anonymous
  • anonymous
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
Mathematics
jamiebookeater
  • jamiebookeater
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campbell_st
  • 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
xapproachesinfinity
  • xapproachesinfinity
just factor as the gentle man said!
xapproachesinfinity
  • xapproachesinfinity
\[x^2-6x-7=(x+1)(x-7)=0\] \[x+1=0~~or~~x-7=0\] and go on

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xapproachesinfinity
  • xapproachesinfinity
once you get the x you plug it in one equation and look for y
anonymous
  • anonymous
i dont know how to factor
anonymous
  • anonymous
whpalmer4
  • 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 \(a=1,\ b=-6,\ c=7\)
anonymous
  • anonymous
im still really confused
whpalmer4
  • 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?
whpalmer4
  • 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.
campbell_st
  • 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
anonymous
  • anonymous
i'm using this website currently it's really helpful, i didn't even think of that, thank you!
anonymous
  • anonymous
whpalmer4
  • 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?
campbell_st
  • 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...

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