Question

Solve the following pairs of equations for x and y:

Answer 1

Answer:

Given Question:-

Solve the following pair of equations for x and y :

$\rm :\longmapsto\:\dfrac{ {a}^{2} }{x} - \dfrac{ {b}^{2} }{y} = 0, \: \: \dfrac{ {a}^{2}b}{x} + \dfrac{ {b}^{2} a}{y} = a + b$

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

Given pair of equations are

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

and

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

On multiply equation (1) by a, we get

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

On adding equation (3) and (2), we get

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

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

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

$\\ \rm\implies \:\boxed{\tt{ \: \: x \: \: = \: \: {a}^{2} \: \: }}$

On substituting the value of x in equation (1), we get

$\rm :\longmapsto\:\dfrac{ {a}^{2} }{ {a}^{2} } - \dfrac{ {b}^{2} }{y} = 0$

$\rm :\longmapsto\:1 - \dfrac{ {b}^{2} }{y} = 0$

$\rm :\longmapsto\: - \dfrac{ {b}^{2} }{y} = - 1$

$\rm :\longmapsto\: \dfrac{ {b}^{2} }{y} = 1$

$\\ \rm\implies \:\boxed{\tt{ \: \: y \: \: = \: \: {b}^{2} \: \: }}$

VERIFICATION:

Consider the first equation

$\rm :\longmapsto\:\dfrac{ {a}^{2} }{x} - \dfrac{ {b}^{2} }{y} = 0$

On substituting the values of x and y, we get

$\rm :\longmapsto\:\dfrac{ {a}^{2} }{ {a}^{2} } - \dfrac{ {b}^{2} }{ {b}^{2} } = 0$

$\rm :\longmapsto\: 1 - 1 = 0$

$\rm\implies \:0 = 0$

Hence, Verified

So, Solution of pair of equations is

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