Answer
{$(1,2),(9,6)$}
Work Step by Step
Two equations, we have $y^2=4x$ and $x=2y-3$
From the second equation, we have
$y^2=4(2y-3)$
or, $y^2-8y+12=0$
or, $(y-6)(y-2)=0$
Thus, $y=${$2,6$}
From the second equation when $y=2$, we have
$x-2(2)+3=0$
or, $x=1$
and
From the second equation when $y=6$, we have
$x-2(6)+3=0$
or, $x=9$
Hence, the solution set: {$(1,2),(9,6)$}
