Math Problem Solver

The system of equations has one solution.

\[x = 4, y = 2\]

Solve \(x^{2} + 2 y^{2} - 24 = 0\) for \(x\). \[x = - \sqrt{24 - 2 y^{2}}, x = \sqrt{24 - 2 y^{2}}\]Substitute \(- \sqrt{24 - 2 y^{2}}\) for \(x\) in \(x + y - 6 = 0\) and simplify. $$\begin{aligned}x + y - 6 &amp= 0 \\ \left(- \sqrt{24 - 2 y^{2}}\right) + y - 6 &= 0 \\ - y + \sqrt{24 - 2 y^{2}} + 6 &= 0 \end{aligned}$$Substitute \(- \sqrt{24 - 2 y^{2}}\) for \(x\) in \(x + y - 6 = 0\) and simplify. $$\begin{aligned}x + y - 6 &amp= 0 \\ \left(- \sqrt{24 - 2 y^{2}}\right) + y - 6 &= 0 \\ - y + \sqrt{24 - 2 y^{2}} + 6 &= 0 \end{aligned}$$Substitute \(\sqrt{24 - 2 y^{2}}\) for \(x\) in \(x + y - 6 = 0\) and simplify. $$\begin{aligned}x + y - 6 &amp= 0 \\ \left(\sqrt{24 - 2 y^{2}}\right) + y - 6 &= 0 \\ y + \sqrt{24 - 2 y^{2}} - 6 &= 0 \end{aligned}$$Substitute \(2\) into \(x^{2} + 2 y^{2} - 24 = 0\) to solve for \(x\). $$\begin{aligned}x^{2} - 24 + 2 \cdot 2^{2} &= 0 \\ x^{2} - 16 &= 0 \\x^{2} - 16 &= 0 \\ \left(x - 4\right) \left(x + 4\right) &= 0 \\ x = -4&, x = 4\end{aligned}$$This yields the following solution. $$\begin{aligned}x = 4,\,y = 2\end{aligned}$$