By cross multiplication method solve bx/a+ay/b=a2b2 …

The given pair of equations are:
$\frac{b}{a}x + \frac{a}{b}y = {a^2} + {b^2}$
So, $\frac{b}{a}x + \frac{a}{b}y -[ {a^2} + {b^2} ] = 0$ ...................(i)
And x + y = 2ab
x + y - 2ab = 0 ....................(ii)
Here,
${a_1} = \frac{b}{a},{b_1} = \frac{a}{b}$, c₁ = -(a² + b²)
a₂ = 1, b₂ = 1, c₂ = -(2ab)
By cross-multiplication method
$\begin{array}{l}\;\frac x{{\displaystyle\frac ab}\times-(2ab)\;-1\lbrack-(a^2\;+\;b^2)\rbrack}=\;\frac y{-(a^2\;+\;b^2)\;-{\displaystyle\frac ba}\lbrack\;-(2ab)\rbrack}=\;\frac1{{\displaystyle\frac ba}-{\displaystyle\frac ab}}\\\frac x{\displaystyle\frac{-2a^2b}b\;\;+(a^2\;+\;b^2)}=\;\frac y{-(a^2\;+\;b^2)\;+{\displaystyle\frac{2ab^2}a}}=\;\frac1{\displaystyle\frac{b^2\;-\;a^2}{ab}\;}\\\end{array}$
$\frac{x}{{{\frac ba - \frac ab} }} = \frac{{ - y}}{{ - {b^2} + {a^2}}} = \frac{1}{{\frac{{{b^2} - {a^2}}}{{ab}}}}$
$\frac{x}{{{b^2} - {a^2}}} = \frac{{ - y}}{{ - {b^2} + {a^2}}} = \frac{1}{{\frac{{{b^2} - {a^2}}}{{ab}}}}$
$\frac{x}{{{b^2} - {a^2}}} = \frac{1}{{\frac{{{b^2} - {a^2}}}{{ab}}}}$
$⇒ x = ab$
And, $\frac{{ - y}}{{ - {b^2} + {a^2}}} = \frac{1}{{\frac{{{b^2} - {a^2}}}{{ab}}}}$
$⇒ y = ab$
The solutions of the given pair of equations is  x= ab and y = ab .