Answer
No solution
Work Step by Step
$x^2+y^2=4$
$x-y=4$
$x-y=4$
$x-y+y=4+y$
$x=4+y$
$x^2+y^2=4$
$(4+y)^2+y^2=4$
$(4+y)(4+y)+y^2=4$
$4*4+4*y+4*y+y^2+y^2=4$
$16+8y+2y^2=4$
$2y^2+8y+16=4$
$2y^2+8y+16-4=4-4$
$2y^2+8y+12=0$
$a=2$,$b=8$, $c=12$
$x=(-b±\sqrt {b^2-4ac})/2a$
$x=(-8±\sqrt {8^2-4*2*12})/2*2$
$x=(-8±\sqrt {64-96})/4$
$x=(-8±\sqrt {-32})/4$
Since we have the square root of a negative number, there is no solution to the system.
