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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

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  1. campbell_st
    • one year ago
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    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
    • one year ago
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    just factor as the gentle man said!

  3. xapproachesinfinity
    • one year ago
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    \[x^2-6x-7=(x+1)(x-7)=0\] \[x+1=0~~or~~x-7=0\] and go on

  4. xapproachesinfinity
    • one year ago
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    once you get the x you plug it in one equation and look for y

  5. anonymous
    • one year ago
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    i dont know how to factor

  6. anonymous
    • one year ago
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    @xapproachesinfinity

  7. whpalmer4
    • one year ago
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    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
    • one year ago
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    im still really confused

  9. whpalmer4
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    i'm using this website currently it's really helpful, i didn't even think of that, thank you!

  13. anonymous
    • one year ago
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    @campbell_st

  14. whpalmer4
    • one year ago
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    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
    • one year ago
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    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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