \\[1em] \Rightarrow n \times 6 - \dfrac{(n - 1)}{2} \times 6 = 1 \times 6 - \dfrac{(n - 2)}{3} \times 6 \\[1em] \Rightarrow 6n - 3(n - 1) = 6 - 2(n - 2) \\[1em] \Rightarrow 6n - 3n + 3 = 6 - 2n + 4 \\[1em] \Rightarrow 3n + 3 = 10 - 2n \\[1em] \Rightarrow 3n + 2n = 10 - 3 \\[1em] \Rightarrow 5n = 7 \\[1em] \Rightarrow n = \dfrac{7}{5} = 1\dfrac{2}{5}$

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

Solve the following equation:

$\dfrac{3t - 2}{3} + \dfrac{2t + 3}{2} = t + \dfrac{7}{6}$

Answer

Given,

$\dfrac{3t - 2}{3} + \dfrac{2t + 3}{2} = t + \dfrac{7}{6}$

L.C.M. of 3, 2 and 6 = 6

Multiplying both sides by 6, we get:

$\Rightarrow \Big[\dfrac{(3t - 2)}{3} + \dfrac{(2t + 3)}{2}\Big] \times 6 = \Big[t + \dfrac{7}{6}\Big] \times 6 \\[1em] \Rightarrow \dfrac{(3t - 2)}{3} \times 6 + \dfrac{(2t + 3)}{2} \times 6 = t \times 6 + \dfrac{7}{6} \times 6 \\[1em] \Rightarrow 2(3t - 2) + 3(2t + 3) = 6t + 7 \\[1em] \Rightarrow 6t - 4 + 6t + 9 = 6t + 7 \\[1em] \Rightarrow 12t + 5 = 6t + 7 \\[1em] \Rightarrow 12t - 6t = 7 - 5 \\[1em] \Rightarrow 6t = 2 \\[1em] \Rightarrow t = \dfrac{2}{6} \\[1em] \Rightarrow t = \dfrac{1}{3}$

Hence, $\bm{t = \dfrac{1}{3}}$.

Solve the following equation:

$4(3x + 2) - 5(6x - 1) = 2(x - 8) - 6(7x - 4)$

Answer

Solving,

$\Rightarrow 4(3x + 2) - 5(6x - 1) = 2(x - 8) - 6(7x - 4) \\[1em] \Rightarrow 12x + 8 - 30x + 5 = 2x - 16 - 42x + 24 \\[1em] \Rightarrow - 18x + 13 = - 40x + 8 \\[1em] \Rightarrow - 18x + 40x = 8 - 13 \\[1em] \Rightarrow 22x = - 5 \\[1em] \Rightarrow x = - \dfrac{5}{22}$

Hence, $\bm{x = -\dfrac{5}{22}}$.

Solve the following equation:

$3(5x + 7) + 5(2x - 11) = 3(8x - 5) - 15$

Answer

Solving,

$\Rightarrow 3(5x + 7) + 5(2x - 11) = 3(8x - 5) - 15 \\[1em] \Rightarrow 15x + 21 + 10x - 55 = 24x - 15 - 15 \\[1em] \Rightarrow 25x - 34 = 24x - 30 \\[1em] \Rightarrow 25x - 24x = - 30 + 34 \\[1em] \Rightarrow x = 4$

Hence, $\bm{x = 4}$.

Solve the following equation:

$\dfrac{3 - 2x}{2x + 5} = -\dfrac{3}{11}$

Answer

Given,

$\dfrac{3 - 2x}{2x + 5} = -\dfrac{3}{11}$

By cross multiplication, we get:

$\Rightarrow 11(3 - 2x) = - 3(2x + 5) \\[1em] \Rightarrow 33 - 22x = - 6x - 15 \\[1em] \Rightarrow 33 + 15 = - 6x + 22x \\[1em] \Rightarrow 48 = 16x \\[1em] \Rightarrow x = \dfrac{48}{16} \\[1em] \Rightarrow x = 3$

Hence, $\bm{x = 3}$.

Solve the following equation:

$\dfrac{5p + 2}{8 - 2p} = \dfrac{7}{6}$

Answer

Given,

$\dfrac{5p + 2}{8 - 2p} = \dfrac{7}{6}$

By cross multiplication, we get:

$\Rightarrow 6(5p + 2) = 7(8 - 2p) \\[1em] \Rightarrow 30p + 12 = 56 - 14p \\[1em] \Rightarrow 30p + 14p = 56 - 12 \\[1em] \Rightarrow 44p = 44 \\[1em] \Rightarrow p = \dfrac{44}{44} \\[1em] \Rightarrow p = 1$

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

Solve the following equation:

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

Answer

Given,

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

By cross multiplication, we get:

$\Rightarrow 5(x - 4) = 7x \\[1em] \Rightarrow 5x - 20 = 7x \\[1em] \Rightarrow - 20 = 7x - 5x \\[1em] \Rightarrow 2x = - 20 \\[1em] \Rightarrow x = - \dfrac{20}{2} \\[1em] \Rightarrow x = - 10$

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

Solve the following equation:

$\dfrac{4}{2x + 3} = \dfrac{5}{x + 4}$

Answer

Given,

$\dfrac{4}{2x + 3} = \dfrac{5}{x + 4}$

By cross multiplication, we get:

$\Rightarrow 4(x + 4) = 5(2x + 3) \\[1em] \Rightarrow 4x + 16 = 10x + 15 \\[1em] \Rightarrow 16 - 15 = 10x - 4x \\[1em] \Rightarrow 1 = 6x \\[1em] \Rightarrow x = \dfrac{1}{6}$

Hence, $\bm{x = \dfrac{1}{6}}$.

