Simultaneous (Linear) Equations — Exercise 6(B)

Solution of equations x + y = 8 and 3x - 5y = 0 is :

1. x = 5, y = -3

2. x = 5, y = 3

3. x = -5, y = 3

4. x = -5, y = -3

Answer

Given, equations :

⇒ x + y = 8 and 3x - 5y = 0

⇒ x + y - 8 = 0 ...........(1)

⇒ 3x - 5y = 0 ..........(2)

By cross-multiplication method :

$\Rightarrow \dfrac{x}{1 \times 0 - (-5) \times (-8)} = \dfrac{y}{(-8) \times 3 - 0 \times 1} = \dfrac{1}{1 \times (-5) - 3 \times 1} \\[1em] \Rightarrow \dfrac{x}{0 - 40} = \dfrac{y}{-24 - 0} = \dfrac{1}{-5 - 3} \\[1em] \Rightarrow \dfrac{x}{-40} = \dfrac{y}{-24} = \dfrac{1}{-8} \\[1em] \Rightarrow \dfrac{x}{-40} = \dfrac{1}{-8} \text{ and } \dfrac{y}{-24} = \dfrac{1}{-8} \\[1em] \Rightarrow x = \dfrac{-40}{-8} \text{ and } y = \dfrac{-24}{-8} \\[1em] \Rightarrow x = 5 \text{ and } y = 3.$

Hence, Option 2 is the correct option.

Solution of equations $\dfrac{4a - 3}{6} + \dfrac{5b - 7}{2}$ = 18 - 5a and a + b = 5 is :

1. a = -3, b = 2

2. a = -3, b = -2

3. a = 3, b = 2

4. a = 3, b = -2

Answer

Given,

1st equation :

$\Rightarrow \dfrac{4a - 3}{6} + \dfrac{5b - 7}{2} = 18 - 5a \\[1em] \Rightarrow \dfrac{4a - 3 + 3(5b - 7)}{6} = 18 - 5a \\[1em] \Rightarrow \dfrac{4a - 3 + 15b - 21}{6} = 18 - 5a \\[1em] \Rightarrow 4a + 15b - 24 = 6(18 - 5a) \\[1em] \Rightarrow 4a + 15b - 24 = 108 - 30a \\[1em] \Rightarrow 4a + 30a + 15b = 108 + 24 \\[1em] \Rightarrow 34a + 15b = 132 \\[1em] \Rightarrow 34a + 15b - 132 = 0$

2nd equation :

⇒ a + b = 5

⇒ a + b - 5 = 0

By cross-multiplication method :

$\Rightarrow \dfrac{a}{15 \times (-5) - 1\times (-132)} = \dfrac{b}{(-132) \times 1 - (-5) \times 34} = \dfrac{1}{34 \times 1 - 1 \times 15} \\[1em] \Rightarrow \dfrac{a}{-75 + 132} = \dfrac{b}{-132 + 170} = \dfrac{1}{34 - 15} \\[1em] \Rightarrow \dfrac{a}{57} = \dfrac{b}{38} = \dfrac{1}{19} \\[1em] \Rightarrow \dfrac{a}{57} = \dfrac{1}{19} \text{ and } \dfrac{b}{38} = \dfrac{1}{19}\\[1em] \Rightarrow a = \dfrac{57}{19} \text{ and } b = \dfrac{38}{19} \\[1em] \Rightarrow a = 3 \text{ and } b = 2.$

Hence, Option 3 is the correct option.

Solution of equations $\dfrac{x}{7} - \dfrac{y}{2} = 2 \text{ and } \dfrac{x - y}{3} = 3$ is :

1. x = 7, y = 2

2. x = -7, y = 2

3. x = -7, y = -2

4. x = 7, y = -2

Answer

Given, equations : $\dfrac{x}{7} - \dfrac{y}{2} = 2 \text{ and } \dfrac{x - y}{3} = 3$

Simplifying first equation :

$\Rightarrow \dfrac{x}{7} - \dfrac{y}{2} = 2 \\[1em] \Rightarrow \dfrac{2x - 7y}{14} = 2 \\[1em] \Rightarrow 2x - 7y = 28 \\[1em] \Rightarrow 2x - 7y - 28 = 0 \text{ ........(1)}$

Simplifying second equation :

$\Rightarrow \dfrac{x - y}{3} = 3 \\[1em] \Rightarrow x - y = 9 \\[1em] \Rightarrow x - y - 9 = 0 \text{ .......(2)}$

By cross-multiplication method :

$\Rightarrow \dfrac{x}{(-7) \times (-9) - (-1) \times (-28)} = \dfrac{y}{(-28) \times 1 - (-9) \times 2} = \dfrac{1}{2 \times (-1) - 1 \times (-7)} \\[1em] \Rightarrow \dfrac{x}{63 - 28} = \dfrac{y}{-28 - (-18)} = \dfrac{1}{-2 + 7} \\[1em] \Rightarrow \dfrac{x}{35} = \dfrac{y}{-10} = \dfrac{1}{5} \\[1em] \Rightarrow \dfrac{x}{35} = \dfrac{1}{5} \text{ and } \dfrac{y}{-10} = \dfrac{1}{5} \\[1em] \Rightarrow x = \dfrac{35}{5} \text{ and } y = \dfrac{-10}{5} \\[1em] \Rightarrow x = 7 \text{ and } y = -2.$

Hence, Option 4 is the correct option.

