How To Solve The Problem 3y/(y+4)(y-2) = 5/y-2 + 2/y+4?


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Oddman Profile
Oddman answered
First of all, you need a few more parentheses for this to make sense. Each denominator that is not a single variable or number should be completely enclosed in parentheses. I believe doing that will make your problem look like this. If I am wrong, please comment or shout.

3y/((y+4)(y-2)) = 5/(y-2) + 2/(y+4)

Multiply both sides of this equation by ((y+4)(y-2)). This makes it somewhat easier to work with.

3y*((y+4)(y-2))/((y+4)(y-2)) =  5(y+4)(y-2)/(y-2) + 2(y-2)(y+4)/(y+4)
3y = 5(y+4) + 2(y-2)    (the expressions in bold are equal to 1, so they can be removed)
3y = 5y + 20 + 2y - 4    (use the distributive property)
3y = 7y + 16    (combine terms)
-4y = 16    (subtract 7y from both sides)
y = 16/-4    (divide both sides by -4)
y = -4    (evaluate)

This is not a helpful answer, because it makes the original expression evaluate to
-12/0 = 5/-6 + 2/0, which is indeterminate.

We can go back a few steps and take another approach. We recognize that the right side of the original equation evaluates to (7y+16)/((y+4)(y-2)). We can subtract the left side of the original equation from both sides to get

0 = 4(y+4)/((y+4)(y-2))

Canceling the (y+4) terms, we get

0 = 4/(y-2)

This can only be made to work for y = Infinity.

So, the answers we have found for this equation are

y = -4  or  y = Infinity.

Neither is very satisfying.
Perhaps you really do mean the problem as
(3y/(y+4))(y-2) = (5/y)-2 + (2/y)+4

Then both sides can be multiplied by the common denominator of y(y+4), and the equation resolves to a cubic.
3y3 - 8y2 - 15y - 28 = 0

This has one real root at approximately y = 4.322794. I found it by iteration using Newton's method after determining that the expression changed sign between y=4 and y=5.

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