Question

Solve the following systems of equations.
$y^2-xy=12,x^2-xy=28$

Answer 1

$y^2-xy=12$

$\Rightarrow y(y-x)=12$

$\Rightarrow y(x-y)=-12⟶\left(1\right)$

and $x^2-xy=28$

$\Rightarrow x(x-y)=28⟶\left(2\right)$

$\frac{y(x-y)}{x(x-y)}=\frac{-12}{28}$

$\Rightarrow \frac{y}{x}=\frac{-3}{7}$

$y=\frac{-3}{7}x$

or $x=\frac{-7y}{3}$

Putting this value in equation $\left(1\right)$

$\Rightarrow y(\frac{-7y}{3}-y)=-12$

$\Rightarrow \frac{-7}{3}{y}^2-y^2=-12$

$7y^2+3y^2=36$

$10y^2=36$

$y^2=\frac{36}{10}$

$y=\frac{6}{\sqrt{10}}$

$\Rightarrow x=\frac{-7}{3}y$

$\frac{-7}{3}\times \frac{6^2}{\sqrt{10}}$

$\frac{-14}{\sqrt{10}}=x$

Hence, solved.