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

Jun 7, 2017

$y \in \left\{1 , - 3\right\}$

#### Explanation:

$\frac{1}{3} + \frac{2}{3 y} = \frac{1}{y} ^ 2$

Now let's get both fractions on the left side to have a common denominator to make it easier for our calculations. We see that $3 y$ is our common denominator and so rewrite the equation as so:

$\frac{y}{3 y} + \frac{2}{3 y} = \frac{1}{y} ^ 2$

Now let's combine the two fractions to create this:

$\frac{y + 2}{3 y} = \frac{1}{y} ^ 2$

Now we can go ahead and use cross multiplication to help simplify the equation, turning it into this:

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

Then we use the distributive property to simplify the left side of the equation:

${y}^{3} + 2 {y}^{2} = 3 y$

$y = 0 \mathmr{and} {y}^{2} + 2 y - 3 = 0$

two numbers: the sum is -2, the product is $\left(- 3\right) \left(+ 1\right)$

you shall not divide by zero.