Originally Posted by

**Quacky** We have:

$\displaystyle \frac{1}{3x}+\frac{14}{3}=2x+2$

Subtracting $\displaystyle \frac{1}{3x}$ from both sides:

$\displaystyle \frac{14}{3}=2x-\frac{1}{3x}+2$

Subtracting $\displaystyle 2$ from both sides:

$\displaystyle \frac{14}{3}-2=2x-\frac{1}{3x}$

We can simplify the left hand side by writing $\displaystyle 2$ as $\displaystyle \frac{6}{3}$, giving:

$\displaystyle \frac{14}{3}-\frac{6}{3}=2x-\frac{1}{3x}$

$\displaystyle \frac{8}{3}=2x-\frac{1}{3x}$

Now, to solve this, we could, as you've suggested, multiply through by x, as long as we remember that $\displaystyle x\neq{0}$.

$\displaystyle \frac{8x}{3}=2x^2-\frac{1}{3}$

I'd then multiply through by 3 for simplicity, giving:

$\displaystyle 8x=6x^2-1$

$\displaystyle 6x^2-8x-1=0$

Go for it!