A. y = x^2 + 2x B. y= 3x +20 Substitute 3x +20 for y in Equation A.
A. (3x + 20) = x^2 + 2x Set equation A up equal to 0 by
-3x -20 -3x -20 subtracting 3x and 20 from both sides.
A. 0 = x^2 - x -20 Factor the quadratic equation and solve.
A. 0 = (x - 5)(x + 4)
A. x - 5= 0 x + 4= 0
+5 +5 -4 -4
A. x = 5, x = -4 Now substitute the values for x back into the equations to find the values for y, and your solutions
y = (5)^2 + 2(5) y = 3(5) +20 y = (-4)^2 + 2(-4) y = 3(-4) +20
y= 25 +10 y = 15 +20 y = 16 + (-8) y = -12 +20
y = 35 y = 35 y = 8 y = 8
So your solutions are (5, 35) and (-4, 8).
