Simultaneous Linear Equations — Exercise 5(A)

Solve the following simultaneous equations:

x + 2y = 1, 3x - y = 17

Answer

Given,

Equations : x + 2y = 1, 3x - y = 17

⇒ x + 2y = 1

⇒ x = 1 - 2y     ....(1)

Substituting value of x from equation (1) in 3x - y = 17, we get :

⇒ 3(1 - 2y) - y = 17

⇒ 3 - 6y - y = 17

⇒ -7y = 17 - 3

⇒ -7y = 14

⇒ y = $-\dfrac{14}{7}$ = -2.

Substituting value of y in equation (1), we get :

⇒ x = 1 - 2y

⇒ x = 1 - 2(-2)

⇒ x = 1 + 4

⇒ x = 5.

Hence, x = 5, y = -2.

Solve the following simultaneous equations:

5x + 4y = 4, x - 12y = 20

Answer

Given,

Equations : 5x + 4y = 4, x - 12y = 20

⇒ x - 12y = 20

⇒ x = 20 + 12y     ....(1)

Substituting value of x from equation (1) in 5x + 4y = 4, we get :

⇒ 5(20 + 12y) + 4y = 4

⇒ 100 + 60y + 4y = 4

⇒ 100 + 64y = 4

⇒ 64y = 4 - 100

⇒ 64y = -96

⇒ y = $-\dfrac{96}{64} = -\dfrac{3}{2}$.

Substituting value of y in equation (1), we get :

⇒ x = 20 + $12 \times -\Big(\dfrac{3}{2}\Big)$

⇒ x = 20 + 6(-3)

⇒ x = 20 - 18

⇒ x = 2.

Hence, x = 2, y = $-\dfrac{3}{2}$.

Solve the following simultaneous equations:

x + 2y + 9 = 0, 3x + 4y + 17 = 0

Answer

Given,

Equations : x + 2y + 9 = 0, 3x + 4y + 17 = 0

⇒ x + 2y + 9 = 0

⇒ x = -9 - 2y     ....(1)

Substituting value of x from equation (1) in 3x + 4y + 17 = 0, we get :

⇒ 3(-9 - 2y) + 4y + 17 = 0

⇒ -27 - 6y + 4y + 17 = 0

⇒ -10 - 2y = 0

⇒ -2y = 10

⇒ y = $-\dfrac{10}{2} = -5$.

Substituting value of y in equation (1), we get :

⇒ x = -9 - 2y

⇒ x = -9 - 2(-5)

⇒ x = -9 + 10

⇒ x = 1.

Hence, x = 1, y = -5.

Solve the following simultaneous equations:

10x + 3y = 75, 6x - 5y = 11

Answer

Given,

Equations : 10x + 3y = 75, 6x - 5y = 11

⇒ 10x + 3y = 75

⇒ 10x = 75 - 3y

⇒ x = $\dfrac{75 - 3y}{10}$     ....(1)

Substituting value of x from equation (1) in 6x - 5y = 11, we get :

$\Rightarrow 6\Big(\dfrac{75 - 3y}{10}\Big) - 5y = 11 \\[1em] \Rightarrow 3\Big(\dfrac{75 - 3y}{5}\Big) - 5y = 11 \\[1em] \Rightarrow \Big(\dfrac{225 - 9y}{5}\Big) - 5y = 11 \\[1em] \Rightarrow \Big(\dfrac{225 - 9y - 25y}{5}\Big) = 11 \\[1em] \Rightarrow \Big(\dfrac{225 - 34y}{5}\Big) = 11 \\[1em] \Rightarrow 225 - 34y = 11 \times 5\\[1em] \Rightarrow -34y = 55 - 225 \\[1em] \Rightarrow -34y = -170 \\[1em] \Rightarrow y = \dfrac{170}{34} \\[1em] \Rightarrow y = 5.$

Substituting value of y in equation (1), we get :

$\Rightarrow x = \dfrac{75 - 3y}{10} \\[1em] \Rightarrow x = \dfrac{75 - 3(5)}{10} \\[1em] \Rightarrow x = \dfrac{75 - 15}{10} \\[1em] \Rightarrow x = \dfrac{60}{10} \\[1em] \Rightarrow x = 6.$

Hence, x = 6, y = 5.

Solve the following simultaneous equations:

7x - 2y = 20, 11x + 15y + 23 = 0

Answer

Given,

Equations : 7x - 2y = 20, 11x + 15y + 23 = 0

⇒ 7x - 2y = 20

⇒ 7x = 20 + 2y

⇒ x = $\dfrac{20 + 2y}{7}$     ....(1)

Substituting value of x from equation (1) in 11x + 15y + 23 = 0, we get :

$\Rightarrow 11\Big(\dfrac{20 + 2y}{7}\Big) + 15y + 23 = 0 \\[1em] \Rightarrow \Big(\dfrac{220 + 22y}{7}\Big) + 15y + 23 = 0 \\[1em] \Rightarrow \Big(\dfrac{220 + 22y + 105y + 161}{7}\Big) = 0 \\[1em] \Rightarrow 381 + 127y = 0 \\[1em] \Rightarrow 127y = - 381 \\[1em] \Rightarrow y = \dfrac{-381}{127}\\[1em] \Rightarrow y = -3.$

Substituting value of y in equation (1), we get :

$\Rightarrow x = \dfrac{20 + 2y}{7} \\[1em] \Rightarrow x = \dfrac{20 + 2(-3)}{7} \\[1em] \Rightarrow x = \dfrac{20 - 6}{7} \\[1em] \Rightarrow x = \dfrac{14}{7} \\[1em] \Rightarrow x = 2.$

Hence, x = 2, y = -3.

