How do you solve $\frac{1}{3} + \frac{2}{3 y} = \frac{1}{y^{2}}$ $1/3 + 2/(3y) = 1/y^2$ ?

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How do you solve $\frac{1}{3} + \frac{2}{3 y} = \frac{1}{y^{2}}$ $1/3 + 2/(3y) = 1/y^2$ ?

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Answer 1 · 2017-06-07T23:08:38

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

Explanation:

Let's start with the original question:

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

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 $3 y$ $3y$ is our common denominator and so rewrite the equation as so:

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

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

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

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

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

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

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

$y = 0 or y^{2} + 2 y - 3 = 0$ $y = 0 or y^2 + 2y - 3 = 0$

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

you shall not divide by zero.

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How do you solve $\frac{1}{3} + \frac{2}{3 y} = \frac{1}{y^{2}}$ $1/3 + 2/(3y) = 1/y^2$ ?

1 Answer

Explanation:

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How do you solve $\frac{1}{3} + \frac{2}{3 y} = \frac{1}{y^{2}}$ $1/3 + 2/(3y) = 1/y^2$ ?