# Thread: Algebra help

1. ## Algebra 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

2. Originally Posted by greghunter
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
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

3. Why did you reject $\displaystyle x=-3$?

Does the question require all positive values of x?

4. it ain't obvious? are those ratios defined when $\displaystyle x=-3$ ?

5. Originally Posted by yeongil
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
Thankyou very much!

6. Oh yeah!!

Sorry! =S