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5 votes
5 votes
Write the partial fraction decomposition
(x )/((6x - 5)(5x + 1))

User Atm
by
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1 Answer

22 votes
22 votes

Since the degree of the denominator is 2 and the degree of the numerator is 1, the form of the partial decomposition is given as


(x)/((6x-5)(5x+1))=(A)/(6x-5)+(B)/(5x+1)

By rewriting the right hand side as a single fraction, we have


(x)/((6x-5)(5x+1))=(A(5x+1)+B(6x-5))/((6x-5)(5x+1))

We can to note that the denominators are equal, so we requiere the equality of the numerator, that is


x=A(5x+1)+B(6x-5)

Then, by distributing A and B into their respective parentheses, we get


x=A5x+A+B6x-5B

Now, by factoring x, we obtain


x=x(5A+6B)+A-5B

The coefficients near the like terms should be equal. Then, by comparing both sides, we can see that


\begin{gathered} 5A+6B=1...(1) \\ and \\ A-5B=0...(2) \end{gathered}

In matrix form, this system can be expressed as


\begin{bmatrix}{5} & {6} \\ {-5} & {25}\end{bmatrix}\begin{bmatrix}{A} & {} \\ {B} & {}\end{bmatrix}=\begin{bmatrix}{1} & {} \\ {0} & {}\end{bmatrix}

So, we need to solve this system of equations. This system is equivalent to


\begin{gathered} 5A+6B=1 \\ -5A+25B=0 \end{gathered}

By adding our last equations, we get


31B=1

So B is given as


B=(1)/(31)

Now, by substituting this result into equation (2), we obtain


A-5((1)/(31))=0

So A is given as


A=(5)/(31)

Therefore, the answer is:


(x)/((6x-5)(5x+1))=((5)/(31))/(6x-5)+((1)/(31))/(5x+1)

User Lajos Arpad
by
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