Chapter 9 - Systems and Matrices - Summary Exercises on Systems of Equations - Exercises - Page 907: 13

$\{(-5,14), (1,2)\}$

1. The second equation gives $y=4-2x$, use it in the first equation to get $4-2x+1=x^2+2x$ or $x^2+4x-5=0$ 2. Factor the equation to get $(x+5)(X-1)=0$, thus $x=-5, 1$ 3. We have $y=14, 2$ 4. Thus the solution set is $\{(-5,14), (1,2)\}$