Solve the following equation:

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

Answer

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

L.C.M. of 2 and $(x - 1)$ = $2(x - 1)$

Multiplying both sides by $2(x - 1)$, we get:

$\Rightarrow \Big[\dfrac{(2x + 5)}{2} - \dfrac{5x}{(x - 1)}\Big] \times 2(x - 1) = x \times 2(x - 1) \\[1em] \Rightarrow \dfrac{(2x + 5)}{2} \times 2(x - 1) - \dfrac{5x}{(x - 1)} \times 2(x - 1) = 2x(x - 1) \\[1em] \Rightarrow (x - 1)(2x + 5) - 10x = 2x(x - 1) \\[1em] \Rightarrow 2x^{2} + 5x - 2x - 5 - 10x = 2x^{2} - 2x \\[1em] \Rightarrow 2x^{2} - 7x - 5 = 2x^{2} - 2x \\[1em] \Rightarrow - 7x + 2x = 5 \\[1em] \Rightarrow - 5x = 5 \\[1em] \Rightarrow x = - \dfrac{5}{5} \\[1em] \Rightarrow x = - 1$

Hence, $\bm{x = -1}$.

Solve the following equation:

$\dfrac{1}{5}\left(\dfrac{1}{3x} - 5\right) = \dfrac{1}{3}\left(3 - \dfrac{1}{x}\right)$

Answer

Solving,

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

By cross multiplication, we get:

$\Rightarrow 6 = 30x \\[1em] \Rightarrow x = \dfrac{6}{30} \\[1em] \Rightarrow x = \dfrac{1}{5}$

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

Solve the following equation:

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

Answer

Given,

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

By cross multiplication, we get:

$\Rightarrow (2x - 3)(3x + 1) = (3x - 1)(2x - 1) \\[1em] \Rightarrow 6x^{2} + 2x - 9x - 3 = 6x^{2} - 3x - 2x + 1 \\[1em] \Rightarrow 6x^{2} - 7x - 3 = 6x^{2} - 5x + 1 \\[1em] \Rightarrow - 7x + 5x = 1 + 3 \\[1em] \Rightarrow - 2x = 4 \\[1em] \Rightarrow x = - \dfrac{4}{2} \\[1em] \Rightarrow x = - 2$

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

Solve the following equation:

$\dfrac{2y + 3}{3y + 2} = \dfrac{4y + 5}{6y + 7}$

Answer

Given,

$\dfrac{2y + 3}{3y + 2} = \dfrac{4y + 5}{6y + 7}$

By cross multiplication, we get:

$\Rightarrow (2y + 3)(6y + 7) = (4y + 5)(3y + 2) \\[1em] \Rightarrow 12y^{2} + 14y + 18y + 21 = 12y^{2} + 8y + 15y + 10 \\[1em] \Rightarrow 12y^{2} + 32y + 21 = 12y^{2} + 23y + 10 \\[1em] \Rightarrow 32y - 23y = 10 - 21 \\[1em] \Rightarrow 9y = - 11 \\[1em] \Rightarrow y = - \dfrac{11}{9}$

Hence, $\bm{y = -\dfrac{11}{9}}$.

If $x = p + 1$, find the value of $p$ from the equation

$\dfrac{1}{2}(5x - 30) - \dfrac{1}{3}(1 + 7p) = \dfrac{1}{4}$

Answer

Given,

$x = p + 1$

Substituting the value of $x$ in the given equation, we get:

$\Rightarrow \dfrac{1}{2}[5(p + 1) - 30] - \dfrac{1}{3}(1 + 7p) = \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{1}{2}(5p + 5 - 30) - \dfrac{1}{3}(1 + 7p) = \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{1}{2}(5p - 25) - \dfrac{1}{3}(1 + 7p) = \dfrac{1}{4}$

L.C.M. of 2, 3 and 4 = 12

Multiplying both sides by 12, we get:

$\Rightarrow \Big[\dfrac{1}{2}(5p - 25) - \dfrac{1}{3}(1 + 7p)\Big] \times 12 = \dfrac{1}{4} \times 12 \\[1em] \Rightarrow \dfrac{1}{2}(5p - 25) \times 12 - \dfrac{1}{3}(1 + 7p) \times 12 = 3 \\[1em] \Rightarrow 6(5p - 25) - 4(1 + 7p) = 3 \\[1em] \Rightarrow 30p - 150 - 4 - 28p = 3 \\[1em] \Rightarrow 2p - 154 = 3 \\[1em] \Rightarrow 2p = 3 + 154 \\[1em] \Rightarrow 2p = 157 \\[1em] \Rightarrow p = \dfrac{157}{2} = 78\dfrac{1}{2}$

Hence, $\bm{p = 78\dfrac{1}{2}}$.

Solve $\dfrac{x + 3}{3} - \dfrac{x - 2}{2} = 1$. Hence find $p$, if $\dfrac{1}{x} + p = 1$

Answer

Given,

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

L.C.M. of 3 and 2 = 6

Multiplying both sides by 6, we get:

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

Substituting $x = 6$ in $\dfrac{1}{x} + p = 1$, we get:

$\Rightarrow \dfrac{1}{6} + p = 1 \\[1em] \Rightarrow p = 1 - \dfrac{1}{6} \\[1em] \Rightarrow p = \dfrac{6 - 1}{6} \\[1em] \Rightarrow p = \dfrac{5}{6}$

Hence, $\bm{x = 6}$ and $\bm{p = \dfrac{5}{6}}$.