System of Equations Question

4x − y = 16
y = 2x2 − 16x + 32

Looking for the solution(s) here? I got (5,4) but I think that is wrong

There are several ways to do it. One function is quadratic and the other is a straight line. You can expect to have two solutions. Draw the lines and you'll see why. The solutions are the intersections of those lines.


Take the first equation
4x − y = 16

solve for y
y = 4x - 16

substitute y on the second eq.
4x - 16 = 2x^2 - 16x + 32

solve for x
2x^2 - 16x + 32 - 4x + 16 = 0
2x^2 - 20x + 48 = 0
x^2 -10x + 24 = 0

Factorize or use the quadratic formula
(x - 6) (x - 4) = 0
x-value of the first solution x = 6 (you still have to find y)
x-value of the second solution x = 4 (you still have to find the other y)

substitute those values on the first eq.
first take x1 and find y1
4(6) - 16 = y1
y1 = 8

now take x2, and find y2
4(4) - 16 = y2
y2 = 0

You have two solutions (two points).
(6, 8) and (4, 0)

If you think a solution is wrong, you can prove it's right or wrong by substituting the solution into the original equations and see if they are satisfied. In this case, there are two solutions so even if (5,4) is one of the solutions, you are missing the other one. The solutions are: (4, 0) and (6, 8)

y = 4x-16 = 2x² - 16x + 32
2x² - 20x + 48 = 0
x² - 10x + 24 = 0
(x-6)(x-4) = 0
x = 4, 6
y = 4*4-16 = 0
y = 6*4-16 = 8

(x,y) = (4, 0), (6, 8)

4x − y = 16
4x -16 = y
4(x-4) = y

y= 2x^2 - 16x+32
y= 2(x^2 - 8x+16)
y= 2(x-4)(x-4)

4(x-4) = y = 2(x-4)(x-4)
4 =2(x-4)
2=x-4
6=x

Now substitute for x in the 1st equation
4x − y = 16
4*6 - y = 16
-y = -8
y = 8

Answer (6,8)