Solve the following simultaneous equations:

$\dfrac{7 - 4x}{3} = y$, 2x + 3y + 1 = 0

Answer

Given,

Equations : $\dfrac{7 - 4x}{3} = y$, 2x + 3y + 1 = 0

⇒ $y = \dfrac{7 - 4x}{3}$     ....(1)

Substituting value of y from equation (1) in 2x + 3y + 1 = 0, we get :

$\Rightarrow 2x + 3\Big(\dfrac{7 - 4x}{3}\Big) + 1 = 0$

⇒ 2x + (7 - 4x) + 1 = 0

⇒ 8 - 2x = 0

⇒ 2x = 8

⇒ x = $\dfrac{8}{2}$

⇒ x = 4.

Substituting value of x in equation (1), we get :

$\Rightarrow y = \dfrac{7 - 4x}{3} \\[1em] \Rightarrow y = \dfrac{7 - 4(4)}{3} \\[1em] \Rightarrow y = \dfrac{7 - 16}{3} \\[1em] \Rightarrow y = -\dfrac{9}{3} \\[1em] \Rightarrow y = -3.$ Hence, x = 4, y = -3.

Solve the following simultaneous equations:

4x - 3y = 8, 18x - 3y = 29

Answer

Given,

Equations : 4x - 3y = 8, 18x - 3y = 29

⇒ 4x - 3y = 8

⇒ 4x = 3y + 8

⇒ x = $\dfrac{3y + 8}{4}$     ....(1)

Substituting value of x from equation (1) in 18x - 3y = 29, we get :

$\Rightarrow 18\Big(\dfrac{3y + 8}{4}\Big) - 3y = 29 \\[1em] \Rightarrow 9\Big(\dfrac{3y + 8}{2}\Big) - 3y = 29 \\[1em] \Rightarrow \Big(\dfrac{27y + 72}{2}\Big) - 3y = 29 \\[1em] \Rightarrow \Big(\dfrac{27y + 72 - 6y}{2}\Big) = 29 \\[1em] \Rightarrow 21y + 72 = 29 \times 2 \\[1em] \Rightarrow 21y + 72 = 58 \\[1em] \Rightarrow 21y = 58 - 72 \\[1em] \Rightarrow 21y = -14 \\[1em] \Rightarrow y = \dfrac{-14}{21} \\[1em] \Rightarrow y = -\dfrac{2}{3}.$

Substituting value of y in equation (1), we get :

$\Rightarrow x = \dfrac{3y + 8}{4} \\[1em] \Rightarrow x = \dfrac{3 \Big(\dfrac{-2}{3}\Big) + 8}{4} \\[1em] \Rightarrow x = \dfrac{-2 + 8}{4} \\[1em] \Rightarrow x = \dfrac{6}{4} \\[1em] \Rightarrow x = \dfrac{3}{2}.$

Hence, $x = \dfrac{3}{2}, y = -\dfrac{2}{3}$.

Solve the following simultaneous equations:

$\dfrac{x + y - 8}{2} = \dfrac{x + 2y - 14}{3} = \dfrac{3x + y - 12}{11}$

Answer

Given,

$\dfrac{x + y - 8}{2} = \dfrac{x + 2y - 14}{3} = \dfrac{3x + y - 12}{11}$

Solving L.H.S. of the given equation,

$\Rightarrow \dfrac{x + y - 8}{2} = \dfrac{x + 2y - 14}{3} \\[1em] \Rightarrow 3(x + y - 8) = 2(x + 2y - 14) \\[1em] \Rightarrow 3x + 3y - 24 = 2x + 4y - 28 \\[1em] \Rightarrow 3x + 3y - 2x - 4y = -28 + 24 \\[1em] \Rightarrow x - y = -4 \\[1em] \Rightarrow x = -4 + y \text{ ....(1)}$

Solving R.H.S. of the given equation,

$\Rightarrow \dfrac{x + 2y - 14}{3} = \dfrac{3x + y - 12}{11} \\[1em] \Rightarrow 11(x + 2y - 14) = 3(3x + y - 12) \\[1em] \Rightarrow 11x + 22y - 154 = 9x + 3y - 36 \\[1em] \Rightarrow 11x + 22y - 9x - 3y = -36 + 154 \\[1em] \Rightarrow 2x + 19y = 118 \text{ ........(2)}$

Substituting value of x from equation (1) in (2), we get :

⇒ 2(-4 + y) + 19y = 118

⇒ -8 + 2y + 19y = 118

⇒ 21y = 118 + 8

⇒ 21y = 126

⇒ y = $\dfrac{126}{21} = 6$.

