How do you solve $y = \frac{1}{2} x^{2} - 4, y = 3 x + 4$ $y=1/2x^2-4, y=3x+4$ using substitution?

Question

1688 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-04T01:22:49

Simply substitute the second equation into the first (both are solved for $y$ $y$ already).

$3 x + 4 = \frac{1}{2} x^{2} - 4$ $3x + 4= 1/2x^2 - 4$

Multiply by a factor of $2$ $2$ so that all terms are integers.

$6 x + 8 = x^{2} - 8$ $6x + 8 = x^2 - 8$

$0 = x^{2} - 6 x - 16$ $0 = x^2 - 6x - 16$

$0 = (x - 8) (x + 2)$ $0 = (x - 8)(x+ 2)$

$x = 8 and - 2$ $x = 8 and -2$

This will mean that $y = 3 (8) + 4 = 28$ $y = 3(8) + 4 = 28$ and $y = 3 (- 2) + 4 = - 2$ $y = 3(-2) + 4 = -2$

Therefore, the intersection points are $(8, 28)$ $(8, 28)$ and $(- 2, - 2)$ $(-2, -2)$ .

Hopefully this helps!

How do you solve y=12x2−4,y=3x+4y=1/2x^2-4, y=3x+4 using substitution?