Solution of equations 11x - 7y = 29 and 7x - 11y = 25 is :

1. x = 2, y = 1

2. x = 2, y = -1

3. x = -2, y = 1

4. x = -2, y = -1

Answer

Given, equations : 11x - 7y = 29 and 7x - 11y = 25

⇒ 11x - 7y - 29 = 0 .......(1)

⇒ 7x - 11y - 25 = 0 .......(2)

By cross-multiplication method we have :

$\Rightarrow \dfrac{x}{(-7) \times (-25) - (-11) \times (-29)} = \dfrac{y}{(-29) \times 7 - (-25) \times 11} = \dfrac{1}{11 \times (-11) - 7 \times (-7)} \\[1em] \Rightarrow \dfrac{x}{175 - 319} = \dfrac{y}{-203 + 275} = \dfrac{1}{-121 + 49} \\[1em] \Rightarrow \dfrac{x}{-144} = \dfrac{y}{72} = \dfrac{1}{-72} \\[1em] \Rightarrow \dfrac{x}{-144} = \dfrac{1}{-72} \text{ and } \dfrac{y}{72} = \dfrac{1}{-72} \\[1em] \Rightarrow x = \dfrac{-144}{-72} \text{ and } y = \dfrac{72}{-72} \\[1em] \Rightarrow x = 2 \text{ and } y = -1.$

Hence, Option 2 is the correct option.

Solution of equations 15x + 6y = 27 and 6x + 15y = 36 is :

1. x = 1, y = -2

2. x = -1, y = 2

3. x = 1, y = 2

4. x = -1, y = -2

Answer

Given, equations can be written as :

⇒ 15x + 6y - 27 = 0 .......(1)

⇒ 6x + 15y - 36 = 0 ........(2)

By cross-multiplication method :

$\Rightarrow \dfrac{x}{6 \times (-36) - 15 \times (-27)} = \dfrac{y}{(-27) \times 6 - (-36) \times 15} = \dfrac{1}{15 \times 15 - 6 \times 6} \\[1em] \Rightarrow \dfrac{x}{-216 + 405} = \dfrac{y}{-162 - (-540)} = \dfrac{1}{225 - 36} \\[1em] \Rightarrow \dfrac{x}{189} = \dfrac{y}{-162 + 540} = \dfrac{1}{189}\\[1em] \Rightarrow \dfrac{x}{189} = \dfrac{y}{378} = \dfrac{1}{189}\\[1em] \Rightarrow \dfrac{x}{189} = \dfrac{1}{189} \text{ and } \dfrac{y}{378} = \dfrac{1}{189} \\[1em] \Rightarrow x = \dfrac{189}{189} \text{ and } y = \dfrac{378}{189} \\[1em] \Rightarrow x = 1 \text{ and } y = 2.$

Hence, Option 3 is the correct option.

Solution of equations $\dfrac{2}{x} - \dfrac{3}{y} = 13 \text{ and } \dfrac{3}{x} + \dfrac{2}{y}$ = 0 is :

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

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

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

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

Answer

Given, equations :

$\dfrac{2}{x} - \dfrac{3}{y} = 13$ .......(1)

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

Multiplying equation (1) by 2, we get :

$\Rightarrow 2\Big(\dfrac{2}{x} - \dfrac{3}{y}\Big) = 2 \times 13 \\[1em] \Rightarrow \dfrac{4}{x} - \dfrac{6}{y} = 26 \text{ .......(3)}$

Multiplying equation (2) by 3, we get :

$\Rightarrow 3\Big(\dfrac{3}{x} + \dfrac{2}{y}\Big) = 3 \times 0 \\[1em] \Rightarrow \dfrac{9}{x} + \dfrac{6}{y} = 0 \text{ .......(4)}$

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

$\Rightarrow \Big(\dfrac{4}{x} - \dfrac{6}{y}\Big) + \Big(\dfrac{9}{x} + \dfrac{6}{y}\Big) = 26 + 0 \\[1em] \Rightarrow \dfrac{4}{x} + \dfrac{9}{x} - \dfrac{6}{y} + \dfrac{6}{y} = 26 \\[1em] \Rightarrow \dfrac{13}{x} = 26 \\[1em] \Rightarrow x = \dfrac{13}{26} = \dfrac{1}{2}.$

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

$\Rightarrow \dfrac{3}{\dfrac{1}{2}} + \dfrac{2}{y} = 0 \\[1em] \Rightarrow 6 + \dfrac{2}{y} = 0 \\[1em] \Rightarrow \dfrac{2}{y} = 0 - 6 \\[1em] \Rightarrow \dfrac{2}{y} = -6 \\[1em] \Rightarrow y = \dfrac{2}{-6} = -\dfrac{1}{3}.$

Hence, Option 3 is the correct option.