Substituting value of y in equation (1), we get :

⇒ x = -4 + y

⇒ x = -4 + 6

⇒ x = 2.

Hence, x = 2, y = 6.

Solve the following simultaneous equations:

$\dfrac{x}{7} + \dfrac{y}{3} = 5, \dfrac{x}{2} - \dfrac{y}{9} = 6$

Answer

Given,

$\dfrac{x}{7} + \dfrac{y}{3} = 5, \dfrac{x}{2} - \dfrac{y}{9} = 6$

Simplifying,

$\Rightarrow \dfrac{x}{7} + \dfrac{y}{3} = 5 \\[1em] \Rightarrow \dfrac{3x + 7y}{21} = 5 \\[1em] \Rightarrow 3x + 7y = 5 \times 21 \\[1em] \Rightarrow 3x + 7y = 105 \\[1em] \Rightarrow 3x = 105 - 7y \\[1em] \Rightarrow x = \dfrac{105 - 7y}{3} \text{ ....(1)}$

Substituting value of x from equation (1) in $\dfrac{x}{2} - \dfrac{y}{9} = 6$, we get :

$\Rightarrow \dfrac{\dfrac{105 - 7y}{3}}{2} - \dfrac{y}{9} = 6 \\[1em] \Rightarrow \dfrac{(105 - 7y)}{6} - \dfrac{y}{9} = 6 \\[1em] \Rightarrow \dfrac{3(105 - 7y) - 2y}{18} = 6 \\[1em] \Rightarrow \dfrac{315 - 21y - 2y}{18} = 6 \\[1em] \Rightarrow 315 - 23y = 6 \times 18 \\[1em] \Rightarrow 315 - 23y = 108 \\[1em] \Rightarrow 23y = 315 - 108 \\[1em] \Rightarrow 23y = 207 \\[1em] \Rightarrow y = \dfrac{207}{23} = 9.$

Substituting value of y in equation (1), we get :

$\Rightarrow x = \dfrac{105 - 7y}{3} \\[1em] \Rightarrow x = \dfrac{105 - 7(9)}{3} \\[1em] \Rightarrow x = \dfrac{105 - 63}{3} \\[1em] \Rightarrow x = \dfrac{42}{3} \\[1em] \Rightarrow x = 14.$

Hence, x = 14, y = 9.

Solve the following simultaneous equations:

$\dfrac{x}{6} + 6 = y, \dfrac{3x}{4} = 1 + y$

Answer

Simplifying, equation : $\dfrac{x}{6} + 6 = y$

$\Rightarrow \dfrac{x}{6} + 6 = y \\[1em] \Rightarrow \dfrac{x + 36}{6} = y \\[1em] \Rightarrow x + 36 = 6y \\[1em] \Rightarrow x = 6y - 36 \text{ ....(1)}$

Substituting value of x from equation (1) in $\dfrac{3x}{4} = 1 + y$, we get :

$\Rightarrow \dfrac{3(6y - 36)}{4} = 1 + y \\[1em] \Rightarrow \dfrac{18y - 108}{4} = 1 + y \\[1em] \Rightarrow 18y - 108 = 4(1 + y) \\[1em] \Rightarrow 18y - 108 = (4 + 4y) \\[1em] \Rightarrow 18y - 4y = 4 + 108 \\[1em] \Rightarrow 14y = 112 \\[1em] \Rightarrow y = \dfrac{112}{14} \\[1em] \Rightarrow y = 8.$

Substituting value of y in equation (1), we get :

⇒ x = 6y - 36

⇒ x = 6(8) - 36

⇒ x = 48 - 36

⇒ x = 12.

Hence, x = 12, y = 8.

Solve the following simultaneous equations:

$4x + \dfrac{x - y}{8} = 17, x + 2y = \dfrac{y - 2}{3} - 2$

Answer

Simplifying, equation : $4x + \dfrac{x - y}{8} = 17$

$\Rightarrow 4x + \dfrac{x - y}{8} = 17 \\[1em] \Rightarrow \dfrac{32x + x - y}{8} = 17 \\[1em] \Rightarrow (33x - y) = 17 \times 8 \\[1em] \Rightarrow (33x - y) = 136 \\[1em] \Rightarrow 33x - y = 136 \\[1em] \Rightarrow y = 33x - 136 \text{ ....(1)}$

Substituting value of y from equation (1) in $x + 2y = \dfrac{y - 2}{3} - 2$, we get :

$\Rightarrow x + 2y = \dfrac{y - 2}{3} - 2 \\[1em] \Rightarrow x + 2(33x - 136) = \dfrac{33x - 136 - 2}{3} - 2 \\[1em] \Rightarrow x + 66x - 272 = \dfrac{33x -138}{3} - 2 \\[1em] \Rightarrow 67x - 272 = \dfrac{3(11x - 46)}{3} - 2 \\[1em] \Rightarrow 67x - 272 = (11x - 46) - 2 \\[1em] \Rightarrow 67x - 272 = (11x - 48) \\[1em] \Rightarrow 67x - 11x = -48 + 272 \\[1em] \Rightarrow 56x = 224 \\[1em] \Rightarrow x = \dfrac{224}{56} = 4.$

Substituting value of y in equation (1), we get :

⇒ y = 33(4) - 136

⇒ y = 132 - 136

⇒ y = -4.

