y = 5x + 35
When x = 0, we substitute x = 0 into the equation above and then solve for y
y = 5(0) + 35
y = 0+ 35
y = 35
when x =3, substitute into the equation and then solve for y
y = 5(3) + 35
y = 15 + 35
y = 50
when x=-7, substitute x=-7 into the equation above and then solve for y
y = 5(-7) + 35
y = -35+ 35
y=0
x y (x,y)
0 35 (0, 35)
3 50 (3, 50)
-7 0 (-7, 0)