#2**+10 **

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

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}$$

Melody Aug 5, 2014

#1**+10 **

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!

NinjaDevo Aug 5, 2014

#2**+10 **

Best Answer

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

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}$$

Melody Aug 5, 2014