Answer
{$(-1,-1),(\dfrac{1}{2},2)$}
Work Step by Step
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: {$(-1,-1),(\dfrac{1}{2},2)$}
