1. ## 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

2. Originally Posted by greghunter
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}
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 $x=-3$?

Does the question require all positive values of x?

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

5. Originally Posted by yeongil
Multiply both sides by $2(x - 4)(x + 3)(x - 1)$:
\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