Answer:
Given Question:-
Solve the following pair of equations for x and y :
[tex]\rm :\longmapsto\:\dfrac{ {a}^{2} }{x} - \dfrac{ {b}^{2} }{y} = 0, \: \: \dfrac{ {a}^{2}b}{x} + \dfrac{ {b}^{2} a}{y} = a + b[/tex]
[tex] \red{\large\underline{\sf{Solution-}}}[/tex]
Given pair of equations are
[tex]\rm :\longmapsto\:\dfrac{ {a}^{2} }{x} - \dfrac{ {b}^{2} }{y} = 0 - - - - (1)[/tex]
and
[tex]\rm :\longmapsto\:\dfrac{ {a}^{2} b}{x} + \dfrac{ {b}^{2}a }{y} = a + b - - - - (2)[/tex]
On multiply equation (1) by a, we get
[tex]\rm :\longmapsto\:\dfrac{ {a}^{3} }{x} - \dfrac{ {b}^{2} a}{y} = 0 - - - (3)[/tex]
On adding equation (3) and (2), we get
[tex]\rm :\longmapsto\:\dfrac{ {a}^{3} }{x} + \dfrac{ {a}^{2} b}{x} = a + b [/tex]
[tex]\rm :\longmapsto\:\dfrac{ {a}^{3} + {a}^{2} b}{x} = a + b [/tex]
[tex]\rm :\longmapsto\:\dfrac{{a}^{2}(a + b)}{x} = a + b [/tex]
[tex] \\ \rm\implies \:\boxed{\tt{ \: \: x \: \: = \: \: {a}^{2} \: \: }} \\ [/tex]
On substituting the value of x in equation (1), we get
[tex]\rm :\longmapsto\:\dfrac{ {a}^{2} }{ {a}^{2} } - \dfrac{ {b}^{2} }{y} = 0[/tex]
[tex]\rm :\longmapsto\:1 - \dfrac{ {b}^{2} }{y} = 0[/tex]
[tex]\rm :\longmapsto\: - \dfrac{ {b}^{2} }{y} = - 1[/tex]
[tex]\rm :\longmapsto\: \dfrac{ {b}^{2} }{y} = 1[/tex]
[tex] \\ \rm\implies \:\boxed{\tt{ \: \: y \: \: = \: \: {b}^{2} \: \: }} \\ [/tex]
VERIFICATION:
Consider the first equation
[tex]\rm :\longmapsto\:\dfrac{ {a}^{2} }{x} - \dfrac{ {b}^{2} }{y} = 0[/tex]
On substituting the values of x and y, we get
[tex]\rm :\longmapsto\:\dfrac{ {a}^{2} }{ {a}^{2} } - \dfrac{ {b}^{2} }{ {b}^{2} } = 0[/tex]
[tex]\rm :\longmapsto\: 1 - 1 = 0[/tex]
[tex]\rm\implies \:0 = 0[/tex]
Hence, Verified
So, Solution of pair of equations is
[tex] \purple{\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\bf{x \: = \: {a}^{2} } \\ \\ &\bf{y \: = \: {b}^{2} } \end{cases}\end{gathered}\end{gathered}}[/tex]
