x-3y=6 x+y=2 this is an substitution problem so first you can do is rewrite the problem by subjection one variable x=3y+6 then substitute this in the other proble x+y=2 (3y+6)+y=2 4y+6=2 4y=2-6 4y=-4 y=-1 then substitute the no. in the original equation. x=3y+6 x=3(-1)+6 x=-3+6 x=3 now you got the intercepts and you draw the line and check. it's in the IV quadrant
