Question

Solve this equations for x and y
x / a + y / b = a + b
x / a² + y / b² = 2 ​

Answer 1

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

Given pair of linear equations are

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

and

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

can be rewritten as

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

$\rm :\longmapsto\:bx + ay = {ba}^{2} + {ab}^{2} - - - (1)$

and

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

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

So, we have two equations as

$\rm :\longmapsto\:bx + ay = {ba}^{2} + {ab}^{2} - - - (1)$

and

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

Now, multiply equation (1), by b, we have

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

On Subtracting equation (3) from (2), we get

$\rm :\longmapsto\: {a}^{2}y - aby = {a}^{2} {b}^{2} - {ab}^{3}$

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

$\bf\implies \:y = {b}^{2}$

On substituting the value of y in equation (2), we get

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

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

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

$\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: Hence-\begin{cases} &\sf{x = {a}^{2} } \\ &\sf{y = {b}^{2} } \end{cases}\end{gathered}\end{gathered}$

Verification :-

Consider, Equation

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

On substituting the values of x and y, we get

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

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

Hence, Verified

Consider, Equation (2)

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

On substituting the values of x and y, we get

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

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

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

Hence, Verified