Solve, using cross-multiplication :

4x + 3y = 17

3x - 4y + 6 = 0

Answer

Given, equations :

⇒ 4x + 3y = 17 and 3x - 4y + 6 = 0

⇒ 4x + 3y - 17 = 0 .......(1)

⇒ 3x - 4y + 6 = 0 ........(2)

By cross-multiplication method :

$\Rightarrow \dfrac{x}{3 \times 6 - (-4) \times (-17)} = \dfrac{y}{(-17) \times 3 - 6 \times 4} = \dfrac{1}{4 \times (-4) - 3 \times 3} \\[1em] \Rightarrow \dfrac{x}{18 - 68} = \dfrac{y}{-51 - 24} = \dfrac{1}{-16 - 9} \\[1em] \Rightarrow \dfrac{x}{-50} = \dfrac{y}{-75} = \dfrac{1}{-25} \\[1em] \Rightarrow \dfrac{x}{-50} = \dfrac{1}{-25} \text{ and } \dfrac{y}{-75} = \dfrac{1}{-25} \\[1em] \Rightarrow x = \dfrac{-50}{-25} \text{ and } y = \dfrac{-75}{-25} \\[1em] \Rightarrow x = 2 \text{ and } y = 3.$

Hence, x = 2 and y = 3.

Solve, using cross-multiplication :

3x + 4y = 11

2x + 3y = 8

Answer

Given, equations :

⇒ 3x + 4y = 11 and 2x + 3y = 8

⇒ 3x + 4y - 11 = 0 .......(1)

⇒ 2x + 3y - 8 = 0 ........(2)

By cross-multiplication method :

$\Rightarrow \dfrac{x}{4 \times (-8) - 3 \times (-11)} = \dfrac{y}{(-11) \times 2 - (-8) \times 3} = \dfrac{1}{3 \times 3 - 2 \times 4} \\[1em] \Rightarrow \dfrac{x}{-32 + 33} = \dfrac{y}{-22 + 24} = \dfrac{1}{9 - 8} \\[1em] \Rightarrow \dfrac{x}{1} = \dfrac{y}{2} = \dfrac{1}{1} \\[1em] \Rightarrow x = \dfrac{y}{2} = 1 \\[1em] \Rightarrow x = 1 \text{ and } \dfrac{y}{2} = 1 \\[1em] \Rightarrow x = 1 \text{ and } y = 2.$

Hence, x = 1 and y = 2.

Solve, using cross-multiplication :

6x + 7y - 11 = 0

5x + 2y = 13

Answer

Given, equations :

⇒ 6x + 7y - 11 = 0 and 5x + 2y = 13

⇒ 6x + 7y - 11 = 0 ........(1)

⇒ 5x + 2y - 13 = 0 ........(2)

By cross-multiplication method :

$\Rightarrow \dfrac{x}{7 \times (-13) - 2 \times (-11)} = \dfrac{y}{(-11) \times 5 - (-13) \times 6} = \dfrac{1}{6 \times 2 - 5 \times 7} \\[1em] \Rightarrow \dfrac{x}{-91 + 22} = \dfrac{y}{-55 + 78} = \dfrac{1}{12 - 35} \\[1em] \Rightarrow \dfrac{x}{-69} = \dfrac{y}{23} = \dfrac{1}{-23} \\[1em] \Rightarrow \dfrac{x}{-69} = \dfrac{1}{-23} \text{ and } \dfrac{y}{23} = \dfrac{1}{-23} \\[1em] \Rightarrow x = \dfrac{-69}{-23} \text{ and } y = \dfrac{23}{-23}\\[1em] \Rightarrow x = 3 \text{ and } y = -1.$

Hence, x = 3 and y = -1.

Solve, using cross-multiplication :

5x + 4y + 14 = 0

3x = -10 - 4y

Answer

Given, equations :

⇒ 5x + 4y + 14 = 0 and 3x = -10 - 4y

⇒ 5x + 4y + 14 = 0 ..........(1)

⇒ 3x + 4y + 10 = 0 ..........(2)

By cross-multiplication method :

$\Rightarrow \dfrac{x}{4 \times 10 - 4 \times 14} = \dfrac{y}{14 \times 3 - 10 \times 5} = \dfrac{1}{5\times 4 - 3 \times 4} \\[1em] \Rightarrow \dfrac{x}{40 - 56} = \dfrac{y}{42 - 50} = \dfrac{1}{20 - 12} \\[1em] \Rightarrow \dfrac{x}{-16} = \dfrac{y}{-8} = \dfrac{1}{8} \\[1em] \Rightarrow \dfrac{x}{-16} = \dfrac{1}{8} \text{ and } \dfrac{y}{-8} = \dfrac{1}{8} \\[1em] \Rightarrow x = \dfrac{-16}{8} \text{ and } y = \dfrac{-8}{8} \\[1em] \Rightarrow x = -2 \text{ and } y = -1.$

Hence, x = -2 and y = -1.

Solve, using cross-multiplication :

x - y + 2 = 0

7x + 9y = 130

Answer

Given, equations :

⇒ x - y + 2 = 0 and 7x + 9y = 130

⇒ x - y + 2 = 0 ........(1)

⇒ 7x + 9y - 130 = 0 ........(2)

By cross-multiplication method :

$\Rightarrow \dfrac{x}{(-1) \times (-130) - 9 \times 2} = \dfrac{y}{2 \times 7 - 1 \times (-130)} = \dfrac{1}{1 \times 9 - 7 \times (-1)} \\[1em] \Rightarrow \dfrac{x}{130 - 18} = \dfrac{y}{14 + 130} = \dfrac{1}{9 + 7} \\[1em] \Rightarrow \dfrac{x}{112} = \dfrac{y}{144} = \dfrac{1}{16} \\[1em] \Rightarrow \dfrac{x}{112} = \dfrac{1}{16} \text{ and } \dfrac{y}{144} = \dfrac{1}{16} \\[1em] \Rightarrow x = \dfrac{112}{16} \text{ and } y = \dfrac{144}{16} \\[1em] \Rightarrow x = 7 \text{ and } y = 9.$

Hence, x = 7 and y = 9.

