Question

how many solutions does the system have? Explain your reasoning. y = -x^2 + 8x + 18y = -2x^2 - 16x - 30

Answer 1

Let:

$\begin{gathered} y=-x^2+8x+18_{\text{ }}(1) \\ y=-2x^2-16x-30_{}(2) \end{gathered}$

Using elimination:

$\begin{gathered} (1)-(2) \\ y-y=-x^2-(-2x^2)+8x-(-16x)+18-(-30) \\ 0=x^2+24x+48 \\ \end{gathered}$

Subtract 48 from both sides:

$\begin{gathered} x^2+24x=-48 \\ \text{add 144 to both sides} \\ x^2+24x+144=96 \\ \text{Factor:} \\ (x+12)^2=96 \\ x=4\sqrt[]{6}-12\approx-2.202 \\ or \\ x=-12-4\sqrt[]{6}\approx-21.798 \end{gathered}$

Let's check the solutions:

$\begin{gathered} y=-(-2.202)^2+8(-2.202)+18=-4.465 \\ y=-2(-2.202)^2-16(-2.202)-30=-4.465 \end{gathered}$

The first solution satisfies the system, therefore, it is a valid solution

$\begin{gathered} y=-(-21.798)^2+8(-21.798)+18=-282.7688 \\ y=-2(-21.798)^2-16(-21.798)-30=-631.53 \\ -282.7688\ne-631.53 \end{gathered}$

This solution doesn't satisfy the system, therefore, it is not a valid solution.

Hence, the system have only one solution and it is:

$(2.202,-4.465)$