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