Question

x-y=a+bax+by=a2-b2by cross multiplication
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Answer 1

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

Given pair of linear equations are

$\rm :\longmapsto\:x - y = a + b$

and

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

So, using Cross multiplication method

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

So,

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

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

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

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

On multiply by a + b, we get

$\rm :\longmapsto\:\dfrac{x}{ - a} = \dfrac{y}{b} = - 1$

$\rm \implies\:\boxed{ \tt{ \: x = a \: }} \: \: and \: \: \boxed{ \tt{ \: y \: = - \: b \: }}$

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Verification :-

Consider first equation

$\rm :\longmapsto\:x - y = a + b$

On substituting the values of x and y, we get

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

$\rm :\longmapsto\:a + b = a + b$

Hence, Verified

Consider second equation

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

On substituting the values of x and y, we get

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

$\rm :\longmapsto\: {a}^{2} - {b}^{2} = {a}^{2} - {b}^{2}$

Hence, Verified