We have the following:
[tex]\begin{gathered} y=x^2+x-12 \\ y=-2x-2 \end{gathered}[/tex]
What we will do to solve the system in the right way, like this
[tex]\begin{gathered} y-y=x^2+x-12-(-2x-2) \\ 0=x^2+3x-10 \end{gathered}[/tex]
the student had two mistakes
[tex]\begin{gathered} x^2+3x=4x\rightarrow\text{ it cannot be added, they are not similar} \\ -2+12=14\rightarrow-2+12=10 \end{gathered}[/tex]
The correct way:
[tex]\begin{gathered} x^2+3x-10=0 \\ (x-2)\cdot(x+5)=0 \\ (x-2)=0\rightarrow x=2 \\ (x+5)=0\rightarrow x=-5 \end{gathered}[/tex]
now, for y:
[tex]\begin{gathered} y=-2x-2=-2\cdot2-2=-6 \\ y=-2x-2=-2\cdot-5-2=8 \end{gathered}[/tex]
The answer is
[tex]\begin{gathered} (2,6) \\ (-5,8) \end{gathered}[/tex]
