Question

$x + y = 2a {}^{2} \\ x(a - b + \frac{ab}{a - b}) = y(a + b - \frac{ab}{a + b})$
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Answer 1

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

Given pair of linear equations are

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

and

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

can be rewritten as

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

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

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

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

$\rm :\longmapsto\:x( {a}^{3} + {b}^{3}) = y( {a}^{3} - {b}^{3})$

$\red{\bigg \{ \because \tt \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} + {y}^{2} - xy \bigg \}}$

$\red{\bigg \{ \because \tt \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + {y}^{2} + xy \bigg \}}$

$\rm :\longmapsto\:x( {a}^{3} + {b}^{3}) + y( {b}^{3} - {a}^{3}) = 0 - - - (2)$

Now, we have 2 equation in simplified form as

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

and

$\rm :\longmapsto\:x( {a}^{3} + {b}^{3}) + y( {b}^{3} - {a}^{3}) = 0 - - - (2)$

Now, using cross multiplication method, we have

$\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf 2 & \bf 3 & \bf 1& \bf 2\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf 1 & \sf {2a}^{2} & \sf 1 & \sf 1\\ \\ \sf {b}^{3} - {a}^{3} & \sf 0 & \sf {a}^{3} + {b}^{3} & \sf {b}^{3} - {a}^{3} \\ \end{array}} \\ \end{gathered}$

So, now

$\rm :\longmapsto\:\dfrac{x}{0 - {2a}^{2}( {b}^{3} - {a}^{3})} = \dfrac{y}{ {2a}^{2} ( {a}^{3} + {b}^{3}) - 0} = \dfrac{ - 1}{ {b}^{3}-{a}^{3} -{a}^{3}-{b}^{3}}$

$\rm :\longmapsto\:\dfrac{x}{{2a}^{2}( {a}^{3} - {b}^{3})} = \dfrac{y}{ {2a}^{2} ( {a}^{3} + {b}^{3} )} = \dfrac{ - 1}{-2{a}^{3}}$

$\rm :\longmapsto\:\dfrac{x}{{2a}^{2}( {a}^{3} - {b}^{3})} = \dfrac{y}{ {2a}^{2} ( {a}^{3} + {b}^{3} )} = \dfrac{ 1}{2{a}^{3}}$

Taking first and third member, we have

$\rm :\longmapsto\:\dfrac{x}{{2a}^{2}( {a}^{3} - {b}^{3})} = \dfrac{ 1}{2{a}^{3}}$

$\bf\implies \:x = \dfrac{ {a}^{3} - {b}^{3} }{a}$

Taking second and third member, we have .

$\rm :\longmapsto\:\dfrac{y}{ {2a}^{2} ( {a}^{3} + {b}^{3} )} = \dfrac{ 1}{2{a}^{3}}$

$\bf\implies \:y = \dfrac{ {a}^{3} + {b}^{3} }{a}$

$\green{\boxed{ \bf{ \: Verification}}}$

Consider, equation (1),

$\rm :\longmapsto\:x + y = {2a}^{2}$

On substituting the values of x and y, we have

$\rm :\longmapsto\:\dfrac{ {a}^{3} - {b}^{3} }{a} + \dfrac{ {a}^{3} + {b}^{3} }{a} = {2a}^{2}$

$\rm :\longmapsto\:\dfrac{ {a}^{3} - {b}^{3} + {a}^{3} + {b}^{3} }{a} = {2a}^{2}$

$\rm :\longmapsto\:\dfrac{ {a}^{3}+ {a}^{3} }{a} = {2a}^{2}$

$\rm :\longmapsto\:\dfrac{2 {a}^{3}}{a} = {2a}^{2}$

$\bf\implies \: {2a}^{2} = {2a}^{2}$

Hence, Verified