Chapter 5, Systems of Equations and Inequalities - Section 5.4 - Systems of Nonlinear Equations - 5.4 Exercises - Page 466: 21

$(-2,-2)$

Step 1. Rewrite the equations. $x=2y+2$ $y^2-x^2=2x+4$ Step 2. Substitute the first equation into the second one. $y^2-(2y+2)^2=2(2y+2)+4$ $y^2-4y^2-8y-4=4y+4+4$ $-3y^2-12y-12=0$ Divide by $-3$ $y^2+4y+4=0$ $(y+2)^2=0$ $y=-2$ Step 3. Solve for $x$. $x=2\cdot (-2)+2$ $x=-4+2$ $x=-2$ Step 4. Conclude the solution, The solution is $(-2,-2)$.