Solve the simultaneous equation by substitution?

Question

the two formulas are as follows

2x-2y=13

y=x^2-6x-13

Below are the steps I have managed to work out, please tell me if I have messed anything up, im stuck now and cant seem to get any further

2x-2(x^2-6x-13)=13

2x-2x^2+12x+26=13

14x-2x^2+26=13

Need help from here.......

Answer 1

2x-2y=13

y=x^2-6x-13

take this,

2x-2y=13

y=x - (13/2)------------[1]

y=x^2-6x-13------------[2]

equating...

x - (13/2) = x^2-6x-13

find x

substitue the value for each function [1] &[2]

Answer 2

Almost there...Since you have an x and an x^2 you now need to use the quadratic equation to find your solutions.

First set the equation equal to 0 by subtracting 13 from both sides

14x-2x^2+13=0

Rewrite as -2x^2+14x+13

a=-2

b=14

c=13

Quadratic formula:

-(b)±sqrt(b^2-4ac)/2a

-14±sqrt(196+104)/-4

x= -14±sqrt(300)/-4

So -14+sqrt(300)/-4 and -14-sqrt(300)/-4 are your 2 solutions

Answer 3

well combining like terms....-2x^2 + 14 x +13 = 0

using quadratic formula   x = -14 +/- SQRT {14^2 -4 ( -2)(13) }/-4

x = {7+/- 10SQRT(3)}/2

Answer 4

That's fine -- now you need to combine like terms and move everything over to the left hand side of the equation so you can set the equation equal to zero and solve it using the quadratic formula:

-2x² + 14x + 13 = 0

You recall the quadratic formula is:

x = (-b ± √ (b² - 4ac)) / 2a

Plugging in the appropriate numbers, you get:

x = (-14 ± √300) / -4, then pull 100 out of the square root:

x = (-14 ± 10√3) / -4, then simplify:

x = (-7 ± 5√3) / -2