# Algebra help

• June 15th 2009, 02:14 PM
greghunter
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
• June 15th 2009, 02:25 PM
yeongil
Quote:

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
• June 15th 2009, 07:10 PM
Showcase_22
Why did you reject $x=-3$?

Does the question require all positive values of x?
• June 15th 2009, 07:19 PM
Krizalid
it ain't obvious? are those ratios defined when $x=-3$ ?
• June 16th 2009, 12:25 AM
greghunter
Quote:

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!
• June 16th 2009, 06:25 AM
Showcase_22
Oh yeah!!

Sorry! =S