Math Problem Solver

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

\[x = -27, y = 31\]\[x = 3, y = 1\]

Solve \(x + y = 4\) for \(x\). \[x = 4 - y\]Substitute \(4 - y\) for \(x\) in \(4 x^{2} - 3 y^{2} = 33\) and simplify. $$\begin{aligned}4 x^{2} - 3 y^{2} &amp= 33 \\ 4 \left(4 - y\right)^{2} - 3 y^{2} &= 33 \\ y^{2} - 32 y &= -31 \\y^{2} - 32 y + 31 &= 0 \\ \left(y - 31\right) \left(y - 1\right) &= 0 \\ y = 1&, y = 31\end{aligned}$$Substitute \(1\) into \(x + y = 4\) to solve for \(x\). \[\begin{aligned}x + 1 &= 4\\x &= 3\end{aligned}\]This yields the following solution. $$\begin{aligned}x = 3,\,y = 1\end{aligned}$$Substitute \(31\) into \(x + y = 4\) to solve for \(x\). \[\begin{aligned}x + 31 &= 4\\x &= -27\end{aligned}\]This yields the following solution. $$\begin{aligned}x = -27,\,y = 31\end{aligned}$$