Answer
Infinitely many solutions with $(x,y)|x+2y-3=0 $ or $(x,y)|12=8y+4x$
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+2y-3=0 $ or $12=8y+4x$
Now, $-4x-8y+12=0 $
or, $4x+8y=12$
or, $12=12$
Hence, the given equation has Infinitely many solutions with $(x,y)|x+2y-3=0 $ or $(x,y)|12=8y+4x$
