# How do you solve \frac { 5} { y } + \frac { 1} { 5} = \frac { 3} { y }?

Nov 6, 2017

$y = - 10$

#### Explanation:

$\text{subtract "3/y" from both sides}$

$\frac{5}{y} - \frac{3}{y} + \frac{1}{5} = \cancel{\frac{3}{y}} \cancel{- \frac{3}{y}}$

$\Rightarrow \frac{2}{y} + \frac{1}{5} = 0$

$\text{subtract "1/5" from both sides}$

$\frac{2}{y} \cancel{+ \frac{1}{5}} \cancel{- \frac{1}{5}} = 0 - \frac{1}{5}$

$\Rightarrow \frac{2}{y} = - \frac{1}{5}$

$\text{using the method of "color(blue)"cross-multiplying}$

•color(white)(x)a/b=c/drArraxxd=bxxc

$\Rightarrow - y = 2 \times 5 = 10$

$\Rightarrow y = - 10$

Nov 6, 2017

$y = - 10$

#### Explanation:

What makes this equation a bit intimidating is that it has all the variables in the denominator, which is not as simple to deal with. Therefore, what I'm going to do first is multiply everything by $y$, so I can get it on the numerator:

$\left(\frac{5}{\cancel{y}}\right) \cancel{y} + \left(\frac{1}{5}\right) y = \left(\frac{3}{\cancel{y}}\right) \cancel{y}$

$5 + \left(\frac{1}{5}\right) y = 3$

Much nicer to look at! Now, there's a bunch of different things you could do from here to solve the equation. What I'm going to do, however, is multiply everything by $5$, in order to eliminate that $\frac{1}{5}$ term:

$25 + \left(\frac{1}{\cancel{5}}\right) y \cdot \cancel{5} = 15$

$25 + y = 15$

Now we just subtract $25$ from both sides to isolate the $y$:

$y = - 10$