Math Problem Solver

The system of equations has \(2\) solutions.

\[x = -2, y = -3\]\[x = 2, y = 5\]

Solve \(y = 2 x + 1\) for \(x\). \[x = \frac{y}{2} - \frac{1}{2}\]Substitute \(\frac{y}{2} - \frac{1}{2}\) for \(x\) in \(y = x^{2} + 2 x - 3\) and simplify. $$\begin{aligned}y &= x^{2} + 2 x - 3 \\ y &= \left(\frac{y}{2} - \frac{1}{2}\right)^{2} + 2 \left(\frac{y}{2} - \frac{1}{2}\right) - 3 \\ y^{2} - 2 y &= 15 \\y^{2} - 2 y - 15 &= 0 \\ \left(y - 5\right) \left(y + 3\right) &= 0 \\ y = -3&, y = 5\end{aligned}$$Substitute \(-3\) into \(y = 2 x + 1\) to solve for \(x\). \[\begin{aligned}-3 &= 2 x + 1\\- 2 x &= 4\\x &= -2\end{aligned}\]This yields the following solution. $$\begin{aligned}x = -2,\,y = -3\end{aligned}$$Substitute \(5\) into \(y = 2 x + 1\) to solve for \(x\). \[\begin{aligned}5 &= 2 x + 1\\- 2 x &= -4\\x &= 2\end{aligned}\]This yields the following solution. $$\begin{aligned}x = 2,\,y = 5\end{aligned}$$