Question

Solve the following using cross multiplication method :

$\frac{x}{a } + \frac{y}{b} = 2 \\ ax - by \: = {a}^{2} - {b}^{2}$
​

Answer 1

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

Given equation are

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

and

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

Now, Equation (1) can be rewritten as

$\rm :\longmapsto\:\dfrac{bx + ay}{ab} = 2$

$\rm :\longmapsto\:bx + ay = 2ab - - - (3)$

So, we have 2 equations as

$\purple{\rm :\longmapsto\:bx + ay = 2ab}$

and

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

So, using Cross multiplication method

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

Now, we have

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

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

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

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

$\bf\implies \:{ \boxed{ \bf{ \:x = a}}} \: \: \: and \: \: \:{ \boxed{ \bf{ \:y = b}}}$

Verification :-

Consider, Equation (1), we have

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

On substituting the values of x and y, we get

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

$\rm :\longmapsto\:1 + 1 = 2$

$\rm :\longmapsto\:2 = 2$

Hence, Verified

Consider Equation (2),

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

On substituting the values of x and y, we get

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

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

Hence, Verified