Hence, x = 4, y = -4.

Solve the following simultaneous equations:

$\dfrac{x}{2} + y = \dfrac{4}{5}, x + \dfrac{y}{2} = \dfrac{7}{10}$

Answer

Simplifying, equation : $\dfrac{x}{2} + y = \dfrac{4}{5}$

$\Rightarrow \dfrac{x}{2} + y = \dfrac{4}{5} \\[1em] \Rightarrow \dfrac{x + 2y}{2} = \dfrac{4}{5} \\[1em] \Rightarrow 5(x + 2y) = 4 \times 2 \\[1em] \Rightarrow 5x + 10y = 8 \\[1em] \Rightarrow 10y = 8 - 5x \\[1em] \Rightarrow y = \dfrac{8 - 5x}{10} \text{ ....(1)}$

Substituting value of y from equation (1) in $x + \dfrac{y}{2} = \dfrac{7}{10}$, we get :

$\Rightarrow x + \dfrac{\dfrac{8 - 5x}{10}}{2} = \dfrac{7}{10} \\[1em] \Rightarrow x + \dfrac{8 - 5x}{20} = \dfrac{7}{10} \\[1em] \Rightarrow \dfrac{20x + 8 - 5x}{20} = \dfrac{7}{10} \\[1em] \Rightarrow 15x + 8 = \dfrac{7}{10} \times 20 \\[1em] \Rightarrow 15x + 8 = 7 \times 2 \\[1em] \Rightarrow 15x = 14 - 8 \\[1em] \Rightarrow 15x = 6 \\[1em] \Rightarrow x = \dfrac{6}{15} = \dfrac{2}{5}.$

Substituting value of x in equation (1), we get :

$\Rightarrow y = \dfrac{8 - 5x}{10} \\[1em] \Rightarrow y = \dfrac{8 - 5 \Big(\dfrac{2}{5}\Big)}{10} \\[1em] \Rightarrow y = \dfrac{8 - 2}{10} \\[1em] \Rightarrow y = \dfrac{6}{10} = \dfrac{3}{5}.$

Hence, $x = \dfrac{2}{5}, y = \dfrac{3}{5}$.

Solve the following simultaneous equations:

$\dfrac{7 + x}{5} - \dfrac{2x - y}{4} = 3y - 5, \dfrac{4x - 3}{6} + \dfrac{5y - 7}{2} = 18 - 5x$

Answer

Simplifying equation : $\dfrac{7 + x}{5} - \dfrac{2x - y}{4} = 3y - 5$

$\Rightarrow \dfrac{7 + x}{5} - \dfrac{2x - y}{4} = 3y - 5 \\[1em] \Rightarrow \dfrac{4(7 + x) - 5(2x - y)}{20} = 3y - 5 \\[1em] \Rightarrow \dfrac{28 + 4x - 10x + 5y}{20} = 3y - 5 \\[1em] \Rightarrow 28 + 4x - 10x + 5y = 20(3y - 5) \\[1em] \Rightarrow 28 - 6x + 5y = 60y - 100 \\[1em] \Rightarrow 60y - 5y + 6x = 28 + 100 \\[1em] \Rightarrow 6x + 55y = 128 \\[1em] \Rightarrow 6x = 128 - 55y \\[1em] \Rightarrow x = \dfrac{128 - 55y}{6} \text{ ....(1)}$

Simplifying equation : $\dfrac{4x - 3}{6} + \dfrac{5y - 7}{2} = 18 - 5x$

$\Rightarrow \dfrac{4x - 3}{6} + \dfrac{5y - 7}{2} = 18 - 5x \\[1em] \Rightarrow \dfrac{2(4x - 3) + 6(5y - 7)}{12} = 18 - 5x \\[1em] \Rightarrow \dfrac{8x - 6 + 30y - 42}{12} = 18 - 5x \\[1em] \Rightarrow 8x - 6 + 30y - 42 = 12(18 - 5x) \\[1em] \Rightarrow 8x + 30y - 48 = 216 - 60x \\[1em] \Rightarrow 8x + 30y + 60x = 216 + 48 \\[1em] \Rightarrow 68x + 30y = 264 \text{ ....(2) }$

Substituting value of x from equation (1) in 68x + 30y = 264, we get :

