Algebra help

greghunter · #1 $Old$ June 15th, 2009, 04:14 PM

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

yeongil · #2 $Old$ June 15th, 2009, 04:25 PM

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 $2(x - 4)(x + 3)(x - 1)$ :
$\begin{aligned}14(x - 1) - 8(x - 4) &= -3(x - 4)(x + 3)(x - 1) \\14x - 14 - 8x + 32 &= -3x^3 + 6x^2 + 33x - 36 \\6x + 18 &= -3x^3 + 6x^2 + 33x - 36 \\ 0 &= -3x^3 + 6x^2 + 27x - 54 \\ 0 &= -3(x^3 - 2x^2 - 9x + 18) \\0 &= -3[x^2(x - 2) - 9(x - 2)] \\0 &= -3(x - 2)(x^2 - 9) \\0 &= -3(x - 2)(x + 3)(x - 3)\end{aligned}$
You have solutions of x = 2, x = -3, and x = 3. Reject the solution x = -3.
x = 2, x = 3

01

Showcase_22 · #3 $Old$ June 15th, 2009, 09:10 PM

Why did you reject $x=-3$ ?

Does the question require all positive values of x?

Krizalid · #4 $Old$ June 15th, 2009, 09:19 PM

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

greghunter · #5 $Old$ June 16th, 2009, 02:25 AM

Quote:

Originally Posted by yeongil $View Post$

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!

Showcase_22 · #6 $Old$ June 16th, 2009, 08:25 AM

Oh yeah!!

Sorry! =S