# How do you solve 5/(3c)=2/3 and find any extraneous solutions?

Jan 29, 2017

$c = \frac{5}{2}$

$c = 0$ is an extraneous solution

#### Explanation:

Extraneous solution is such that we are not able to divide by 0.

So $3 c \ne 0 \implies c = 0$ is the extraneous solution.

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$\textcolor{b l u e}{\text{First principle method}}$

$\textcolor{g r e e n}{\frac{5}{3 c} = \frac{2}{3} \textcolor{red}{\times 1}}$

$\textcolor{g r e e n}{\frac{5}{3 c} = \frac{2}{3} \textcolor{red}{\times \frac{c}{c}}}$

$\textcolor{g r e e n}{\frac{5}{3 c} = \frac{2 c}{3 c}}$

As the denominators (bottom values) are both the same we only need compare the numerators (top values).

Thus $5 = 2 c$

$\implies c = \frac{5}{2}$
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$\textcolor{b l u e}{\text{Short cut method}}$

$\frac{5}{3 c} = \frac{2}{3}$

Cross multiply

$5 \times 3 = 2 \times 3 c$

$15 = 6 c$

$c = \frac{15}{6} = \frac{15 \div 3}{6 \div 3} = \frac{5}{2}$