If , then

$$\frac{4}{3}x-\frac{1}{2}=0$$

I would do this question by multiplying both sides by the lowest common denominator which is 6.

That way you get rid of all the fractions right from the begining.  

$$\begin{array}{rll}
6\times \left(\frac{4}{3}x-\frac{1}{2}\right)&=&0 \times 6\\\\
\frac{6\times 4}{3}x-\frac{6}{2}&=&0\\\\
8x-3&=&0\\\\
8x&=&3\\\\
x&=&\frac{3}{8}\\\\
\end{array}$$

.

Here, we just need to get x alone, or in other words, x = something.

$$\begin{array}{l}
\frac{4}{3}x-\frac{1}{2}=0\qquad\mbox{Add $\frac{1}{2}$ to both sides}\\
\frac{4}{3}x=\frac{1}{2}\qquad\mbox{Divide both sides by $\frac{4}{3}$}\\
x=\frac{1}{2}\div\frac{4}{3}\qquad\mbox{Remember, when diving two fractions, multiply by the reciprocal (flip-flop the second fraction)}\\
x=\frac{1}{2}\times\frac{3}{4}\qquad\mbox{Multiply the left side out}\\
x=\frac{3}{8}

\end{array}$$

To check this, put in $$\frac{3}{8}$$ for x!

$$\begin{array}{l}
\frac{4}{3}x-\frac{1}{2}=0\\
\frac{4}{3}\times\frac{3}{8}-\frac{1}{2}=0\\
\frac{12}{24}-\frac{1}{2}=0\qquad\mbox{Reduce the first fraction}\\
\frac{1}{2}-\frac{1}{2}=0\\
0=0

\end{array}$$

It works, because 0 does equal 0!