$\Rightarrow 68\Big(\dfrac{128 - 55y}{6}\Big) + 30y = 264 \\[1em] \Rightarrow 34\Big(\dfrac{128 - 55y}{3}\Big) + 30y = 264 \\[1em] \Rightarrow \Big(\dfrac{4352 - 1870y}{3}\Big) + 30y = 264 \\[1em] \Rightarrow \Big(\dfrac{4352 - 1870y + 90y}{3}\Big) = 264 \\[1em] \Rightarrow (4352 - 1780y) = 264 \times 3 \\[1em] \Rightarrow (4352 - 1780y) = 792 \\[1em] \Rightarrow 1780y = 4352 - 792 \\[1em] \Rightarrow 1780y = 3560 \\[1em] \Rightarrow y = \dfrac{3560}{1780} = 2.$

Substituting value of y in equation (1), we get :

$\Rightarrow x = \dfrac{128 - 55y}{6} \\[1em] \Rightarrow x = \dfrac{128 - 55(2)}{6} \\[1em] \Rightarrow x = \dfrac{128 - 110}{6} \\[1em] \Rightarrow x = \dfrac{18}{6} \\[1em] \Rightarrow x = 3.$

Hence, x = 3, y = 2.

Solve the following simultaneous equations:

$4x + \dfrac{6}{y} = 15, 3x - \dfrac{4}{y} = 7$

Answer

Given,

Equations:

$\Rightarrow 4x + \dfrac{6}{y} = 15 \text{ ....(1)} \\[1em] \Rightarrow 3x - \dfrac{4}{y} = 7 \text{ ....(2)}$

Multiplying equation (1) by 4, we get :

$\Rightarrow 4\Big(4x + \dfrac{6}{y}\Big) = 15 \times 4 \\[1em] \Rightarrow 4\Big(4x + \dfrac{6}{y}\Big) = 15 \times 4 \\[1em] \Rightarrow 16x + \dfrac{24}{y} = 60\text{ ....(3)}$

Multiplying equation (2) by 6, we get :

$\Rightarrow 6\Big(3x - \dfrac{4}{y}\Big) = 7 \times 6 \\[1em] \Rightarrow 18x - \dfrac{24}{y} = 42\text{ ....(4)}$

Adding equation (3) and (4), we get:

$\Rightarrow \Big(16x + \dfrac{24}{y}\Big) + \Big(18x - \dfrac{24}{y}\Big) = 60 + 42 \\[1em] \Rightarrow 18x + \dfrac{24}{y} + 16x - \dfrac{24}{y} = 102 \\[1em] \Rightarrow 34x = 102 \\[1em] \Rightarrow x = \dfrac{102}{34} \\[1em] \Rightarrow x = 3.$

Substituting value of x in equation (1), we get :

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

Hence, x = 3, y = 2.

Solve the following simultaneous equations:

$5x - 9 = \dfrac{1}{y}, x + \dfrac{1}{y} = 3$

Answer

Equations:

$\Rightarrow 5x - 9 = \dfrac{1}{y} \text{ ....(1)} \\[1em] \Rightarrow x + \dfrac{1}{y} = 3 \\[1em] \Rightarrow \dfrac{1}{y} = 3 - x \text{ ....(2)}$

From equation (1) and (2), we get :

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

Substituting value of x in equation (2), we get :

$\Rightarrow \dfrac{1}{y} = 3 - 2 \\[1em] \Rightarrow \dfrac{1}{y} = 1 \\[1em] \Rightarrow y = \dfrac{1}{1} \\[1em] \Rightarrow y = 1.$

Hence, x = 2, y = 1.

Solve the following simultaneous equations:

$\dfrac{2}{x} + \dfrac{2}{3y} = \dfrac{1}{6},;\dfrac{3}{x} + \dfrac{2}{y} = 0$

Answer

Given,

Equations:

$\dfrac{2}{x} + \dfrac{2}{3y} = \dfrac{1}{6} \text{ ....(1) }$,

$\dfrac{3}{x} + \dfrac{2}{y} = 0 \text{ ....(2) }$

Multiplying equation (1) by 3, we get:

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

Multiplying equation (2) by 2, we get:

$\Rightarrow 2\Big(\dfrac{3}{x} + \dfrac{2}{y} \Big) = 0 \times 2 \\[1em] \Rightarrow \Big(\dfrac{6}{x} + \dfrac{4}{y}\Big) = 0 \text{ ....(4) }$

Subtracting equation (3) from (4), we get:

$\Rightarrow \Big(\dfrac{6}{x} + \dfrac{4}{y}\Big) - \Big(\dfrac{6}{x} + \dfrac{2}{y}\Big) = 0 - \dfrac{1}{2} \\[1em] \Rightarrow \Big(\dfrac{6}{x} + \dfrac{4}{y} - \dfrac{6}{x} - \dfrac{2}{y}\Big) = -\dfrac{1}{2} \\[1em] \Rightarrow \Big(\dfrac{4}{y} - \dfrac{2}{y}\Big) = -\dfrac{1}{2} \\[1em] \Rightarrow \Big(\dfrac{2}{y}\Big) = -\dfrac{1}{2} \\[1em] \Rightarrow y = \dfrac{2 \times 2}{-1} \\[1em] \Rightarrow y = -4.$

Substituting value of y in equation (1),

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

Hence, x = 6, y = -4.

