Simultaneous Linear Equations — Exercise 5(B)

Solve the following system of equations by using the method of cross multiplication:

2x − 5y + 8 = 0, x − 4y + 7 = 0

Answer

Given,

Equations:

⇒ 2x − 5y + 8 = 0

⇒ x − 4y + 7 = 0

By cross-multiplication method,

$\Rightarrow \dfrac{x}{(-5) \times (7) - (-4) \times (8)} = \dfrac{y}{(8) \times (1) - (7) \times (2)} = \dfrac{1}{(2) \times (-4) - (1) \times (-5)} \\[1em] \Rightarrow \dfrac{x}{(-35) + 32} = \dfrac{y}{8 - 14} = \dfrac{1}{-8 + 5} \\[1em] \Rightarrow \dfrac{x}{-3} = \dfrac{y}{-6} = \dfrac{1}{-3} \\[1em] \Rightarrow \dfrac{x}{-3} = \dfrac{1}{-3} \text{ and } \dfrac{y}{-6} = \dfrac{1}{-3} \\[1em] \Rightarrow x = \dfrac{-3}{-3} \text{ and } y = \dfrac{-6}{-3} \\[1em] \Rightarrow x = 1 \text{ and } y = 2.$

Hence, x = 1 and y = 2.

Solve the following system of equations by using the method of cross multiplication:

5x − 4y + 2 = 0, 2x + 3y = 13

Answer

Given,

Equations :

⇒ 5x − 4y + 2 = 0

⇒ 2x + 3y - 13 = 0

By cross-multiplication method,

$\Rightarrow \dfrac{x}{(-4) \times (-13) - (3) \times (2)} = \dfrac{y}{(2) \times (2) - (-13) \times (5)} = \dfrac{1}{(5) \times (3) - (2) \times (-4)} \\[1em] \Rightarrow \dfrac{x}{(52) - 6} = \dfrac{y}{4 + 65} = \dfrac{1}{15 + 8} \\[1em] \Rightarrow \dfrac{x}{46} = \dfrac{y}{69} = \dfrac{1}{23} \\[1em] \Rightarrow \dfrac{x}{46} = \dfrac{1}{23} \text{ and } \dfrac{y}{69} = \dfrac{1}{23} \\[1em] \Rightarrow x = \dfrac{46}{23} \text{ and } y = \dfrac{69}{23} \\[1em] \Rightarrow x = 2 \text{ and } y = 3.$

Hence, x = 2 and y = 3.

Solve the following system of equations by using the method of cross multiplication:

3x − 5y = 19, 7x − 3y = 1

Answer

Given,

Equations:

⇒ 3x − 5y - 19 = 0

⇒ 7x − 3y - 1 = 0

By cross-multiplication method,

$\Rightarrow \dfrac{x}{(-5) \times (-1) - (-3) \times (-19)} = \dfrac{y}{(-19) \times (7) - (-1) \times (3)} = \dfrac{1}{(3) \times (-3) - (7) \times (-5)} \\[1em] \Rightarrow \dfrac{x}{5 - 57} = \dfrac{y}{-133 + 3} = \dfrac{1}{-9 + 35} \\[1em] \Rightarrow \dfrac{x}{-52} = \dfrac{y}{-130} = \dfrac{1}{26} \\[1em] \Rightarrow \dfrac{x}{-52} = \dfrac{1}{26} \text{ and } \dfrac{y}{-130} = \dfrac{1}{26} \\[1em] \Rightarrow x = \dfrac{-52}{26} \text{ and } y = \dfrac{-130}{26} \\[1em] \Rightarrow x = -2 \text{ and } y = -5.$

Hence, x = -2 and y = -5.

