Question

Solve the following systems of equations by using cross multiplication method. ax+by-(a-b)=0 ; bx-ay-(a-b)=0.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given pair of equations are

$\rm :\longmapsto\:ax + by - (a - b) = 0$

can be rewritten as

$\rm :\longmapsto\:ax + by = a - b - - - (1)$

and

$\rm :\longmapsto\:bx - ay - (a - b) = 0$

$\rm :\longmapsto\:bx - ay = a - b - - - (2)$

Now,

Using Cross Multiplication Method,

$\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf 2 & \bf 3 & \bf 1& \bf 2\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf b & \sf a - b & \sf a & \sf b\\ \\ \sf - a & \sf {a} - {b} & \sf b & \sf - a\\ \end{array}} \\ \end{gathered}$

So,

$\rm :\longmapsto\:\dfrac{x}{b(a - b) + a(a - b)} = \dfrac{y}{b(a - b) - a((a - b)} = \dfrac{ - 1}{ - {a}^{2} - {b}^{2} }$

$\rm :\longmapsto\:\dfrac{x}{(a - b)(a + b)} = \dfrac{y}{ - (a - b)(a - b)} = \dfrac{ 1}{{a}^{2} + {b}^{2} }$

$\rm :\longmapsto\:\dfrac{x}{ {a}^{2} - {b}^{2} } = \dfrac{y}{ - (a - b)^{2} } = \dfrac{ 1}{{a}^{2} + {b}^{2} }$

So,

Taking first and third member, we get

$\bf\implies \:x = \dfrac{ {a}^{2} - {b}^{2} }{ {a}^{2} + {b}^{2} }$

and

On taking second and third member, we get

$\bf\implies \:y = - \: \dfrac{ {(a - b)}^{2} }{ {a}^{2} + {b}^{2} }$

Additional Information : -

Solution by Elimination Method

$\rm :\longmapsto\:ax + by = a - b - - - (1)$

and

$\rm :\longmapsto\:bx - ay = a - b - - - (2)$

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

$\rm :\longmapsto\: {a}^{2} x + aby = {a}^{2} - ab - - - (3)$

and

$\rm :\longmapsto\: {b}^{2} x - aby = ab - {b}^{2} - - - (4)$

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

$\rm :\longmapsto\:x( {a}^{2} + {b}^{2}) = {a}^{2} - {b}^{2}$

$\bf\implies \:x = \dfrac{ {a}^{2} - {b}^{2} }{ {a}^{2} + {b}^{2} }$

On substituting the value of x in equation (1) we get

$\rm :\longmapsto\:a \times \dfrac{ {a}^{2} - {b}^{2} }{ {a}^{2} + {b}^{2} } + by = a - b$

$\rm :\longmapsto\:by = a - b - \dfrac{a(a + b)(a - b)}{ {a}^{2} +{b}^{2} }$

$\rm :\longmapsto\:by = (a - b)\bigg(1 - \dfrac{a(a + b)}{ {a}^{2} +{b}^{2} }\bigg)$

$\rm :\longmapsto\:by = (a - b)\bigg(\dfrac{ {a}^{2} + {b}^{2} - {a}^{2} - ab}{ {a}^{2} +{b}^{2} }\bigg)$

$\rm :\longmapsto\:by = (a - b)\bigg(\dfrac{ {b}^{2} - ab}{ {a}^{2} +{b}^{2} }\bigg)$

$\rm :\longmapsto\:by = b(a - b)\bigg(\dfrac{ {b} - a}{ {a}^{2} +{b}^{2} }\bigg)$

$\rm :\longmapsto\:y = (a - b)\bigg(\dfrac{ {b} - a}{ {a}^{2} +{b}^{2} }\bigg)$

$\bf\implies \:y = - \: \dfrac{ {(a - b)}^{2} }{ {a}^{2} + {b}^{2} }$