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# If , then

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If , then Aug 5, 2014

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

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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!

Aug 5, 2014
#2
+10
$$\frac{4}{3}x-\frac{1}{2}=0$$
That way you get rid of all the fractions right from the begining. $$\begin{array}{rll} 6\times \left(\frac{4}{3}x-\frac{1}{2}\right)&=&0 \times 6\\\\ \frac{6\times 4}{3}x-\frac{6}{2}&=&0\\\\ 8x-3&=&0\\\\ 8x&=&3\\\\ x&=&\frac{3}{8}\\\\ \end{array}$$