Solve the following system of equations by using the method of cross multiplication:

2x + 3y = 17, 3x − 2y = 6

Answer

Given,

Equations:

⇒ 2x + 3y - 17 = 0

⇒ 3x - 2y - 6 = 0

By cross-multiplication method,

$\Rightarrow \dfrac{x}{(3) \times (-6) - (-2) \times (-17)} = \dfrac{y}{(-17) \times (3) - (-6) \times (2)} = \dfrac{1}{(2) \times (-2) - (3) \times (3)} \\[1em] \Rightarrow \dfrac{x}{(-18) - 34} = \dfrac{y}{-51 + 12} = \dfrac{1}{-4 - 9} \\[1em] \Rightarrow \dfrac{x}{-52} = \dfrac{y}{-39} = \dfrac{1}{-13} \\[1em] \Rightarrow \dfrac{x}{-52} = \dfrac{1}{-13} \text{ and } \dfrac{y}{-39} = \dfrac{1}{-13} \\[1em] \Rightarrow x = \dfrac{-52}{-13} \text{ and } y = \dfrac{-39}{-13} \\[1em] \Rightarrow x = 4 \text{ and } y = 3.$

Hence, x = 4 and y = 3.

Solve the following system of equations by using the method of cross multiplication:

x + 2y + 1 = 0, 2x − 3y = 12

Answer

Given,

Equations:

⇒ x + 2y + 1 = 0

⇒ 2x − 3y - 12 = 0

By cross-multiplication method,

$\Rightarrow \dfrac{x}{(2) \times (-12) - (-3) \times (1)} = \dfrac{y}{(1) \times (2) - (-12) \times (1)} = \dfrac{1}{(1) \times (-3) - (2) \times (2)} \\[1em] \Rightarrow \dfrac{x}{(-24) + 3} = \dfrac{y}{2 + 12} = \dfrac{1}{-3 - 4} \\[1em] \Rightarrow \dfrac{x}{-21} = \dfrac{y}{14} = \dfrac{1}{-7} \\[1em] \Rightarrow \dfrac{x}{-21} = \dfrac{1}{-7} \text{ and } \dfrac{y}{14} = \dfrac{1}{-7} \\[1em] \Rightarrow x = \dfrac{-21}{-7} \text{ and } y = \dfrac{14}{-7} \\[1em] \Rightarrow x = 3 \text{ and } y = -2.$

Hence, x = 3 and y = -2.

Solve the following system of equations by using the method of cross multiplication:

2x + 5y = 1, 2x + 3y = 3

Answer

Given,

Equations:

⇒ 2x + 5y - 1 = 0

⇒ 2x + 3y - 3 = 0

By cross-multiplication method,

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

Hence, x = 3 and y = -1.

Solve the following system of equations by using the method of cross multiplication:

8x − 3y = 12, 5x = 2y + 7

Answer

Given,

Equations:

⇒ 8x − 3y - 12 = 0

⇒ 5x - 2y - 7 = 0

By cross-multiplication method,

$\Rightarrow \dfrac{x}{(-3) \times (-7) - (-2) \times (-12)} = \dfrac{y}{(-12) \times (5) - (-7) \times (8)} = \dfrac{1}{(8) \times (-2) - (5) \times (-3)} \\[1em] \Rightarrow \dfrac{x}{21 - 24} = \dfrac{y}{-60 + 56} = \dfrac{1}{-16 + 15} \\[1em] \Rightarrow \dfrac{x}{-3} = \dfrac{y}{-4} = \dfrac{1}{-1} \\[1em] \Rightarrow \dfrac{x}{-3} = \dfrac{1}{-1} \text{ and } \dfrac{y}{-4} = \dfrac{1}{-1} \\[1em] \Rightarrow x = \dfrac{-3}{-1} \text{ and } y = \dfrac{-4}{-1} \\[1em] \Rightarrow x = 3 \text{ and } y = 4.$

Hence, x = 3 and y = 4.

