## Intermediate Algebra for College Students (7th Edition)

$(x,y)=${$(-3,-4),(4,3)$}
Re-write the second equation as: $x=y+1$ First equation yield: $(y+1)^2+y^2=25$ or, $y^2+y-12=0$ or, $(y+4)(y-3)=0$ or, $y=${$-4,3$} From first equation when $y=-4$, we have $x=-3$ From first equation when $y=3$, we have $x=4$ Thus, solution set is $(x,y)=${$(-3,-4),(4,3)$}