Invent a system and solve it with Cramer's rule

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Invent a linear system of two equations in two unknowns whose solution is $x=-3$ and $y=1$, then solve it using Cramer’s rule. Both equations must contain both $x$ and $y$ with non-zero coefficients.

Solution

Build two equations by requiring $(-3;1)$ to satisfy them. For instance, with $x+y$ and $x-y$: $x+y=-3+1=-2, \qquad x-y=-3-1=-4.$ System: $\begin{cases} x+y=-2\\ x-y=-4 \end{cases}$ Solution with Cramer. Determinant of the coefficients: $\det A=\begin{vmatrix} 1 & 1\\ 1 & -1\end{vmatrix}=-2.$ $x=\frac{\begin{vmatrix} -2 & 1\\ -4 & -1\end{vmatrix}}{-2}=\frac{2+4}{-2}=\frac{6}{-2}=-3,\qquad y=\frac{\begin{vmatrix} 1 & -2\\ 1 & -4\end{vmatrix}}{-2}=\frac{-4+2}{-2}=\frac{-2}{-2}=1.$ $\boxed{\begin{cases} x+y=-2\\ x-y=-4 \end{cases}\ \Rightarrow\ x=-3,\ y=1}$ (Any other system with non-zero determinant satisfied by $(-3;1)$ is equally correct.)