Solve the following system of equations by using the method of cross multiplication:

7x − 2y = 20, 11x + 15y + 23 = 0

Answer

Given,

Equations:

⇒ 7x − 2y - 20 = 0

⇒ 11x + 15y + 23 = 0

By cross-multiplication method,

$\Rightarrow \dfrac{x}{(-2) \times (23) - (15) \times (-20)} = \dfrac{y}{(-20) \times (11) - (23) \times (7)} = \dfrac{1}{(7) \times (15) - (11) \times (-2)} \\[1em] \Rightarrow \dfrac{x}{(-46) + 300} = \dfrac{y}{-220 - 161} = \dfrac{1}{105 + 22} \\[1em] \Rightarrow \dfrac{x}{254} = \dfrac{y}{-381} = \dfrac{1}{127} \\[1em] \Rightarrow \dfrac{x}{254} = \dfrac{1}{127} \text{ and } \dfrac{y}{-381} = \dfrac{1}{127} \\[1em] \Rightarrow x = \dfrac{254}{127} \text{ and } y = \dfrac{-381}{127} \\[1em] \Rightarrow x = 2 \text{ and } y = -3.$

Hence, x = 2 and y = -3.

Solve the following system of equations by using the method of cross multiplication:

ax + by = (a − b), bx − ay = (a + b)

Answer

Given,

Equations:

⇒ ax + by - (a − b) = 0

⇒ bx − ay - (a + b) = 0

By cross-multiplication method,

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

Hence, x = 1 and y = -1.

Solve the following system of equations by using the method of cross multiplication:

3x + 2y + 25 = 0, 2x + y + 10 = 0

Answer

Given,

Equations :

⇒ 3x + 2y + 25 = 0

⇒ 2x + y + 10 = 0

By cross-multiplication method,

$\Rightarrow \dfrac{x}{(2) \times (10) - (1) \times (25)} = \dfrac{y}{(25) \times (2) - (10) \times (3)} = \dfrac{1}{(3) \times (1) - (2) \times (2)} \\[1em] \Rightarrow \dfrac{x}{20 - 25} = \dfrac{y}{50 - 30} = \dfrac{1}{3 - 4} \\[1em] \Rightarrow \dfrac{x}{-5} = \dfrac{y}{20} = \dfrac{1}{-1} \\[1em] \Rightarrow \dfrac{x}{-5} = \dfrac{1}{-1} \text{ and } \dfrac{y}{20} = \dfrac{1}{-1} \\[1em] \Rightarrow x = \dfrac{-5}{-1} \text{ and } y = \dfrac{20}{-1} \\[1em] \Rightarrow x = 5 \text{ and } y = -20.$

Hence, x = 5 and y = -20.

Solve the following system of equations by using the method of cross multiplication:

$\dfrac{5}{x} - \dfrac{4}{y} + 2 = 0, \dfrac{2}{x} + \dfrac{3}{y} = 13$, (x ≠ 0, y ≠ 0)

Answer

Substituting $\dfrac{1}{x}= a, \dfrac{1}{y} = b$ in $\dfrac{5}{x} - \dfrac{4}{y} + 2 = 0$, we get:

⇒ 5a - 4b + 2 = 0     ..........(1)

Substituting $\dfrac{1}{x} = a, \dfrac{1}{y} = b$ in $\dfrac{2}{x} + \dfrac{3}{y} = 13$, we get :

⇒ 2a + 3b = 13

⇒ 2a + 3b - 13 = 0     ........(2)

Applying cross-multiplication method for solving equations (1) and (2), we get :

$\Rightarrow \dfrac{a}{(-4) \times (-13) - (3) \times (2)} = \dfrac{b}{(2) \times (2) - (-13) \times (5)} = \dfrac{1}{(5) \times (3) - (2) \times (-4)} \\[1em] \Rightarrow \dfrac{a}{52 - 6} = \dfrac{b}{4 + 65} = \dfrac{1}{15 + 8} \\[1em] \Rightarrow \dfrac{a}{46} = \dfrac{b}{69} = \dfrac{1}{23} \\[1em] \Rightarrow \dfrac{a}{46} = \dfrac{1}{23} \text{ and } \dfrac{b}{69} = \dfrac{1}{23} \\[1em] \Rightarrow a = \dfrac{46}{23} \text{ and } b = \dfrac{69}{23} \\[1em] \Rightarrow a = 2 \text{ and } b = 3.$

