Math Problem Solver

The system of equations has $2$ solutions.

$$x = -8, y = -4$$$$x = 2, y = 1$$

Solve $x y + 3 x = 8$ for $x$. $$x = \frac{8}{y + 3}$$Substitute $\frac{8}{y + 3}$ for $x$ in $x - 2 y = 0$ and simplify. $$\begin{aligned}x - 2 y &= 0 \\ \left(\frac{8}{y + 3}\right) - 2 y &= 0 \\ \frac{2 \left(- y \left(y + 3\right) + 4\right)}{y + 3} &= 0 \end{aligned}$$Substitute $-4$ into $x y + 3 x = 8$ to solve for $x$. $$\begin{aligned}- x &= 8\\x &= -8\end{aligned}$$This yields the following solution. $$\begin{aligned}x = -8,\,y = -4\end{aligned}$$Substitute $1$ into $x y + 3 x = 8$ to solve for $x$. $$\begin{aligned}4 x &= 8\\x &= 2\end{aligned}$$This yields the following solution. $$\begin{aligned}x = 2,\,y = 1\end{aligned}$$