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NEXUS

When finding solutions with equations that include at least one quadratic, you may get multiple answers. It is no longer possible to eliminate, so we will only be using the substitution method.

\[\begin{align} y &= x^2 - 4 \quad \text{(quadratic equation)} \\ y &= 2x + 1 \quad \text{(linear equation)} \end{align}\]

Both equations are equal to \(y\) so get them equal to each other.

\[2x + 1 = x^2 - 4\]

Rearrange to solve.

\[x^2 - 2x - 5 = 0\]

Solve using the quadratic formula.

\[x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-5)}}{2(1)}\]

\[= \frac{2 \pm \sqrt{4 + 20}}{2}\]

\[= \frac{2 \pm \sqrt{24}}{2}\]

\[= \frac{2 \pm 2\sqrt{6}}{2}\]

\[= 1 \pm \sqrt{6}\]

\[x = 1 + \sqrt{6}\text{ and }1 - \sqrt{6}\]

Find the corresponding \(y\) values by substituting \(x\) back into the linear equation.

\[y = 2(1 + \sqrt{6}) + 1 = 3 + 2\sqrt{6}\]

and

\[y = 2(1 - \sqrt{6}) + 1 = 3 - 2\sqrt{6}\]

So the solutions are:

\[\left(1 + \sqrt{6}, 3 + 2\sqrt{6}\right) \text{ and } \left(1 - \sqrt{6}, 3 - 2\sqrt{6}\right)\]

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