Solve the equation:

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

The answers are 2,3 but I can't work out a method.

Could really do with some help.

Thanks

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- Jun 15th 2009, 02:14 PMgreghunterAlgebra help
Solve the equation:

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

The answers are 2,3 but I can't work out a method.

Could really do with some help.

Thanks - Jun 15th 2009, 02:25 PMyeongil
Multiply both sides by $\displaystyle 2(x - 4)(x + 3)(x - 1)$:

$\displaystyle \begin{aligned}

14(x - 1) - 8(x - 4) &= -3(x - 4)(x + 3)(x - 1) \\

14x - 14 - 8x + 32 &= -3x^3 + 6x^2 + 33x - 36 \\

6x + 18 &= -3x^3 + 6x^2 + 33x - 36 \\

0 &= -3x^3 + 6x^2 + 27x - 54 \\

0 &= -3(x^3 - 2x^2 - 9x + 18) \\

0 &= -3[x^2(x - 2) - 9(x - 2)] \\

0 &= -3(x - 2)(x^2 - 9) \\

0 &= -3(x - 2)(x + 3)(x - 3)

\end{aligned}$

You have solutions of x = 2, x = -3, and x = 3. Reject the solution x = -3.

**x = 2, x = 3**

01 - Jun 15th 2009, 07:10 PMShowcase_22
Why did you reject $\displaystyle x=-3$?

Does the question require all positive values of x? - Jun 15th 2009, 07:19 PMKrizalid
it ain't obvious? are those ratios defined when $\displaystyle x=-3$ ?

- Jun 16th 2009, 12:25 AMgreghunter
- Jun 16th 2009, 06:25 AMShowcase_22
Oh yeah!!

Sorry! =S