Question

Solve for $x$ and $y$, if $\left(\begin{array}{c}{x}^2\\ y^2\end{array}\right)+3\left(\begin{array}{c}2x\\ -y\end{array}\right)=\left(\begin{array}{c}-9\\ 4\end{array}\right)$

Answer 1

Given, $\left(\begin{array}{c}{x}^2\\ y^2\end{array}\right)+3\left(\begin{array}{c}2x\\ -y\end{array}\right)=\left(\begin{array}{c}-9\\ 4\end{array}\right)$
$\Rightarrow \left(\begin{array}{c}{x}^2\\ y^2\end{array}\right)+\left(\begin{array}{c}6x\\ -3y\end{array}\right)=\left(\begin{array}{c}-9\\ 4\end{array}\right)$
$\Rightarrow \left(\begin{array}{c}{x}^2+6x\\ y^2-3y\end{array}\right)=\left(\begin{array}{c}-9\\ 4\end{array}\right)$
$\Rightarrow x^2+6x=-9$
$\Rightarrow x^2+6x+9=0$
$\Rightarrow (x+3)(x+3)=0$
$\Rightarrow x=-3,-3$
$\Rightarrow y^2-3y=4$
$\Rightarrow y^2-3y-4=0$
$\Rightarrow (y+1)(y-4)=0$
$\Rightarrow y=-1,4$
Thus, $x=-3,-3,y=-1,4$