Now we have a = 2 and b = 3,

$\Rightarrow \dfrac{1}{x} = a \\[1em] \Rightarrow \dfrac{1}{x} = 2 \\[1em] \Rightarrow x = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{1}{y} = b \\[1em] \Rightarrow \dfrac{1}{y} = 3 \\[1em] \Rightarrow y = \dfrac{1}{3}.$

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

Solve the following system of equations by using the method of cross multiplication:

$\dfrac{1}{x} + \dfrac{1}{y} = 7, \dfrac{2}{x} + \dfrac{3}{y} = 17$ (x ≠ 0, y ≠ 0)

Answer

Substituting $\dfrac{1}{x}= a, \dfrac{1}{y} = b$ in $\dfrac{1}{x} + \dfrac{1}{y} = 7$, we get :

⇒ a + b = 7

⇒ a + b - 7 = 0     .........(1)

Substituting $\dfrac{1}{x} = a, \dfrac{1}{y} = b$ in $\dfrac{2}{x} + \dfrac{3}{y} = 17$, we get :

⇒ 2a + 3b = 17

⇒ 2a + 3b - 17 = 0     .........(2)

Applying cross-multiplication method for solving equations (1) and (2), we get :

$\Rightarrow \dfrac{a}{(1) \times (-17) - (3) \times (-7)} = \dfrac{b}{(-7) \times (2) - (-17) \times (1)} = \dfrac{1}{(1) \times (3) - (2) \times (1)} \\[1em] \Rightarrow \dfrac{a}{(-17) + 21} = \dfrac{b}{-14 + 17} = \dfrac{1}{3 - 2} \\[1em] \Rightarrow \dfrac{a}{4} = \dfrac{b}{3} = \dfrac{1}{1} \\[1em] \Rightarrow \dfrac{a}{4} = 1 \text{ and } \dfrac{b}{3} = \dfrac{1}{1} \\[1em] \Rightarrow a = \dfrac{4}{1} \text{ and } b = \dfrac{3}{1} \\[1em] \Rightarrow a = 4 \text{ and } b = 3.$

Now we have a = 4 and b = 3,

$\Rightarrow \dfrac{1}{x} = a \\[1em] \Rightarrow \dfrac{1}{x} = 4 \\[1em] \Rightarrow x = \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{1}{y} = b \\[1em] \Rightarrow \dfrac{1}{y} = 3 \\[1em] \Rightarrow y = \dfrac{1}{3}.$

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

Solve the following system of equations by using the method of cross multiplication:

$\dfrac{10}{x + y} + \dfrac{2}{x - y} = 4, \dfrac{15}{x + y} - \dfrac{5}{x - y} + 2 = 0$, where x ≠ -y and x ≠ y

Answer

Substituting $\dfrac{1}{x + y}= a, \dfrac{1}{x - y} = b$ in $\dfrac{10}{x + y} + \dfrac{2}{x - y} = 4$, we get:

⇒ 10a + 2b = 4

⇒ 10a + 2b - 4 = 0     .....(1)

Substituting $\dfrac{1}{x + y}= a, \dfrac{1}{x - y} = b$ in $\dfrac{15}{x + y} - \dfrac{5}{x - y} + 2 = 0$, we get:

⇒ 15a - 5b + 2 = 0     ....(2)

Applying cross-multiplication method for solving equations (1) and (2), we get :

$\Rightarrow \dfrac{a}{(2) \times (2) - (-5) \times (-4)} = \dfrac{b}{(-4) \times (15) - (2) \times (10)} = \dfrac{1}{(10) \times (-5) - (15) \times (2)} \\[1em] \Rightarrow \dfrac{a}{(4) - 20} = \dfrac{b}{-60 - 20} = \dfrac{1}{-50 - 30} \\[1em] \Rightarrow \dfrac{a}{-16} = \dfrac{b}{-80} = \dfrac{1}{-80} \\[1em] \Rightarrow \dfrac{a}{-16} = \dfrac{1}{-80} \text{ and } \dfrac{b}{-80} = \dfrac{1}{-80} \\[1em] \Rightarrow a = \dfrac{-16}{-80} \text{ and } b = \dfrac{-80}{-80} \\[1em] \Rightarrow a = \dfrac{1}{5} \text{ and } b = 1.$

Now we have a = $\dfrac{1}{5}$ and b = 1,

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

On adding equations (3) and (4) we get,

⇒ (x + y) + (x - y) = 5 + 1

⇒ 2x = 6

⇒ x = $\dfrac{6}{2} = 3$.

