Question

Solve the following system of equations in $x$ and $y$,

$ax+by=1$ and $bx+ay=\frac{(a+b{)}^2}{a^2+b^2}-1$ or, $bx+ay=\frac{2ab}{a^2+b^2}$.

Answer 1

The given system of equations may be written as

$ax+by-1=0$

$bx+ay-\frac{2ab}{a^2+b^2}=0$

By cross-multiplication, we have

$\frac{x}{b\times -\frac{2ab}{a^2+b^2}-a\times -1}=\frac{-y}{a\times -\frac{2ab}{a^2+b^2}-b\times -1}=\frac{1}{a\times a-b\times b}$

$\Rightarrow \frac{x}{\frac{-2a{b}^2}{a^2+b^2}+a}=\frac{-y}{\frac{-2a^{2}b}{a^2+b^2}+b}=\frac{1}{a^2-b^2}$

$\Rightarrow \frac{x}{\frac{-2a{b}^2+a^3+a{b}^2}{a^2+b^2}}=\frac{-y}{\frac{-2a^{2}b+a^{2}b+b^3}{a^2+b^2}}=\frac{1}{a^2-b^2}$

$\Rightarrow \frac{x}{\frac{a^3-a{b}^2}{a^2+b^2}}=\frac{-y}{\frac{-a^{2}b+b^3}{a^2+b^2}}=\frac{1}{a^2-b^2}$

$\Rightarrow \frac{x}{\frac{a(a^2-b^2)}{a^2+b^2}}=\frac{-y}{\frac{b(a^2-b^2)}{a^2+b^2}}=\frac{1}{a^2-b^2}$

$\Rightarrow x=\frac{a(a^2-b^2)}{a^2+b^2}\times \frac{1}{a^2-b^2}$ and $y=\frac{b(a^2-b^2)}{a^2+b^2}\times \frac{1}{a^2-b^2}$

$\Rightarrow x=\frac{a}{a^2+b^2}$ and $y=\frac{b}{a^2+b^2}$

Hence, the solution of the given system of equations is $x=\frac{a}{a^2+b^2},y=\frac{b}{a^2+b^2}$.