How do you solve $\frac{5}{3 c} = \frac{2}{3}$ $5/(3c)=2/3$ and find any extraneous solutions?

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Answer 1 · 2017-01-29T15:07:09

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

$c = 0$ $c=0$ is an extraneous solution

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

So $3 c \neq 0 \Rightarrow c = 0$ $3c !=0=> c=0$ is the extraneous solution.

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$First principle method$ $color(blue)("First principle method")$

$\frac{5}{3 c} = \frac{2}{3} \times 1$ $color(green)(5/(3c)=2/3color(red)(xx1))$

$\frac{5}{3 c} = \frac{2}{3} \times \frac{c}{c}$ $color(green)(5/(3c)=2/3color(red)(xxc/c))$

$\frac{5}{3 c} = \frac{2 c}{3 c}$ $color(green)(5/(3c)=(2c)/(3c))$

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

Thus $5 = 2 c$ $5=2c$

$\Rightarrow c = \frac{5}{2}$ $=>c=5/2$
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$Short cut method$ $color(blue)("Short cut method")$

$\frac{5}{3 c} = \frac{2}{3}$ $5/(3c)=2/3$

Cross multiply

$5 \times 3 = 2 \times 3 c$ $5xx3=2xx3c$

$15 = 6 c$ $15=6c$

$c = \frac{15}{6} = \frac{15 \div 3}{6 \div 3} = \frac{5}{2}$ $c=15/6 = (15-:3)/(6-:3)=5/2$

How do you solve 5/(3c)=2/353c=235/(3c)=2/3 and find any extraneous solutions?