## College Algebra (10th Edition)

$y=2,x=5$
Our equations according to the exercise are: $x+y=7, x^2-y^2=21$. So $x=7-y$, thus $(7-y)^2-y^2=21\\y^2-14y+49-y^2=21\\-14y+28=0\\-y+2=0\\y=2$ Thus $y=2,x=5$.