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Math Help - Algebra help

  1. #1
    Junior Member
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    Algebra help

    Solve the equation:

    \frac {7}{(x-4)(x+3)} - \frac {4}{(x+3)(x-1)} = \frac {-3}{2}

    The answers are 2,3 but I can't work out a method.
    Could really do with some help.
    Thanks
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  2. #2
    Super Member
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    Quote Originally Posted by greghunter View Post
    Solve the equation:

    \frac {7}{(x-4)(x+3)} - \frac {4}{(x+3)(x-1)} = \frac {-3}{2}

    The answers are 2,3 but I can't work out a method.
    Could really do with some help.
    Thanks
    Multiply both sides by 2(x - 4)(x + 3)(x - 1):
    \begin{aligned}<br />
14(x - 1) - 8(x - 4) &= -3(x - 4)(x + 3)(x - 1) \\<br />
14x - 14 - 8x + 32 &= -3x^3 + 6x^2 + 33x - 36 \\<br />
6x + 18 &= -3x^3 + 6x^2 + 33x - 36 \\<br />
 0 &= -3x^3 + 6x^2 + 27x - 54 \\<br />
  0 &= -3(x^3 - 2x^2 - 9x + 18) \\<br />
0 &= -3[x^2(x - 2) - 9(x - 2)] \\<br />
0 &= -3(x - 2)(x^2 - 9) \\<br />
0 &= -3(x - 2)(x + 3)(x - 3)<br />
\end{aligned}
    You have solutions of x = 2, x = -3, and x = 3. Reject the solution x = -3.
    x = 2, x = 3


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  3. #3
    Super Member Showcase_22's Avatar
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    Why did you reject x=-3?

    Does the question require all positive values of x?
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    it ain't obvious? are those ratios defined when x=-3 ?
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  5. #5
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    Quote Originally Posted by yeongil View Post
    Multiply both sides by 2(x - 4)(x + 3)(x - 1):
    \begin{aligned}<br />
14(x - 1) - 8(x - 4) &= -3(x - 4)(x + 3)(x - 1) \\<br />
14x - 14 - 8x + 32 &= -3x^3 + 6x^2 + 33x - 36 \\<br />
6x + 18 &= -3x^3 + 6x^2 + 33x - 36 \\<br />
 0 &= -3x^3 + 6x^2 + 27x - 54 \\<br />
  0 &= -3(x^3 - 2x^2 - 9x + 18) \\<br />
0 &= -3[x^2(x - 2) - 9(x - 2)] \\<br />
0 &= -3(x - 2)(x^2 - 9) \\<br />
0 &= -3(x - 2)(x + 3)(x - 3)<br />
\end{aligned}
    You have solutions of x = 2, x = -3, and x = 3. Reject the solution x = -3.
    x = 2, x = 3


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    Thankyou very much!
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  6. #6
    Super Member Showcase_22's Avatar
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    Oh yeah!!

    Sorry! =S
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