Solve the following simultaneous equations:

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

Answer

Given,

Equations:

$\dfrac{3}{2x} + \dfrac{2}{3y} = 5 \text{ ....(1) },$

$\dfrac{5}{x} - \dfrac{3}{y} = 1 \text{ ....(2) }$

Multiplying equation (1) by 9, we get:

$\Rightarrow 9\Big(\dfrac{3}{2x} + \dfrac{2}{3y}\Big) = 5 \times 9 \\[1em] \Rightarrow \Big(\dfrac{27}{2x} + \dfrac{6}{y}\Big) = 45 \text{ ..........(3) }$

Multiplying equation (2) by 2, we get:

$\Rightarrow 2\Big(\dfrac{5}{x} - \dfrac{3}{y}\Big) = 1 \times 2 \\[1em] \Rightarrow \Big(\dfrac{10}{x} - \dfrac{6}{y}\Big) = 2 \text{ .........(4) }$

Adding equation (3) from (4), we get:

$\Rightarrow \Big(\dfrac{27}{2x} + \dfrac{6}{y}\Big) + \Big(\dfrac{10}{x} - \dfrac{6}{y}\Big) = 45 + 2 \\[1em] \Rightarrow \Big(\dfrac{27}{2x} + \dfrac{10}{x}\Big) = 47 \\[1em] \Rightarrow \Big(\dfrac{27 + 20}{2x}\Big) = 47 \\[1em] \Rightarrow \Big(\dfrac{47}{2x}\Big) = 47 \\[1em] \Rightarrow 2x = \dfrac{47}{47} \\[1em] \Rightarrow 2x = 1 \\[1em] \Rightarrow x = \dfrac{1}{2}.$

Substituting value of x in equation (1),

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

Hence, $x = \dfrac{1}{2}, y = \dfrac{1}{3}$.

Solve the following simultaneous equations:

x + y = 2xy, x − y = 6xy

Answer

Given,

Equations :

⇒ x + y = 2xy     ....(1)

⇒ x - y = 6xy     ....(2)

Adding equations (1) and (2), we get :

⇒ x + y + (x - y) = 2xy + 6xy

⇒ x + y + x - y = 8xy

⇒ 2x = 8xy

⇒ 2 = 8y

⇒ y = $\dfrac{2}{8} = \dfrac{1}{4}$.

Substituting $y = \dfrac{1}{4}$ in equation (1), we get :

$\Rightarrow x + y = 2xy \\[1em] \Rightarrow x + \dfrac{1}{4} = 2 \times x \times \Big(\dfrac{1}{4}\Big) \\[1em] \Rightarrow x + \dfrac{1}{4} = \dfrac{x}{2} \\[1em] \Rightarrow x - \dfrac{x}{2} = \dfrac{-1}{4} \\[1em] \Rightarrow \dfrac{x}{2} = \dfrac{-1}{4} \\[1em] \Rightarrow x = -\dfrac{2}{4} \\[1em] \Rightarrow x = -\dfrac{1}{2}.$

Hence, $x = -\dfrac{1}{2}, y = \dfrac{1}{4}$.

Solve the following simultaneous equations:

$\dfrac{3}{x + y} + \dfrac{2}{x - y} = 3,;\dfrac{2}{x + y} + \dfrac{3}{x - y} = \dfrac{11}{3}$

Answer

Given,

Equations:

$\dfrac{3}{x + y} + \dfrac{2}{x - y} = 3$ ........(1)

$\dfrac{2}{x + y} + \dfrac{3}{x - y} = \dfrac{11}{3}$ ..........(2)

Multiplying equation (1) by 2, we get:

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

Multiplying equation (2) by 3, we get:

$\Rightarrow 3\Big(\dfrac{2}{x + y} + \dfrac{3}{x - y}\Big) = \dfrac{11}{3} \times 3 \\[1em] \Rightarrow \Big(\dfrac{6}{x + y} + \dfrac{9}{x - y}\Big) = 11 \text{ .........(4) }$

Subtracting equation (3) from (4), we get:

$\Rightarrow \Big(\dfrac{6}{x + y} + \dfrac{9}{x - y}\Big) - \Big(\dfrac{6}{x + y} + \dfrac{4}{x - y}\Big) = 11 - 6 \\[1em] \Rightarrow \Big(\dfrac{6}{x + y} + \dfrac{9}{x - y} - \dfrac{6}{x + y} - \dfrac{4}{x - y}\Big) = 5 \\[1em] \Rightarrow \Big(\dfrac{9}{x - y} - \dfrac{4}{x - y}\Big) = 5 \\[1em] \Rightarrow \Big(\dfrac{5}{x - y}\Big) = 5 \\[1em] \Rightarrow 5 = 5(x - y) \\[1em] \Rightarrow x - y = 1 \text{ .........(5) }$

Substituting value of x - y from equation (5) in equation (1), we get:

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

Adding equations (5) and (6), we get:

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

Substituting value of x in equation (6), we get:

⇒ x + y = 3

⇒ 2 + y = 3

⇒ y = 3 - 2

⇒ y = 1.

