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

{$(-1,-1),(\dfrac{1}{2},2)$}
Two equations, we have $xy=1$ and $y=2x+1$ From the second equation, we have $x(2x+1)=1$ or, $2x^2+x-1=0$ or, $(2x-1)(x+1)=0$ Thus, $x=${$-1,\dfrac{1}{2}$} From the second equation when $x=\dfrac{1}{2}$, we have $y=2(\dfrac{1}{2})+1$ or, $y=2$ and From the second equation when $x=-1$, we have $y=2(-1)+1$ or, $y=-1$ Hence, the solution set: {$(-1,-1),(\dfrac{1}{2},2)$}