Substituting value of x in equation (4) we get,

⇒ x - y = 1

⇒ 3 - y = 1

⇒ y = 3 - 1 = 2.

Hence, x = 3, y = 2.

Solve the following system of equations by using the method of cross multiplication:

$\dfrac{5}{x + 1} - \dfrac{2}{y - 1} = \dfrac{1}{2},\dfrac{10}{x + 1} + \dfrac{2}{y - 1} = \dfrac{5}{2}$, where x ≠ -1 and x ≠ 1.

Answer

$\dfrac{5}{x + 1} - \dfrac{2}{y - 1} = \dfrac{1}{2},\dfrac{10}{x + 1} + \dfrac{2}{y - 1} = \dfrac{5}{2}$

Substituting $\dfrac{1}{x + 1}= u, \dfrac{1}{y - 1} = v$ in $\dfrac{5}{x + 1} - \dfrac{2}{y - 1} = \dfrac{1}{2}$, we get:

⇒ 5u - 2v = $\dfrac{1}{2}$

⇒ 5u - 2v - $\dfrac{1}{2}$ = 0     ....(1)

Substituting $\dfrac{1}{x + 1}= u, \dfrac{1}{y - 1} = v$ in $\dfrac{10}{x + 1} + \dfrac{2}{y - 1} = \dfrac{5}{2}$, we get:

⇒ 10u + 2v = $\dfrac{5}{2}$

⇒ 10u + 2v - $\dfrac{5}{2}$ = 0     ....(2)

Multiply equation (1) and (2) by 2, we get,

⇒ $2\Big(5u - 2v - \dfrac{1}{2}\Big)$= 0

⇒ 10u - 4v - 1 = 0     .......(3)

⇒ $2\Big(10u + 2v - \dfrac{5}{2}\Big)$ = 0

⇒ 20u + 4v - 5 = 0     .........(4)

Applying cross-multiplication method for solving equations (3) and (4), we get :

$\Rightarrow \dfrac{u}{(-4) \times (-5) - (4) \times (-1)} = \dfrac{v}{(-1) \times (20) - (-5) \times (10)} = \dfrac{1}{(10) \times (4) - (20) \times (-4)} \\[1em] \Rightarrow \dfrac{u}{(20) + 4} = \dfrac{v}{-20 + 50} = \dfrac{1}{40 + 80} \\[1em] \Rightarrow \dfrac{u}{24} = \dfrac{v}{30} = \dfrac{1}{120} \\[1em] \Rightarrow \dfrac{u}{24} = \dfrac{1}{120} \text{ and } \dfrac{v}{30} = \dfrac{1}{120} \\[1em] \Rightarrow u = \dfrac{24}{120} \text{ and } v = \dfrac{30}{120} \\[1em] \Rightarrow u = \dfrac{1}{5} \text{ and } v = \dfrac{1}{4}.$

Now we have u = $\dfrac{1}{5} \text{ and } v = \dfrac{1}{4}$,

$\Rightarrow \dfrac{1}{x + 1} = u \\[1em] \Rightarrow \dfrac{1}{x + 1} = \dfrac{1}{5} \\[1em] \Rightarrow x + 1 = 5 \\[1em] \Rightarrow x = 5 - 1 = 4 \\[1em] \Rightarrow \dfrac{1}{y - 1} = v \\[1em] \Rightarrow \dfrac{1}{y - 1} = \dfrac{1}{4} \\[1em] \Rightarrow y - 1 = 4 \\[1em] \Rightarrow y = 4 + 1 = 5.$

Hence, x = 4, y = 5.