Solve, using cross-multiplication :

4x - y = 5

5y - 4x = 7

Answer

Given, equations :

⇒ 4x - y = 5 and 5y - 4x = 7

⇒ 4x - y - 5 = 0 ..........(1)

⇒ -4x + 5y - 7 = 0 .........(2)

By cross-multiplication method :

$\Rightarrow \dfrac{x}{(-1) \times (-7) - 5 \times (-5)} = \dfrac{y}{(-5) \times (-4) - (-7) \times 4} = \dfrac{1}{4 \times 5 - (-4) \times (-1)} \\[1em] \Rightarrow \dfrac{x}{7 + 25} = \dfrac{y}{20 + 28} = \dfrac{1}{20 - 4} \\[1em] \Rightarrow \dfrac{x}{32} = \dfrac{y}{48} = \dfrac{1}{16} \\[1em] \Rightarrow \dfrac{x}{32} = \dfrac{1}{16} \text{ and } \dfrac{y}{48} = \dfrac{1}{16} \\[1em] \Rightarrow x = \dfrac{32}{16} \text{ and } y = \dfrac{48}{16} \\[1em] \Rightarrow x = 2 \text{ and } y = 3.$

Hence, x = 2 and y = 3.

Solve, using cross-multiplication :

4x - 3y = 0

2x + 3y = 18

Answer

Given, equations :

⇒ 4x - 3y = 0 and 2x + 3y = 18

⇒ 4x - 3y = 0 .........(1)

⇒ 2x + 3y - 18 = 0 .....(2)

By cross-multiplication method :

$\Rightarrow \dfrac{x}{(-3) \times (-18) - 3 \times 0} = \dfrac{y}{0 \times 2 - (-18) \times 4} = \dfrac{1}{4 \times 3 - 2 \times (-3)} \\[1em] \Rightarrow \dfrac{x}{54 - 0} = \dfrac{y}{0 + 72} = \dfrac{1}{12 + 6} \\[1em] \Rightarrow \dfrac{x}{54} = \dfrac{y}{72} = \dfrac{1}{18} \\[1em] \Rightarrow \dfrac{x}{54} = \dfrac{1}{18} \text{ and } \dfrac{y}{72} = \dfrac{1}{18} \\[1em] \Rightarrow x = \dfrac{54}{18} \text{ and } y = \dfrac{72}{18}\\[1em] \Rightarrow x = 3 \text{ and } y = 4.$

Hence, x = 3 and y = 4.

Solve, using cross-multiplication :

8x + 5y = 9

3x + 2y = 4

Answer

Given, equations :

⇒ 8x + 5y = 9 and 3x + 2y = 4

⇒ 8x + 5y - 9 = 0 ..........(1)

⇒ 3x + 2y - 4 = 0 .........(2)

By cross-multiplication method we have :

$\Rightarrow \dfrac{x}{5 \times (-4) - 2 \times (-9)} = \dfrac{y}{(-9) \times 3 - (-4) \times 8} = \dfrac{1}{8 \times 2 - 3 \times 5} \\[1em] \Rightarrow \dfrac{x}{-20 + 18} = \dfrac{y}{-27 + 32} = \dfrac{1}{16 - 15} \\[1em] \Rightarrow \dfrac{x}{-2} = \dfrac{y}{5} = \dfrac{1}{1} \\[1em] \Rightarrow \dfrac{x}{-2} = \dfrac{1}{1} \text{ and } \dfrac{y}{5} = \dfrac{1}{1} \\[1em] \Rightarrow x = \dfrac{-2}{1} \text{ and } y = \dfrac{5}{1} \\[1em] \Rightarrow x = -2 \text{ and } y = 5.$

Hence, x = -2 and y = 5.

Solve, using cross-multiplication :

4x - 3y - 11 = 0

6x + 7y - 5 = 0

Answer

Given, equations :

⇒ 4x - 3y - 11 = 0 ............(1)

⇒ 6x + 7y - 5 = 0 ............(2)

By cross-multiplication method :

$\Rightarrow \dfrac{x}{(-3) \times (-5) - 7 \times (-11)} = \dfrac{y}{(-11) \times 6 - (-5) \times 4} = \dfrac{1}{4 \times 7 - 6 \times (-3)} \\[1em] \Rightarrow \dfrac{x}{15 + 77} = \dfrac{y}{-66 + 20} = \dfrac{1}{28+ 18} \\[1em] \Rightarrow \dfrac{x}{92} = \dfrac{y}{-46} = \dfrac{1}{46} \\[1em] \Rightarrow \dfrac{x}{92} = \dfrac{1}{46} \text{ and } \dfrac{y}{-46} = \dfrac{1}{46} \\[1em] \Rightarrow x = \dfrac{92}{46} \text{ and } y = \dfrac{-46}{46} \\[1em] \Rightarrow x = 2 \text{ and } y = -1.$

Hence, x = 2 and y = -1.

