Chapter 13 - Section 13.3 - Solving Nonlinear Systems of Equations - Exercise Set - Page 948: 31

$(3, \sqrt 3)$

$y=\sqrt x$ $x^2+y^2=12$ $y=\sqrt x$ $y^2=(\sqrt x)^2$ $y^2=x$ $x^2+y^2=12$ $x^2+x=12$ $x^2+x-12=12-12$ $x^2+x-12=0$ $(x+4)(x-3)=0$ $x+4=0$ $x+4-4=0-4$ $x=-4$ $x-3=0$ $x-3+3=0+3$ $x=3$ $x=-4$ $y=\sqrt x$ $y=\sqrt -4$ Since we want the square root of a negative number, we find that $x$ cannot equal $-4$. $x=3$ $y=\sqrt x$ $y=\sqrt 3$