Answer
Infinitely many solutions with $(x,y)|x=4y-1 $ or $(x,y)|2x-8y=-2$
Work Step by Step
To solve the given problem we will have to plug the simplest equation into the other equation and then solve for the other variable.
$x=4y-1 $ and $2x-8y=-2$
Now, $2(4y-1)-8y=-2 $
or, $8y-2=8y-2$
or, $8y=8y$ or, $0=0$
Hence, the given equation has Infinitely many solutions with $(x,y)|x=4y-1 $ or $(x,y)|2x-8y=-2$