Solve :

$\dfrac{9}{x} - \dfrac{4}{y} = 8$

$\dfrac{13}{x} + \dfrac{7}{y} = 101$

Answer

Given, equations :

$\Rightarrow \dfrac{9}{x} - \dfrac{4}{y} = 8$ .......(1)

$\Rightarrow \dfrac{13}{x} + \dfrac{7}{y} = 101$ .....(2)

Multiplying equation (1) by 7, we get :

$\Rightarrow 7\Big(\dfrac{9}{x} - \dfrac{4}{y}\Big) = 7 \times 8 \\[1em] \Rightarrow \dfrac{63}{x} - \dfrac{28}{y} = 56 \text{ ......(3)}$

Multiplying equation by (2) by 4, we get :

$\Rightarrow 4\Big(\dfrac{13}{x} + \dfrac{7}{y}\Big) = 4 \times 101 \\[1em] \Rightarrow \dfrac{52}{x} + \dfrac{28}{y} = 404 \text{ .......(4)}$

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

$\Rightarrow \Big(\dfrac{63}{x} - \dfrac{28}{y}\Big) + \Big(\dfrac{52}{x} + \dfrac{28}{y} \Big) = 56 + 404 \\[1em] \Rightarrow \dfrac{63 + 52}{x} = 460 \\[1em] \Rightarrow \dfrac{115}{x} = 460 \\[1em] \Rightarrow x = \dfrac{115}{460} = \dfrac{1}{4}.$

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

$\Rightarrow \dfrac{9}{\dfrac{1}{4}} - \dfrac{4}{y} = 8 \\[1em] \Rightarrow 36 - \dfrac{4}{y} = 8 \\[1em] \Rightarrow \dfrac{4}{y} = 36 - 8 \\[1em] \Rightarrow \dfrac{4}{y} = 28 \\[1em] \Rightarrow y = \dfrac{4}{28} = \dfrac{1}{7}.$

Hence, x = $\dfrac{1}{4} \text{ and } y = \dfrac{1}{7}$.

Solve :

$\dfrac{3}{x} + \dfrac{2}{y} = 10$

$\dfrac{9}{x} - \dfrac{7}{y} = 10.5$

Answer

Given, equations :

$\Rightarrow \dfrac{3}{x} + \dfrac{2}{y} = 10$ ............(1)

$\Rightarrow \dfrac{9}{x} - \dfrac{7}{y} = 10.5$ ..........(2)

Multiplying equation (1) by 3, we get :

$\Rightarrow 3\Big(\dfrac{3}{x} + \dfrac{2}{y}\Big) = 3 \times 10 \\[1em] \Rightarrow \dfrac{9}{x} + \dfrac{6}{y} = 30 ..........(3)$

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

$\Rightarrow \Big(\dfrac{9}{x} + \dfrac{6}{y}\Big) - \Big(\dfrac{9}{x} - \dfrac{7}{y}\Big) = 30 - 10.5 \\[1em] \Rightarrow \dfrac{9}{x} - \dfrac{9}{x} + \dfrac{6}{y} + \dfrac{7}{y} = 19.5 \\[1em] \Rightarrow \dfrac{13}{y} = 19.5 \\[1em] \Rightarrow y = \dfrac{13}{19.5} \\[1em] \Rightarrow y = \dfrac{1}{1.5} = \dfrac{2}{3}.$

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

$\Rightarrow \dfrac{3}{x} + \dfrac{2}{\dfrac{2}{3}} = 10 \\[1em] \Rightarrow \dfrac{3}{x} + \dfrac{6}{2} = 10 \\[1em] \Rightarrow \dfrac{3}{x} + 3 = 10 \\[1em] \Rightarrow \dfrac{3}{x} = 7 \\[1em] \Rightarrow x = \dfrac{3}{7}.$

Hence, x = $\dfrac{3}{7} \text{ and } y = \dfrac{2}{3}$.

Solve :

$5x + \dfrac{8}{y} = 19$

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

Answer

Given, equations :

$\Rightarrow 5x + \dfrac{8}{y} = 19$ .........(1)

$\Rightarrow 3x - \dfrac{4}{y} = 7$ ........(2)

Multiplying equation (2) by 2, we get :

$\Rightarrow 2\Big(3x - \dfrac{4}{y}\Big) = 2 \times 7 \\[1em] \Rightarrow 6x - \dfrac{8}{y} = 14 \text{ ........(3)}$

Adding equation (1) and (3), we get :

$\Rightarrow \Big(5x + \dfrac{8}{y}\Big) + \Big(6x - \dfrac{8}{y}\Big) = 19 + 14 \\[1em] \Rightarrow 5x + 6x + \dfrac{8}{y} - \dfrac{8}{y} = 33 \\[1em] \Rightarrow 11x = 33 \\[1em] \Rightarrow x = \dfrac{33}{11} = 3.$

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

$\Rightarrow 5 \times 3 + \dfrac{8}{y} = 19 \\[1em] \Rightarrow 15 + \dfrac{8}{y} = 19 \\[1em] \Rightarrow \dfrac{8}{y} = 19 - 15 \\[1em] \Rightarrow \dfrac{8}{y} = 4 \\[1em] \Rightarrow y = \dfrac{8}{4} = 2.$

Hence, x = 3 and y = 2.

