Respuesta :
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
[tex]\frac{x}{(6x-5)(5x+1)}=\frac{A}{6x-5}+\frac{B}{5x+1}[/tex]By rewriting the right hand side as a single fraction, we have
[tex]\frac{x}{(6x-5)(5x+1)}=\frac{A(5x+1)+B(6x-5)}{(6x-5)(5x+1)}[/tex]We can to note that the denominators are equal, so we requiere the equality of the numerator, that is
[tex]x=A(5x+1)+B(6x-5)[/tex]Then, by distributing A and B into their respective parentheses, we get
[tex]x=A5x+A+B6x-5B[/tex]Now, by factoring x, we obtain
[tex]x=x(5A+6B)+A-5B[/tex]The coefficients near the like terms should be equal. Then, by comparing both sides, we can see that
[tex]\begin{gathered} 5A+6B=1...(1) \\ and \\ A-5B=0...(2) \end{gathered}[/tex]In matrix form, this system can be expressed as
[tex]\begin{bmatrix}{5} & {6} \\ {-5} & {25}\end{bmatrix}\begin{bmatrix}{A} & {} \\ {B} & {}\end{bmatrix}=\begin{bmatrix}{1} & {} \\ {0} & {}\end{bmatrix}[/tex]So, we need to solve this system of equations. This system is equivalent to
[tex]\begin{gathered} 5A+6B=1 \\ -5A+25B=0 \end{gathered}[/tex]By adding our last equations, we get
[tex]31B=1[/tex]So B is given as
[tex]B=\frac{1}{31}[/tex]Now, by substituting this result into equation (2), we obtain
[tex]A-5(\frac{1}{31})=0[/tex]So A is given as
[tex]A=\frac{5}{31}[/tex]Therefore, the answer is:
[tex]\frac{x}{(6x-5)(5x+1)}=\frac{\frac{5}{31}}{6x-5}+\frac{\frac{1}{31}}{5x+1}[/tex]