Answer
$(1,2), (9,6)$
Work Step by Step
We are given the system:
$\begin{cases}
x-2y+3=0\\
y^2=4x
\end{cases}$
We will use the substitution method. Solve Equation 1 for $x$ and substitute the expression of $x$ in Equation 2 to eliminate $x$ and determine $y$:
$\begin{cases}
x=2y-3\\
y^2=4(2y-3)
\end{cases}$
$y^2=4(2y-3)$
$y^2=8y-12$
$y^2-8y+12=0$
$y^2-2y-6y+12=0$
$y(y-2)-6(y-2)=0$
$(y-2)(y-6)=0$
$y-2=0\Rightarrow y_1=2$
$y-6=0\Rightarrow y_2=6$
Substitute each of the values of $y$ in the expression of $x$ to determine $x$:
$x=2y-3$
$y_1=2\Rightarrow x_1=2(2)-3=1$
$y_2=6\Rightarrow x_2=2(6)-3=9$
The system's solutions are:
$(1,2), (9,6)$