Solve : $4x + \dfrac{6}{y} = 15 \text{ and } 3x - \dfrac{4}{y} = 7$.

Hence, find 'a' if y = ax - 2.

Answer

Given, equations :

$\Rightarrow 4x + \dfrac{6}{y} = 15$ ..........(1)

$\Rightarrow 3x - \dfrac{4}{y} = 7$ ..........(2)

Multiplying equation (1) by 2, we get :

$\Rightarrow 2\Big(4x + \dfrac{6}{y}\Big) = 2 \times 15 \\[1em] \Rightarrow 8x + \dfrac{12}{y} = 30 \text{ .........(3)}$

Multiplying equation (2) by 3, we get :

$\Rightarrow 3\Big(3x - \dfrac{4}{y}\Big) = 3 \times 7 \\[1em] \Rightarrow 9x - \dfrac{12}{y} = 21 \text{ .........(4)}$

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

$\Rightarrow 8x + \dfrac{12}{y} + 9x - \dfrac{12}{y} = 30 + 21 \\[1em] \Rightarrow 17x = 51 \\[1em] \Rightarrow x = \dfrac{51}{17} = 3.$

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

$\Rightarrow 3 \times 3 - \dfrac{4}{y} = 7 \\[1em] \Rightarrow 9 - \dfrac{4}{y} = 7 \\[1em] \Rightarrow \dfrac{4}{y} = 9 - 7 \\[1em] \Rightarrow \dfrac{4}{y} = 2 \\[1em] \Rightarrow y = \dfrac{4}{2} = 2.$

Given,

⇒ y = ax - 2

⇒ 2 = 3a - 2

⇒ 3a = 2 + 2

⇒ 3a = 4

⇒ a = $\dfrac{4}{3} = 1\dfrac{1}{3}$.

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

Solve :

$\dfrac{3}{x} - \dfrac{2}{y} = 0 \text{ and } \dfrac{2}{x} + \dfrac{5}{y} = 19$.

Hence, find 'a' if y = ax + 3,

Answer

Given, equations :

$\Rightarrow \dfrac{3}{x} - \dfrac{2}{y} = 0$ .......(1)

$\Rightarrow \dfrac{2}{x} + \dfrac{5}{y} = 19$ .......(2)

Multiplying equation (1) by 2, we get :

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

Multiplying equation (2) by 3, we get :

$\Rightarrow 3\Big(\dfrac{2}{x} + \dfrac{5}{y}\Big) = 3 \times 19 \\[1em] \Rightarrow \dfrac{6}{x} + \dfrac{15}{y} = 57 \text{ .........(4)}$

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

$\Rightarrow \Big(\dfrac{6}{x} + \dfrac{15}{y}\Big) - \Big(\dfrac{6}{x} - \dfrac{4}{y}\Big) = 57 - 0 \\[1em] \Rightarrow \dfrac{6}{x} - \dfrac{6}{x} + \dfrac{15}{y} + \dfrac{4}{y} = 57 \\[1em] \Rightarrow \dfrac{19}{y} = 57 \\[1em] \Rightarrow y = \dfrac{19}{57} = \dfrac{1}{3}.$

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

$\Rightarrow \dfrac{3}{x} - \dfrac{2}{\dfrac{1}{3}} = 0 \\[1em] \Rightarrow \dfrac{3}{x} - 6 = 0 \\[1em] \Rightarrow \dfrac{3}{x} = 6 \\[1em] \Rightarrow x = \dfrac{3}{6} = \dfrac{1}{2}.$

Given,

$\Rightarrow y = ax + 3 \\[1em] \Rightarrow \dfrac{1}{3} = a \times \dfrac{1}{2} + 3 \\[1em] \Rightarrow \dfrac{1}{3} = \dfrac{a}{2} + 3 \\[1em] \Rightarrow \dfrac{a}{2} = \dfrac{1}{3} - 3 \\[1em] \Rightarrow \dfrac{a}{2} = \dfrac{1 - 9}{3} \\[1em] \Rightarrow \dfrac{a}{2} = \dfrac{-8}{3} \\[1em] \Rightarrow a = -\dfrac{16}{3} = -5\dfrac{1}{3}.$

Hence, $x = \dfrac{1}{2}, y = \dfrac{1}{3}, a = -5\dfrac{1}{3}$.

