Question

Hello, I need some assistance with this precalculus homework question, please?HW Q8

Answer 1

Given:

`(x)/((6x-7)(5x+1))`

Required:

We need to find the partial decomposition of the given rational expression.

Step-by-step explanation:

Use the partial decomposition formula.

`(x)/((6x-7)(5x+1))=(A)/((6x-7))+(B)/((5x+1))`

`(x)/((6x-7)(5x+1))=(A(5x+1))/((6x-7)(5x+1))+(B(6x-7))/((5x+1)(6x-7))`

`(x)/((6x-7)(5x+1))=(A(5x+1)+B\left(6x-7\right))/((6x-7)(5x+1))`

Equate the numerator of both sides since the denominator of both sides of the equation is equal.

`x=A(5x+1)+B\lparen6x-7)`

Set x =-1/5 and substitute in the equation to find the value of B.

`-(1)/(5)=A(5(-(1)/(5))+1)+B\lparen6(-(1)/(5))-7)`

`-(1)/(5)=A(-1+1)+B\lparen-(6)/(5)-7)`

`-(1)/(5)=B\lparen-(6)/(5)-7*(5)/(5))`

`-(1)/(5)=B\lparen(-6)/(5)-(35)/(5))`

`-(1)/(5)=B\lparen(=6-35)/(5))`

`-(1)/(5)=B\lparen(-41)/(5))`

Solve for B.

`-(1)/(5)*((-5)/(41))=B\lparen(-41)/(5))*((-5)/(41))`

`(1)/(41)=B`

We get B=1/41.

Set x =0 and substitute x=0 and b =1/41 in the equation to find the value A.

`0=A(5(0)+1)+(1)/(41)\lparen6(0)-7)`

`0=A+(1)/(41)\lparen-7)`

`0=A-(7)/(41)`

Solve for A.

`0+(7)/(41)=A-(7)/(41)+(7)/(41)`
`A=(7)/(41)`

We get A=7/41.

`Substitute\text{ A =}(7)/(41)\text{ and B=}(1)/(41)\text{ in the equation }(x)/((6x-7)(5x+1))=(A)/((6x-7))+(B)/((5x+1)).`

`(x)/((6x-7)(5x+1))=(7)/(41(6x-7))+(1)/(41(5x+1))`

Final answer:

`(7)/(41(6x-7))+(1)/(41(5x+1))`