Question

Solving a General System of Equations Find xand yin

terms of aand b.

$\left\{\begin{array}{l}ax+by=0\\ a^{2}x+b^{2}y=1\end{array}\right.(a\ne 0,b\ne 0,a\ne b)$

Short answer

$x=-\frac{b}{a(b^2-ab)}$

$y=\frac{1}{b^2-ab}$

Step-by-step solution

Step 1. Write down the concept.

The given system of equation can be written as,

ax + by = 0 …..(1)

$a^{2}x+b^{2}y=1$ …...(2)

Multiplying (1) with a, we get

$a^{2}x+aby=0$ ……(3)

Subtracting (3) from (2) we get,

$b^{2}y-aby=1$

$y(b^2-ab)=1$

$y=\frac{1}{b^2-ab}$

Substituting y in (1) we get

$ax+b\left(\frac{1}{b^2-ab}\right)=0$

$ax=-\frac{b}{b^2-ab}$

$x=-\frac{b}{a(b^2-ab)}$