Answer
$(4,0),(0,-2)$
Work Step by Step
$y^{2}=4-x$ Equation $(1)$
$x-2y = 4 $ Equation $(2)$
From Equation $(2)$
$x = 4+2y $ Equation $(3)$
Substituting $x = 4+2y $ in Equation $(1)$
$y^{2}=4-x$
$y^{2}=4-(4+2y)$
$y^{2}=4-4-2y$
$y^{2}=-2y$
$y^{2}+2y = 0$
$y(y+2)=0$
$y=0$ or $y=-2$
Substitute $ y$ values in Equation $(3)$
Let $y=0$
$x = 4+2y $
$x = 4+0$
$x = 4$
Let $y=-2$
$x = 4+2(-2)$
$x=4-4$
$x=0$
$(4,0),(0,-2)$ both satisfies the equations $(1)$ and $(2)$.
So, solutions are $(4,0),(0,-2)$
