Given:
[tex]\frac{x}{(6x-7)(5x+1)}[/tex]
Required:
We need to find the partial decomposition of the given rational expression.
Explanation:
Use the partial decomposition formula.
[tex]\frac{x}{(6x-7)(5x+1)}=\frac{A}{(6x-7)}+\frac{B}{(5x+1)}[/tex]
[tex]\frac{x}{(6x-7)(5x+1)}=\frac{A(5x+1)}{(6x-7)(5x+1)}+\frac{B(6x-7)}{(5x+1)(6x-7)}[/tex]
[tex]\frac{x}{(6x-7)(5x+1)}=\frac{A(5x+1)+B\left(6x-7\right)}{(6x-7)(5x+1)}[/tex]
Equate the numerator of both sides since the denominator of both sides of the equation is equal.
[tex]x=A(5x+1)+B\lparen6x-7)[/tex]
Set x =-1/5 and substitute in the equation to find the value of B.
[tex]-\frac{1}{5}=A(5(-\frac{1}{5})+1)+B\lparen6(-\frac{1}{5})-7)[/tex]
[tex]-\frac{1}{5}=A(-1+1)+B\lparen-\frac{6}{5}-7)[/tex]
[tex]-\frac{1}{5}=B\lparen-\frac{6}{5}-7\times\frac{5}{5})[/tex]
[tex]-\frac{1}{5}=B\lparen\frac{-6}{5}-\frac{35}{5})[/tex]
[tex]-\frac{1}{5}=B\lparen\frac{=6-35}{5})[/tex]
[tex]-\frac{1}{5}=B\lparen\frac{-41}{5})[/tex]
Solve for B.
[tex]-\frac{1}{5}\times(\frac{-5}{41})=B\lparen\frac{-41}{5})\times(\frac{-5}{41})[/tex]
[tex]\frac{1}{41}=B[/tex]
We get B=1/41.
Set x =0 and substitute x=0 and b =1/41 in the equation to find the value A.
[tex]0=A(5(0)+1)+\frac{1}{41}\lparen6(0)-7)[/tex]
[tex]0=A+\frac{1}{41}\lparen-7)[/tex]
[tex]0=A-\frac{7}{41}[/tex]
Solve for A.
[tex]0+\frac{7}{41}=A-\frac{7}{41}+\frac{7}{41}[/tex][tex]A=\frac{7}{41}[/tex]
We get A=7/41.
[tex]Substitute\text{ A =}\frac{7}{41}\text{ and B=}\frac{1}{41}\text{ in the equation }\frac{x}{(6x-7)(5x+1)}=\frac{A}{(6x-7)}+\frac{B}{(5x+1)}.[/tex]
[tex]\frac{x}{(6x-7)(5x+1)}=\frac{7}{41(6x-7)}+\frac{1}{41(5x+1)}[/tex]
Final answer:
[tex]\frac{7}{41(6x-7)}+\frac{1}{41(5x+1)}[/tex]
