## Algebra 1: Common Core (15th Edition)

The solutions of the system are $(6, 10)$ and $(-7,192)$.
Given: $y=-14x +94$ $y=x^2-13x+52$ $\rightarrow -14x +94=x^2-13x+52$ $x^2+x-42=0$ $(x-6)(x+7)=0$ $x=6$ or $x=-7$ Solve for y: $y=-14(6) +94$ or $y=-14(-7) +94$ $y=10$ or $x=192$ The solutions of the system are $(6, 10)$ and $(-7,192)$.