Chapter 13 - Section 13.3 - Solving Nonlinear Systems of Equations - Exercise Set - Page 948: 11

${(1,-2), (3,6)}$

$y=x^2-3$ $4x-y=6$ $4x-y=6$ $4x-(x^2-3)=6$ $4x-x^2+3=6$ $-x^2+4x+3=6$ $-1*(-x^2+4x+3=6)$ $x^2-4x-3=-6$ $x^2-4x-3+6=-6+6$ $x^2-4x+3=0$ $(x-1)(x-3)=0$ $x-1=0$ $x-1+1=0+1$ $x=1$ $x-3=0$ $x-3+3=0+3$ $x=3$ $x=1$ $y=x^2-3$ $y=1^2-3$ $y=1-3$ $y=-2$ $x=3$ $y=x^2-3$ $y=3^2-3$ $y=9-3$ $y=6$