Solve the system y^2=16x x=y-4

Answer 1

y² = 16x and x = y - 4

So, y² = 16(y - 4)

y² - 16y + 64 = 0

(y - 8)² = 0

y = 8 and x = y - 4 = 4

The graphs intersect only at the point (4, 8).