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

1 Answer

Answer:

#y in {1, -3}#

Explanation:

Let's start with the original question:

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

#y/(3y)+2/(3y)=1/y^2#

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

#(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)=3y#

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

#y^3+2y^2=3y#

#y = 0 or y^2 + 2y - 3 = 0#

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

you shall not divide by zero.