Question

The only directions I was given was to solve the system. I am really unclear on how to do this. I would love to learn the process for this so I can better understand how to do this with similar problems. y = -x^2 - 2x + 8y = x^2 - 8x - 12Looking for clear, step by step instructions on the process.

Answer 1

`\begin{gathered} y=-x^2-2x+8 \\ y=x^2-8x-12 \end{gathered}`

start by making both expressions equal

`x^2-8x-12=-x^2-2x+8`

bring all to one side

`\begin{gathered} x^2+x^2-8x+2x-12-8=0 \\ 2x^2-6x-20=0 \end{gathered}`

use the quadratic formula to solve the quadratic equation using a as 2, b as -6, and c as -20.

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

then,

`\begin{gathered} x=\frac{-(-6)\pm\sqrt[]{(-6)^2-4\cdot(2)(-20)}}{2\cdot2} \\ x=\frac{6\pm\sqrt[]{36+160}}{4} \\ x=\frac{6\pm\sqrt[]{196}}{4} \\ x=(6\pm14)/(4) \\ \end{gathered}`

since the expression is a quadratic equation there are going to be two possible solutions

`\begin{gathered} x_1=(6+14)/(4) \\ x_1=(20)/(4)=5 \\ x_2=(6-14)/(4) \\ x_2=(-8)/(4)=-2 \end{gathered}`

then, replace both -2 and 5 into one of the equations

`\begin{gathered} \text{for x=-2} \\ y=-(-2)^2-2(-2)+8 \\ y=-4+4+8 \\ y=8 \\ \text{for x=5} \\ y=5^2-8\cdot5-12 \\ y=25-40-12 \\ y=-27 \end{gathered}`

the solutions to the system are (-2,8) and (5,-27)