Question

If any one of you know how to solve this algebra problem, please help. Thanks ​

Answer 1

`a=-5,\: b=-1`

Explanation:

we are given

`3-(x+1)/(x^2-1)/(3x+2)=(a)/(x+b)`

first, we can take the reciprocal of a fraction when we do division, and turn the operation to multiplication. lets focus on the left hand side.

`3-(x+1)/(x^2-1)/(3x+2)=3-(x+1)*(3x+2)/(x^2-1)`

next, recall that
`a^2-b^2=(a+b)(a-b).` here, we have
`x^2-1=(x+1)(x-1)`, and so we can cancel out
`(x+1)`. the result is

`3-(3x+2)/(x-1)`.

now, we express 3 as a fraction with denominator
`x-1`.

`3-(3x+2)/(x-1)=(3(x-1))/(x-1)-(3x+2)/(x-1)`.

next, we simplify this:

`(3(x-1))/(x-1)-(3x+2)/(x-1)=(3x-3)/(x-1)-(3x+2)/(x-1)=(3x-3-3x-2)/(x-1)`

finally, we have

`(-5)/(x-1)`

as our solution. so, we have
`a=-5` and
`b=-1`.