Hence, x = 2, y = 1.

Solve the following simultaneous equations:

$\dfrac{22}{x + y} + \dfrac{15}{x - y} = 5,\dfrac{55}{x + y} + \dfrac{40}{x - y} = 13$

Answer

Given,

Equations:

$\dfrac{22}{x + y} + \dfrac{15}{x - y} = 5$ ..........(1)

$\dfrac{55}{x + y} + \dfrac{40}{x - y} = 13$ ..........(2)

Multiplying equation (1) by 5, we get:

$\Rightarrow 5\Big(\dfrac{22}{x + y} + \dfrac{15}{x - y}\Big) = 5 \times 5 \\[1em] \Rightarrow \Big(\dfrac{110}{x + y} + \dfrac{75}{x - y}\Big) = 25 \text{ ..........(3) }$

Multiplying equation (2) by 2, we get:

$\Rightarrow 2\Big(\dfrac{55}{x + y} + \dfrac{40}{x - y}\Big) = 13 \times 2 \\[1em] \Rightarrow \Big(\dfrac{110}{x + y} + \dfrac{80}{x - y}\Big) = 26 \text{ ..........(4) }$

Subtracting equation (3) from (4), we get:

$\Rightarrow \Big(\dfrac{110}{x + y} + \dfrac{80}{x - y}\Big) - \Big(\dfrac{110}{x + y} + \dfrac{75}{x - y}\Big) = 26 - 25 \\[1em] \Rightarrow \Big(\dfrac{110}{x + y} + \dfrac{80}{x - y} - \dfrac{110}{x + y} - \dfrac{75}{x - y}\Big) = 1 \\[1em] \Rightarrow \Big(\dfrac{80 - 75}{x - y}\Big) = 1 \\[1em] \Rightarrow \Big(\dfrac{5}{x - y}\Big) = 1 \\[1em] \Rightarrow 5 = (x - y) \\[1em] \Rightarrow x - y = 5 \text{ .........(5) }$

Substituting value of x - y from equation (5) in equation (1), we get:

$\Rightarrow \dfrac{22}{x + y} + \dfrac{15}{x - y} = 5 \\[1em] \Rightarrow \dfrac{22}{x + y} + \dfrac{15}{5} = 5 \\[1em] \Rightarrow \dfrac{22}{x + y} + 3 = 5 \\[1em] \Rightarrow \dfrac{22}{x + y} = 5 - 3 \\[1em] \Rightarrow \dfrac{22}{x + y} = 2 \\[1em] \Rightarrow x + y = \dfrac{22}{2} \\[1em] \Rightarrow x + y = 11 \text{ .......(6) }$

Adding equations (5) and (6), we get:

$\Rightarrow x + y + x - y = 11 + 5 \\[1em] \Rightarrow 2x = 16 \\[1em] \Rightarrow x = \dfrac{16}{2} = 8.$

Substituting value of x in equation (6), we get:

⇒ x + y = 11

⇒ 8 + y = 11

⇒ y = 11 - 8

⇒ y = 3

Hence, x = 8, y = 3.

Solve the following simultaneous equations:

103x + 51y = 617, 97x + 49y = 583

Answer

Given,

Equations :

103x + 51y = 617     .......(1),

97x + 49y = 583     .......(2)

Adding equations (1) and (2),

⇒ 103x + 51y + (97x + 49y) = 617 + 583

⇒ 103x + 51y + 97x + 49y = 617 + 583

⇒ 200x + 100y = 1200

⇒ 100(2x + y) = 1200

⇒ 2x + y = $\dfrac{1200}{100}$

⇒ 2x + y = 12     ........(3)

Subtracting equation (2) from (1),

⇒ 103x + 51y - (97x + 49y) = 617 - 583

⇒ 103x + 51y - 97x - 49y = 34

⇒ 6x + 2y = 34

⇒ 2(3x + y) = 34

⇒ 3x + y = $\dfrac{34}{2}$

⇒ 3x + y = 17     ........(4)

Subtracting equation 4 from 3,

⇒ 2x + y - (3x + y) = 12 - 17

⇒ 2x + y - 3x - y = -5

⇒ -x = -5

⇒ x = 5.

Substituting value of x in equation 1,

⇒ 103x + 51y = 617

⇒ 103(5) + 51y = 617

⇒ 515 + 51y = 617

⇒ 51y = 617 - 515

⇒ 51y = 102

⇒ y = $\dfrac{102}{51}$ = 2.

Hence, x = 5, y = 2.

Solve the following simultaneous equations:

23x − 29y = 98, 29x − 23y = 110

Answer

Given,

Equations :

23x − 29y = 98     .........(1),

29x − 23y = 110     ........(2)

Adding equations 1 and 2,

⇒ 23x − 29y + (29x − 23y) = 98 + 110

⇒ 23x − 29y + 29x − 23y = 98 + 110

⇒ 52x − 52y = 208

⇒ 52(x - y) = 208

⇒ (x - y) = $\dfrac{208}{52}$

⇒ x - y = 4     .........(3)

Subtracting equation 2 from 1,

⇒ 23x − 29y - (29x − 23y) = 98 - 110

⇒ 23x − 29y - 29x + 23y = 98 - 110

⇒ -6x - 6y = -12

⇒ -6(x + y) = -12

⇒ x + y = $\dfrac{-12}{-6}$

⇒ x + y = 2     .......(4)

Adding equation 4 and 5,

⇒ x + y + x - y = 2 + 4

⇒ 2x = 6

⇒ $x = \dfrac{6}{2}$

⇒ x = 3.

