## A community for students. Sign up today

Here's the question you clicked on:

## anonymous one year ago find the solutions to the system y=x^2-2x-2 y=4x+5 a.(-1.1) (-7.-23) b.(-1.1) (7,33) c.(-1,33) (7,1) d. no solution

• This Question is Closed
1. campbell_st

well since both equations are equal to, you can equate them $4x + 5 = x^2 -2x -2$ which can be rewritten as $x^2 - 6x - 7 = 0$ now solve for x by factoring

2. xapproachesinfinity

just factor as the gentle man said!

3. xapproachesinfinity

$x^2-6x-7=(x+1)(x-7)=0$ $x+1=0~~or~~x-7=0$ and go on

4. xapproachesinfinity

once you get the x you plug it in one equation and look for y

5. anonymous

i dont know how to factor

6. anonymous

@xapproachesinfinity

7. whpalmer4

Then use the formula for the solutions of the quadratic $$ax^2 + bx + c = 0$$: $x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$ Your quadratic is $x^2-6x+7=0$so you have $$a=1,\ b=-6,\ c=7$$

8. anonymous

im still really confused

9. whpalmer4

Okay, you'll have to do a better job of describing your confusion. I gave you a formula and 3 numbers to plug into the formula. Are you able to do that?

10. whpalmer4

if you are hoping that I will just give you the answer instead of helping you through the process of finding it yourself, today is not your lucky day.

11. campbell_st

If you can't do this algebraically then you can graph the curves use this site https://www.desmos.com/calculator then find where the 2 curves intersect

12. anonymous

i'm using this website currently it's really helpful, i didn't even think of that, thank you!

13. anonymous

@campbell_st

14. whpalmer4

It is troubling to hear you say that you haven't learned this, when you clearly are expected to know it...this is all material you will use many, many times in the future, except perhaps in the unlikely case this is the last math class you ever take. Do you have a live teacher to ask for help on this material?

15. campbell_st

I agree with @whpalmer4 , I'm sure this work on solving simultaneous equations would be covered in your notes the original method I posted is basically substitution.... then you would need to solve the quadratic equation... anyway, the graphing method will work...

#### Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy