## Intermediate Algebra (6th Edition)

$(2,\sqrt 2)$
$y= \sqrt x$ Equation $(1)$ $x^{2}+y^{2}= 6$ Equation $(2)$ Substitute $y= \sqrt x$ in Equation $(2)$ we get, $x^{2}+y^{2}= 6$ $x^{2}+(\sqrt x)^{2}= 6$ $x^{2}+x= 6$ $x^{2}+x- 6=0$ By factoring, $(x-2)(x+3)=0$ $x=2$ or $x=-3$ Substitute $x$ values in Equation $(1)$ to get $y$, Let $x=2$ $y= \sqrt 2$ Let $x=-3$ $y= \sqrt (-3)$ is not a real number. $(2,\sqrt 2)$ satisfies both equations. So, $(2,\sqrt 2)$ is the only solution.