Answer
$(\frac{1}{a+b}, \frac{1}{a+b})$
Work Step by Step
Given system is-
$ax +by$ = $1$
i.e. $ax $ = $1-by$
i.e. $x $ = $\frac{1-by}{a}$ __ eq.1
$bx +ay$ = $1$ __ eq.2
Substituting for $x$ in eq.2 from eq.1
$b(\frac{1-by}{a})+ay$ = $1$
i.e. $b(\frac{1}{a}-\frac{by}{a})+ay$ = $1$
i.e. $\frac{b}{a}-\frac{b^{2}y}{a}+ay$ = $1$
i.e. $ay-\frac{b^{2}y}{a}$ = $1-\frac{b}{a}$
i.e. $y(a-\frac{b^{2}}{a})$ = $1-\frac{b}{a}$
i.e. $y(\frac{a^{2}-b^{2}}{a})$ = $\frac{a-b}{a}$
i.e. $y$ = $\frac{a(a-b)}{a(a^{2}-b^{2})}$
i.e. $y$ = $\frac{a(a-b)}{a(a^{2}-b^{2})}$, Given $a^{2}- b^{2}\ne 0$
i.e. $y$ = $\frac{a(a-b)}{a(a-b)(a+b)}$
i.e. $y$ = $\frac{1}{(a+b)}$
Substituting for $y$ in eq.2
$bx +a(\frac{1}{a+b})$ = $1$
i.e. $bx $ = $1-\frac{a}{a+b}$
i.e. $bx $ = $\frac{a+b-a}{a+b}$
i.e. $bx $ = $\frac{b}{a+b}$
i.e. $x $ = $\frac{1}{a+b}$
Thus $(\frac{1}{a+b}, \frac{1}{a+b})$ is the solution of given system.
