Answer
When $\Rightarrow$ $x=4 , y=4$ or $x=1 , y=-2$
Work Step by Step
Given:
$y=2x-4$
$y^{2}=4x$
Substituting value of $y=2x-4$ in $y^{2}=4x$,
we get: $(2x-4)^{2}=4x$
Expanding,
$4x^{2}-16x+16=4x$
$\Rightarrow$ $4x^{2}-20x+16=0$
$\Rightarrow$ $x^{2}-5x+4=0$ (Taking 4 common from both sides)
$\Rightarrow$ $x^{2}-4x-x+4=0$ ....... (because $-5x=-4x-x)$
$\Rightarrow$ $x(x-4)-1(x-4)=0$
$\Rightarrow$ $(x-4)(x-1)=0$
$\Rightarrow$ $x=4$ or $x=1$
Thus, $y=2x-4$
When $x=4$ $\Rightarrow$ $y=8-4$
When $x=1$ $\Rightarrow$ $y=2-4$
$\Rightarrow$ $x=4 , y=4$ or $x=1 , y=-2$
