# How do you solve 1/y + 1/(y+2) = 1/3?

Nov 5, 2015

You have two solutions: $y = 2 \pm \sqrt{10}$.

#### Explanation:

Sum the two fractions, obtaining only one denominator:

$\frac{1}{y} + \frac{1}{y + 2} = \frac{\left(y + 2\right) + y}{y \left(y + 2\right)} = \frac{2 y + 2}{y \left(y + 2\right)}$

So, the expression becomes

$\frac{2 y + 2}{y \left(y + 2\right)} = \frac{1}{3}$

Multiply both sides by $y \left(y + 2\right)$ and then by $3$:

$3 \cancel{y \left(y + 2\right)} \frac{2 y + 2}{\cancel{y \left(y + 2\right)}} = \frac{1}{\cancel{3}} \cancel{3} y \left(y + 2\right)$

The equation becomes

$3 \left(2 y + 2\right) = y \left(y + 2\right)$

Expand both terms:

$6 y + 6 = {y}^{2} + 2 y$

Bring everything to one side:

${y}^{2} + 2 y - 6 y - 6 = 0 \iff {y}^{2} - 4 y - 6 = 0$

Complete the square:

${y}^{2} - 4 y - 6 = \left({y}^{2} - 4 y + 4\right) - 10 = {\left(y - 2\right)}^{2} - 10$

So, the equation can be rewritten as

${\left(y - 2\right)}^{2} - 10 = 0 \iff {\left(y - 2\right)}^{2} = 10 \setminus \iff y - 2 = \pm \sqrt{10}$

And thus $y = 2 \pm \sqrt{10}$.