\(y = x^{2} + 3x - 7\)

\(3x - y = -2\)

I'll plug the top into the bottom since it is already y=

\(3x - (x^{2} + 3x - 7) = -2\)

distribute the negative and get rid of parenthesis

\(3x - x^{2} - 3x + 7 = -2\)

condense \(-x^{2} + 7 = -2\)

re-write as quadratic with positive \(x^{2}\)

\(x^{2} - 9 = 0\)

solve for x

\(x = -3\),

\(x = 3\)

plug in each x to get its y

\(y = -7\),

\(y = 11\)

solutions: (-3,-7) and (3,11)