Question

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

Answer 1

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

Answer 2

just factor as the gentle man said!

Answer 3

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

Answer 4

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

Answer 5

i dont know how to factor

Answer 6

@xapproachesinfinity

Answer 7

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 \(a=1,\ b=-6,\ c=7\)

Answer 8

im still really confused

Answer 9

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?

Answer 10

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.

Answer 11

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

Answer 12

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

Answer 13

@campbell_st

Answer 14

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?

Answer 15

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...