## Big Ideas Math - Algebra 1, A Common Core Curriculum

The solutions are $(-3+\sqrt{15},6-2\sqrt{15})$ and $(-3-\sqrt{15},6+2\sqrt{15})$.
The given system is $y+2x=0$ ...... (1) $y=x^2+4x-6$ ...... (2) Subtract equation (1) from equation (2). $\Rightarrow y-(y+2x)=x^2+4x-6$ $\Rightarrow y-y-2x=x^2+4x-6$ Simplify. $\Rightarrow -2x=x^2+4x-6$ Add $2x$ to each side. $\Rightarrow -2x+2x=x^2+4x-6+2x$ Simplify. $\Rightarrow 0=x^2+6x-6$ Add $15$ to each side. $\Rightarrow 0+15=x^2+6x-6+15$ Simplify. $\Rightarrow 15=x^2+6x+9$ Write the right terms as square of the polynomial. $\Rightarrow 15=(x+3)^2$ Take square root on each side. $\Rightarrow \pm\sqrt{15}=x+3$ Subtract $3$ from each side. $\Rightarrow \pm\sqrt{15}-3=x+3-3$ Simplify. $\Rightarrow -3\pm\sqrt{15}=x$ Substitute $-3+\sqrt{15}$ for $x$ in equation (1). $\Rightarrow y+2(-3+\sqrt{15})=0$ Simplify. $\Rightarrow y-6+2\sqrt{15}=0$ Solve for $y$. $\Rightarrow y=6-2\sqrt{15}$ Substitute $-3-\sqrt{15}$ for $x$ in equation (1). $\Rightarrow y+2(-3-\sqrt{15})=0$ Simplify. $\Rightarrow y-6-2\sqrt{15}=0$ Solve for $y$. $\Rightarrow y=6+2\sqrt{15}$ Hence, the solutions are $(-3+\sqrt{15},6-2\sqrt{15})$ and $(-3-\sqrt{15},6+2\sqrt{15})$.