Solve :

$\dfrac{20}{x + y} + \dfrac{3}{x - y} = 7$

$\dfrac{8}{x - y} - \dfrac{15}{x + y} = 5$

Answer

Given, equations :

$\dfrac{20}{x + y} + \dfrac{3}{x - y} = 7$ .......(1)

$\dfrac{8}{x - y} - \dfrac{15}{x + y} = 5$ .............(2)

Multiplying equation (1) by 8, we get :

$\Rightarrow 8\Big(\dfrac{20}{x + y} + \dfrac{3}{x - y}\Big) = 8 \times 7 \\[1em] \Rightarrow \dfrac{160}{x + y} + \dfrac{24}{x - y} = 56 \text{ ..........(3)}$

Multiplying equation (2) by 3, we get :

$\Rightarrow 3\Big(\dfrac{8}{x - y} - \dfrac{15}{x + y}\Big) = 3 \times 5 \\[1em] \Rightarrow \dfrac{24}{x - y} - \dfrac{45}{x + y} = 15 \text{ .........(4)}$

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

$\Rightarrow \dfrac{160}{x + y} + \dfrac{24}{x - y} - \Big(\dfrac{24}{x - y} - \dfrac{45}{x + y}\Big) = 56 - 15 \\[1em] \Rightarrow \dfrac{160}{x + y} + \dfrac{45}{x + y} + \dfrac{24}{x - y} - \dfrac{24}{x - y} = 41 \\[1em] \Rightarrow \dfrac{205}{x + y} = 41 \\[1em] \Rightarrow x + y = \dfrac{205}{41} \\[1em] \Rightarrow x + y = 5 \text{ ........(5)}$

Substituting value of x + y from equation 5 in equation 1, we get :

$\Rightarrow \dfrac{20}{x + y} + \dfrac{3}{x - y} = 7 \\[1em] \Rightarrow \dfrac{20}{5} + \dfrac{3}{x - y} = 7 \\[1em] \Rightarrow 4 + \dfrac{3}{x - y} = 7 \\[1em] \Rightarrow \dfrac{3}{x - y} = 7 - 4 \\[1em] \Rightarrow \dfrac{3}{x - y} = 3 \\[1em] \Rightarrow x - y = \dfrac{3}{3} \\[1em] \Rightarrow x - y = 1 \text{ ........(6)}$

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

⇒ (x + y) + (x - y) = 5 + 1

⇒ 2x = 6

⇒ x = $\dfrac{6}{2}$

⇒ x = 3.

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

⇒ 3 - y = 1

⇒ y = 3 - 1 = 2.

Hence, x = 3 and y = 2.

Solve :

$\dfrac{34}{3x + 4y} + \dfrac{15}{3x - 2y} = 5$

$\dfrac{25}{3x - 2y} - \dfrac{8.50}{3x + 4y}$ = 4.5

Answer

Given, equations :

$\dfrac{34}{3x + 4y} + \dfrac{15}{3x - 2y} = 5$ .......(1)

$\dfrac{25}{3x - 2y} - \dfrac{8.50}{3x + 4y}$ = 4.5 .........(2)

Multiplying equation (2) by 4, we get :

$\Rightarrow 4\Big(\dfrac{25}{3x - 2y} - \dfrac{8.50}{3x + 4y}\Big) = 4 \times 4.5 \\[1em] \Rightarrow \dfrac{100}{3x - 2y} - \dfrac{34}{3x + 4y} = 18 \text{ .........(3)}$

Adding equation (1) and (3), we get :

$\Rightarrow \dfrac{34}{3x + 4y} + \dfrac{15}{3x - 2y} + \Big(\dfrac{100}{3x - 2y} - \dfrac{34}{3x + 4y}\Big) = 5 + 18 \\[1em] \Rightarrow \dfrac{115}{3x - 2y} = 23 \\[1em] \Rightarrow 3x- 2y = \dfrac{115}{23} \\[1em] \Rightarrow 3x - 2y = 5 \text{ .........(4)} \\[1em] \Rightarrow 3x = 5 + 2y \\[1em] \Rightarrow x = \dfrac{5 + 2y}{3} \text{ ......(5)}$

Substituting value of 3x - 2y from equation (4) in (2), we get :

$\Rightarrow \dfrac{25}{5} - \dfrac{8.50}{3x + 4y} = 4.5 \\[1em] \Rightarrow \dfrac{25(3x + 4y) - 42.50}{5(3x + 4y)} = \dfrac{45}{10} \\[1em] \Rightarrow \dfrac{75x + 100y - 42.50}{15x + 20y} = \dfrac{9}{2} \\[1em] \Rightarrow 2(75x + 100y - 42.50) = 9(15x + 20y) \\[1em] \Rightarrow 150x + 200y - 85 = 135x + 180y \\[1em] \Rightarrow 150x - 135x + 200y - 180y = 85 \\[1em] \Rightarrow 15x + 20y = 85 \text{ ........(6)}$

Substituting value of x from equation (5) in (6), we get :

$\Rightarrow 15 \times \dfrac{5 + 2y}{3} + 20y = 85 \\[1em] \Rightarrow 5(5 + 2y) + 20y = 85 \\[1em] \Rightarrow 25 + 10y + 20y = 85 \\[1em] \Rightarrow 30y = 85 - 25 \\[1em] \Rightarrow 30y = 60 \\[1em] \Rightarrow y = \dfrac{60}{30} = 2.$

Substituting value of y in equation (5), we get :

$\Rightarrow x = \dfrac{5 + 2y}{3} \\[1em] = \dfrac{5 + 2 \times 2}{3} \\[1em] = \dfrac{5 + 4}{3} \\[1em] = \dfrac{9}{3} \\[1em] = 3.$

Hence, x = 3 and y = 2.

