# How do you solve (r+4)/3=r/5 and find any extraneous solutions?

Dec 24, 2016

$r = - 10$

#### Explanation:

To ' eliminate' the fractions on both sides of the equation, multiply by the lowest common multiple (LCM ) of 3 and 5 which is 15

${\cancel{15}}^{5} \times \frac{r + 4}{\cancel{3}} ^ 1 = {\cancel{15}}^{3} \times \frac{r}{\cancel{5}} ^ 1$

$\Rightarrow 5 \left(r + 4\right) = 3 r \leftarrow \text{ no fractions}$

$\Rightarrow 5 r + 20 = 3 r$

subtract 3r from both sides.

$5 r - 3 r + 20 = \cancel{3 r} \cancel{- 3 r}$

$\Rightarrow 2 r + 20 = 0$

subtract 20 from both sides.

$2 r \cancel{+ 20} \cancel{- 20} = 0 - 20$

$\Rightarrow 2 r = - 20$

To solve for r, divide both sides by 2

$\frac{\cancel{2} r}{\cancel{2}} = \frac{- 20}{2}$

$\Rightarrow r = - 10 \text{ is the only solution}$