# How do you solve the following linear system: x/a+y/b=2, bx-ay=0?

Dec 28, 2015

$\left\{\begin{matrix}x = a \\ y = b\end{matrix}\right.$

#### Explanation:

$\left\{\begin{matrix}\frac{1}{a} x + \frac{1}{b} y = 2 \\ b x - a y = 0\end{matrix}\right.$

Multiply the second equation by $\frac{1}{a b}$

$\left\{\begin{matrix}\frac{1}{a} x + \frac{1}{b} y = 2 \\ \frac{1}{a} x - \frac{1}{b} y = 0\end{matrix}\right.$

Add the second equation to the first

$\frac{1}{a} x + \frac{1}{b} y + \left(\frac{1}{a} x - \frac{1}{b} y\right) = 2 + 0$

$\implies \frac{2}{a} x = 2$

$\implies x = a$

Substitute this back into $\frac{1}{a} x + \frac{1}{b} y = 2$

$\implies \frac{1}{a} a + \frac{1}{b} y = 2$

$\implies 1 + \frac{1}{b} y = 2$

$\implies \frac{1}{b} y = 1$

$\implies y = b$

Thus we have the solution

$\left\{\begin{matrix}x = a \\ y = b\end{matrix}\right.$