Solve :

x + y = 2xy

x - y = 6xy

Answer

Given, equations : x + y = 2xy and x - y = 6xy

Dividing both the sides of first equation by xy, we get :

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

Dividing both the sides of second equation by xy, we get :

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

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

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

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

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

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

Solve :

x + y = 7xy

2x - 3y = -xy

Answer

Given, equations : x + y = 7xy and 2x - 3y = -xy

Dividing both the sides of first equation by xy, we get :

$\Rightarrow \dfrac{x + y}{xy} = \dfrac{7xy}{xy} \\[1em] \Rightarrow \dfrac{x}{xy} + \dfrac{y}{xy} = 7 \\[1em] \Rightarrow \dfrac{1}{y} + \dfrac{1}{x} = 7 \\[1em]$

Multiplying both sides of the above equation by 3, we get :

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

Dividing both the sides of second equation by xy, we get :

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

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

$\Rightarrow \Big(\dfrac{3}{y} + \dfrac{3}{x}\Big) + \Big(\dfrac{2}{y} - \dfrac{3}{x}\Big) = 21 + (-1) \\[1em] \Rightarrow \dfrac{5}{y} = 20 \\[1em] \Rightarrow y = \dfrac{5}{20} = \dfrac{1}{4}.$

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

$\Rightarrow \dfrac{3}{\dfrac{1}{4}} + \dfrac{3}{x} = 21 \\[1em] \Rightarrow 12 + \dfrac{3}{x} = 21 \\[1em] \Rightarrow \dfrac{3}{x} = 21 - 12 \\[1em] \Rightarrow \dfrac{3}{x} = 9 \\[1em] \Rightarrow x = \dfrac{3}{9} = \dfrac{1}{3}.$

Hence, $x = \dfrac{1}{3} \text{ and } y = \dfrac{1}{4}$.

Solve :

$\dfrac{a}{x} - \dfrac{b}{y} = 0$

$\dfrac{ab^2}{x} + \dfrac{a^2b}{y} = a^2 + b^2$

Answer

Let $\dfrac{1}{x} = p \text{ and } \dfrac{1}{y} = q$. Substituting in equations, we get :

⇒ ap - bq = 0 ..........(1)

⇒ ab²p + a²bq = a² + b²

⇒ ab²p + a²bq - (a² + b²) = 0 .........(2)

By cross-multiplication method we have :

$\Rightarrow \dfrac{p}{-b \times -(a^2 + b^2) - a^2b \times 0} = \dfrac{q}{0 \times ab^2 - [-(a^2 + b^2)] \times a} = \dfrac{1}{a \times a^2b - ab^2 \times (-b)} \\[1em] \Rightarrow \dfrac{p}{b(a^2 + b^2)} = \dfrac{q}{a(a^2 + b^2)} = \dfrac{1}{a^3b + ab^3} \\[1em] \Rightarrow \dfrac{p}{b(a^2 + b^2)} = \dfrac{1}{a^3b + ab^3} \text{ and } \dfrac{q}{a(a^2 + b^2)} = \dfrac{1}{a^3b + ab^3} \\[1em] \Rightarrow \dfrac{p}{b(a^2 + b^2)} = \dfrac{1}{ab(a^2 + b^2)} \text{ and } \dfrac{q}{a(a^2 + b^2)} = \dfrac{1}{ab(a^2 + b^2)} \\[1em] \Rightarrow p = \dfrac{b(a^2 + b^2)}{ab(a^2 + b^2)} \text{ and } q = \dfrac{a(a^2 + b^2)}{ab(a^2 + b^2)} \\[1em] \Rightarrow p = \dfrac{1}{a} \text{ and } q = \dfrac{1}{b} \\[1em] \Rightarrow \dfrac{1}{x} = p \text{ and } \dfrac{1}{y} = q \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{1}{a} \text{ and } \dfrac{1}{y} = \dfrac{1}{b} \\[1em] \Rightarrow x = a \text{ and } y = b.$

Hence, x = a and y = b.

Solve :

$\dfrac{2xy}{x + y} = \dfrac{3}{2}$

$\dfrac{xy}{2x - y} = -\dfrac{3}{10}$

x + y ≠ 0 and 2x - y ≠ 0

Answer

Simplifying first equation :

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

Simplifying second equation :

$\Rightarrow \dfrac{xy}{2x - y} = -\dfrac{3}{10} \\[1em] \Rightarrow \dfrac{2x - y}{xy} = -\dfrac{10}{3} \\[1em] \Rightarrow \dfrac{2x}{xy} - \dfrac{y}{xy} = -\dfrac{10}{3} \\[1em] \Rightarrow \dfrac{2}{y} - \dfrac{1}{x} = -\dfrac{10}{3} \text{ ..........(2)}$

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

$\Rightarrow \dfrac{1}{y} + \dfrac{1}{x} + \Big(\dfrac{2}{y} - \dfrac{1}{x}\Big) = \dfrac{4}{3} + \Big(-\dfrac{10}{3}\Big) \\[1em] \Rightarrow \dfrac{1}{y} + \dfrac{2}{y} + \dfrac{1}{x} - \dfrac{1}{x} = \dfrac{4 - 10}{3} \\[1em] \Rightarrow \dfrac{3}{y} = \dfrac{-6}{3} \\[1em] \Rightarrow y = \dfrac{3 \times 3}{-6} \\[1em] \Rightarrow y = -\dfrac{9}{6} = -\dfrac{3}{2}.$

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

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

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