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Math Help - System of Equations HELP

  1. #1
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    System of Equations HELP

    Solve the system:

    x^2 + y^2 = 100
    -x + y = 2

    I don't know where to start because my teacher never taught us how to solve a system of equations where the variables are squared. If someone could show me how to do this then that'd be great. THANKS!
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  2. #2
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    Re: System of Equations HELP

    Write y in terms of x (from the second equation), substitute it into the first equation, then solve the quadratic.
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  3. #3
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    Re: System of Equations HELP

    Okay. I got:

    x^2 + y^2 = 100
    -x + y = 2 ------> y = x + 2

    x^2 + (x + 2)^2 = 100
    x^2 + x^2 + 4x + 4 = 100
    2x^2 + 4x = 96
    2x^2 + 4x - 96 = 0
    (2x + 12)(x - 8)

    2x + 12 = 0
    2x = -12
    x = -6

    x - 8 = 0
    x = 8

    So from here would I plug both x-values in to my 'y = x + 2' equation? Thus giving me 2 solutions to the system of equations? Idk if I'm doing this right.
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  4. #4
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    Re: System of Equations HELP

    Quote Originally Posted by Prove It View Post
    Write y in terms of x (from the second equation), substitute it into the first equation, then solve the quadratic.
    Okay. I got:

    x^2 + y^2 = 100
    -x + y = 2 ------> y = x + 2

    x^2 + (x + 2)^2 = 100
    x^2 + x^2 + 4x + 4 = 100
    2x^2 + 4x = 96
    2x^2 + 4x - 96 = 0
    (2x + 12)(x - 8)

    2x + 12 = 0
    2x = -12
    x = -6

    x - 8 = 0
    x = 8

    So from here would I plug both x-values in to my 'y = x + 2' equation? Thus giving me 2 solutions to the system of equations? Idk if I'm doing this right.
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  5. #5
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    Re: System of Equations HELP

    (2x+12)(x-8) = 2x^2+12x-16x-96 = 2x^2-4x-96, so you have the signs backwards. It should be 2(x-6)(x+8) = 0 (note: I factored out the 2). Hence x = 6, x = -8. Now, plug those values in for y = x+2.

    Finally, plug in both sets of values you found for x and y to the first equation to make sure it is correct.
    Thanks from StephenSkinner
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  6. #6
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    Re: System of Equations HELP

    Quote Originally Posted by SlipEternal View Post
    (2x+12)(x-8) = 2x^2+12x-16x-96 = 2x^2-4x-96, so you have the signs backwards. It should be 2(x-6)(x+8) = 0 (note: I factored out the 2). Hence x = 6, x = -8. Now, plug those values in for y = x+2.

    Finally, plug in both sets of values you found for x and y to the first equation to make sure it is correct.
    So therefore the solutions would be: (6, 8) & (-8, -6).

    Thanks so much!
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