# How do you solve the system 2x - 3y = -11 and 3x + 2y = 29?

Jun 12, 2015

Let $M = \left(\begin{matrix}2 & 3 \\ - 3 & 2\end{matrix}\right)$. Then $\det \left(M\right) = {2}^{2} + {3}^{2} = 13$

$\frac{1}{13} \left(\begin{matrix}2 & 3 \\ - 3 & 2\end{matrix}\right) \left(\begin{matrix}- 11 \\ 29\end{matrix}\right) = \frac{1}{13} \left(\begin{matrix}65 \\ 91\end{matrix}\right) = \left(\begin{matrix}5 \\ 7\end{matrix}\right)$

$x = 5$ and $y = 7$

#### Explanation:

Since the second equation swaps the coefficients of $x$ and $y$ with exactly one sign change, we can tell that the second equation is of a line perpendicular to that described by the first.

Hence all we need to do to get $x$ and $y$ is to multiply $\left(\begin{matrix}- 11 \\ 29\end{matrix}\right)$ by a rotation matrix constructed from the coefficients of $x$ and $y$.

graph{(2x-3y+11)(3x+2y-29) = 0 [-16.18, 23.82, -4.16, 15.84]}