ax+by =a2 , bx +ay=b2 . …

We can write the given system of  equations  as
ax + by - a² = 0........(1)
bx + ay - b² = 0........(2)
From equation (1) & (2), we have
a₁ = a, b₁ = b, c₁ = -a²
a₂ = b, b₂ = a, and c₂ = -b²
Therefore, by cross-multiplication, we get
{tex}\Rightarrow \frac{x}{{b \times ( - {b^2}) - ( - {a^2}) \times a}}{/tex}{tex} = \frac{{ - y}}{{a \times ( - {b^2}) - ( - {a^2}) \times b}}{/tex}{tex}= \frac{1}{{a \times a - b \times b}}{/tex}
{tex}\Rightarrow \frac{x}{{ - {b^3} + {a^3}}} = \frac{{ - y}}{{ - a{b^2} + {a^2}b}} = \frac{1}{{{a^2} - {b^2}}}{/tex}
Now,
{tex}\frac{x}{{ - {b^3} + {a^3}}} = \frac{1}{{{a^2} - {b^2}}}{/tex}
{tex}\Rightarrow x = \frac{{{a^3} - {b^3}}}{{{a^2} - {b^2}}}{/tex}
{tex} = \frac{{(a - b)\left( {{a^2} + ab + {b^2}} \right)}}{{(a - b)(a + b)}}{/tex}
{tex} = \frac{{{a^2} + ab + {b^2}}}{{a + b}}{/tex}
also
{tex}\frac{{ - y}}{{ - a{b^2} + {a^2}b}} = \frac{1}{{{a^2} - {b^2}}}{/tex}
{tex}\Rightarrow - y = \frac{{{a^2}b - a{b^2}}}{{{a^2} - {b^2}}}{/tex}
{tex}\Rightarrow y = \frac{{a{b^2} - {a^2}b}}{{{a^2} - {b^2}}}{/tex}
{tex} = \frac{{ab(b - a)}}{{(a - b)(a + b)}}{/tex}
{tex} = \frac{{ - ab(a - b)}}{{(a - b)(a + b)}}{/tex}
{tex} = \frac{{ - ab}}{{a + b}}{/tex}
Therefore, {tex}x = \frac{{{a^2} + ab + {b^2}}}{{a + b}},\;y = \frac{{ - ab}}{{a + b}}{/tex} is the solution of the given system of the equations.