Y=x2−3x−4 x=y+8
solution set
a: {(-1,0), (4,0)}
b: {(8,0), (0,8)}
c: {(0,0)}
d: {(2,-6)}
e: there is no real solution

Plug in x into first equation
Y=(y+8)^2-3 (y+8)-4
Y=y2+16y+64-3y-24-4
Y=y2+13y+36
Y=(y+9) (y+4)
Y =-9,-4 plug in -9 to find x n plug in -4 to find its x
X=y+8
X=-9+8 x=-1 Plug in -1 into first equation
Y=x2-3x-4
Y=-1^2+3-4
Y =1+3-4
Y=0
(-1,0) do the same for -4
X=y+8
X=-4+8
X=4
Y=x2-3x-4
Y=4^2-12-4
Y=16-16
Y=0
(4,0)
Solutions are (-1,0)(4,0)

Google it that might work