a. At this point, we know three methods: Equal Values Mwthod, Substitution Method, and Elimination Method.

b. Since both equations have y isolated, we can use the Equal Values Method then solve for x: \(\displaystyle{x}^{{2}}-{3}{x}-{10}=-{2}{x}+{2}\)

Write in standart form: \(\displaystyle{a}{x}^{{2}}+{b}{x}+{c}={0}\)

\(\displaystyle{x}^{{2}}-{3}-{12}={0}\)

Factor the left side: \((x+3)(x-4)=0\)

By Zero Product Property, \(x+3=0\)

\(x=-3\)

\(x-4=0\)

\(x=4\)

Solve for the corresponding y-values. I used the first equation. When x=-3,

\(y=-2(-3)+2\)

\(y=6+2\)

\(y=8\)

When \(x=4\), \(y=-2(4)+2\)

\(y=-8+2\)

\(y=-6\)

So, the points of intersection are: (-3.8) and (4,-6)