Question

ax+by=a^2; bx+ay=b^2 solve it by cross multiplication method​

Answer 1

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

Given pair of equations are

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

and

$\rm :\longmapsto\:bx + ay = {b}^{2}$

Now, using Cross multiplication method, we have

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

Now, we have

$\rm :\longmapsto\:\dfrac{x}{ {b}^{3} - {a}^{3} } = \dfrac{y}{ {ba}^{2} - {ab}^{2} } = \dfrac{ - 1}{ {a}^{2} - {b}^{2} }$

can be rewritten as

$\rm :\longmapsto\:\dfrac{x}{ {b}^{3} - {a}^{3} } = \dfrac{y}{ {ba}^{2} - {ab}^{2} } = \dfrac{1}{ {b}^{2} - {a}^{2} }$

Also, we know that

$\boxed{ \bf{ \: {b}^{3} - {a}^{3} = (b - a)( {b}^{2} + ab + {a}^{2}}}$

and

$\boxed{ \bf{ \: {a}^{2} - {b}^{2} = (a - b)(a + b)}}$

So, using these, we get

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

Multiply each term by b - a, we get

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

Taking first and second member, we get

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

$\bf:\longmapsto\:x= \dfrac{ {b}^{2} + ab + {a}^{2} }{b + a}$

Taking second and third member, we get

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

$\bf :\longmapsto\: y = \dfrac{ - ab}{b + a}$

VERIFICATION :-

Consider first equation, we have

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

On substituting the values of a and b, we get

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

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

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

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

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

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

Hence, Verified