Substituting value of x in equation 1,

⇒ 23x − 29y = 98

⇒ 23(3) − 29y = 98

⇒ 69 − 29y = 98

⇒ −29y = 98 − 69

⇒ −29y = 29

⇒ y = $\dfrac{29}{-29}$ = -1.

Hence, x = 3, y = -1.

Solve the following simultaneous equations:

$\dfrac{a}{x} - \dfrac{b}{y} = 0, \dfrac{ab^2}{x} + \dfrac{a^2 b}{y} = (a^2 + b^2)$

Answer

Substituting $\dfrac{1}{x} = m \text{ and } \dfrac{1}{y} = n$ in $\dfrac{a}{x} - \dfrac{b}{y} = 0$,

⇒ am - bn = 0

⇒ am = bn

⇒ m = $\dfrac{bn}{a}$     ........(1)

Substituting $\dfrac{1}{x} = m \text{ and } \dfrac{1}{y} = n$ in $\dfrac{ab^2}{x} + \dfrac{a^2b}{y} = (a^2 + b^2)$

⇒ ab²m + a²bn = a² + b²     .........(2)

Substituting value of m from equation (1) in (2), we get :

⇒ ab² $\Big(\dfrac{bn}{a}\Big)$ + a²bn = a² + b²

⇒ b³n + a²bn = a² + b²

⇒ bn(b² + a²) = a² + b²

⇒ bn = $\dfrac{(a^2 + b^2)}{(a^2 + b^2)}$

⇒ bn = 1

⇒ n = $\dfrac{1}{b}$

Substituting value of n in equation (1), we get:

⇒ m = $\dfrac{b}{a} \times \dfrac{1}{b}$

⇒ m = $\dfrac{1}{a}$

$\Rightarrow \dfrac{1}{x} = m \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{1}{a} \\[1em] \Rightarrow x = a \\[1em] \Rightarrow \dfrac{1}{y} = n \\[1em] \Rightarrow \dfrac{1}{y} = \dfrac{1}{b} \\[1em] \Rightarrow y = b.$

Hence, x = a, y = b.

If 2x + y = 32 and 3x + 4y = 68, find the value of $\dfrac{x}{y}$.

Answer

Given,

Equations :

2x + y = 32     .........(1),

3x + 4y = 68     .........(2)

Multiplying equation (1) by 4, we get :

⇒ 4(2x + y) = 32 × 4

⇒ 8x + 4y = 128     .........(3)

Subtracting equation (2) from (3), we get :

⇒ 8x + 4y - (3x + 4y) = 128 - 68

⇒ 8x + 4y - 3x - 4y = 60

⇒ 5x = 60

⇒ x = $\dfrac{60}{5}$

⇒ x = 12.

Substituting value of x in equation (2), we get :

⇒ 3x + 4y = 68

⇒ 3 × 12 + 4y = 68

⇒ 36 + 4y = 68

⇒ 4y = 68 - 36

⇒ 4y = 32

⇒ y = $\dfrac{32}{4}$

⇒ y = 8.

Substituting value of x and y in $\dfrac{x}{y}$, we get :

$\Rightarrow \dfrac{x}{y} = \dfrac{12}{8} = \dfrac{3}{2}$.

Hence, $\dfrac{x}{y} = \dfrac{3}{2}$.

The sides of an equilateral triangle are (x + 3y) cm, (3x + 2y − 2) cm and $\Big(4x + \dfrac{y}{2} + 1\Big)$ cm. Find the length of each side.

Answer

Given,

In an equilateral triangle, all three sides are equal.

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

Solving L.H.S of the above equation, we get :

⇒ x + 3y = 3x + 2y - 2

⇒ x - 3x + 3y - 2y = -2

⇒ -2x + y = -2

⇒ y = 2x - 2     .....(1)

Solving R.H.S of the above equation, we get :

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

Substituting value of y from equation (1) in (2), we get :

⇒ 3(2x - 2) - 2x = 6

⇒ 6x - 6 - 2x = 6

⇒ 4x - 6 = 6

⇒ 4x = 6 + 6

⇒ 4x = 12

⇒ x = $\dfrac{12}{4}$

⇒ x = 3.

Substituting value of x in equation (1), we get :

⇒ y = 2(3) - 2

⇒ y = 6 - 2

⇒ y = 4.

Substituting value of x and y in x + 3y, we get :

⇒ x + 3y = 3 + 3(4) = 3 + 12 = 15 cm.

Since, all sides are equal.

Hence, the length of each side is 15 cm.