I have been presented with the following equation:

$\displaystyle \frac{x}{x + 3} + \frac{1}{x - 3} = 1$

According to the text, the solution is6. I have been unable to reach this answer.

Here is the method I've used:

The LCD of $\displaystyle \frac{x}{x + 3} + \frac{1}{x - 3} = 1$ is $\displaystyle (x + 3)(x - 3)$.

I multiply each side of the equation by the LCD and get the following:

$\displaystyle x(x - 3) + x + 3 = (x - 3)(x + 3)$

which is then simplified into

$\displaystyle x^2 - 3x + x + 3 = (x - 3)(x + 3)$

$\displaystyle x^2 -2x + 3 = (x - 3)(x + 3)$

From there I try to factor the left side of the equation, but it turns out being prime after using the FOIL method.

$\displaystyle (x+1)(x-3) \neq x^2 -2x + 3$

Yet somehow my textbook lists the solution as6. What am I doing wrong?

I encounter the same situation in another similar equation who's solution set is supposed to be 2, 3 :

$\displaystyle \frac{6}{t} - \frac{2}{t-1} = 1$

Again, I determine the LCD is $\displaystyle t(t-1)$ and multiply each side by this.

$\displaystyle 6(t-1) - 2t = t(t-1)$

simplified to

$\displaystyle 6t-6 - 2t = t^2-t$

$\displaystyle 0 = t^2-5t+6$

I once again try to factor, but just as before

$\displaystyle (t+1)(t -6) \neq t^2-5t+6$

The solution set is supposedly2, 3. Grrrr...