Chapter 10 - Section 10.3 - Solving Nonlinear Systems of Equations - Exercise Set - Page 624: 18

$(-2,1)$

$x= y^{2}-3$ Equation $(1)$ $x= y^{2}-3y$ Equation $(2)$ From Equation $(1)$ and Equation $(2)$ $y^{2}-3 = y^{2}-3y$ $y^{2}-3 - y^{2}+3y = 0$ $3y-3=0$ $3(y-1)=0$ $y-1=0$ $y=1$ Substitute $y$ value in Equation $(1)$ to get $x$ $x= y^{2}-3$ $x=1^{2}-3$ $x=1-3$ $x=-2$ $(-2,1)$ satisfy both the equations. So, $(-2,1)$ is the solution.