1. Some binomial fractional equations

I have been presented with the following equation:
$\frac{x}{x + 3} + \frac{1}{x - 3} = 1$

According to the text, the solution is 6. I have been unable to reach this answer.

Here is the method I've used:
The LCD of $\frac{x}{x + 3} + \frac{1}{x - 3} = 1$ is $(x + 3)(x - 3)$.
I multiply each side of the equation by the LCD and get the following:
$x(x - 3) + x + 3 = (x - 3)(x + 3)$
which is then simplified into
$x^2 - 3x + x + 3 = (x - 3)(x + 3)$
$x^2 -2x + 3 = (x - 3)(x + 3)$

From there I try to factor the left side of the equation, but it turns out being prime after using the FOIL method.
$(x+1)(x-3) \neq x^2 -2x + 3$

Yet somehow my textbook lists the solution as 6. What am I doing wrong?
I encounter the same situation in another similar equation who's solution set is supposed to be 2, 3 :
$\frac{6}{t} - \frac{2}{t-1} = 1$
Again, I determine the LCD is $t(t-1)$ and multiply each side by this.
$6(t-1) - 2t = t(t-1)$
simplified to
$6t-6 - 2t = t^2-t$
$0 = t^2-5t+6$

I once again try to factor, but just as before
$(t+1)(t -6) \neq t^2-5t+6$

The solution set is supposedly 2, 3. Grrrr...

2. $\begin{gathered}
x\left( {x - 3} \right) + \left( {x + 3} \right) = \left( {x - 3} \right)\left( {x + 3} \right) \hfill \\
x^2 - 2x + 3 = x^2 - 9 \hfill \\
\end{gathered}$

3. D'oh

Amusingly enough I just realized that my had professor advised me a few weeks ago that when in doubt, thou shall FOIL ... but what of the second equation?

4. Originally Posted by Darlyn
D'oh

Amusingly enough I just realized that my had professor advised me a few weeks ago that when in doubt, thou shall FOIL ... but what of the second equation?
what second equation? you just want to solve the last equation that Plato posted

5. For you second equation:

Where you end up with $t^2 - 5t + 6 = 0$

You should be able to see the factorisation is $(t - 3)(t - 2) = 0$

This of course giving the required answers...