Question

(100 points) Can someone help me understand how to solve this problem? Not just the answer to the problem but the process too please. Thank you!

Answer 1

`\displaystyle (3x+4)/((x-5)(x-1)) = (19)/(4(x-5)) +(-7)/(4(x-1))`

Explanation:

We are given the rational expression:

`\displaystyle (3x+4)/(x^2 -6x +5)`

And we want to find two rational expressions that sum to the above expression.

This technique is known as partial fraction decomposition. First, factor the denominator into linear factors:

`\displaystyle = (3x+4)/((x-5)(x-1))`

Let A and B be two unknown constants. We can let:

`\displaystyle (3x+4)/((x-5)(x-1)) = (A)/(x-5) + (B)/(x-1)`

Find A and B. Multiply the entire equation by the denominator:

`\displaystyle \displaystyle (3x+4)/((x-5)(x-1))\left((x-5)(x-1)\right) = \left((A)/(x-5) + (B)/(x-1)\right)\left( (x-5)(x-1)\right)`

Simplify:

`3x + 4 = A(x-1) + B(x-5)`

To find A and B, let x equal some value such that it will cancel out one variable. First, let x = 1. Then:

`\displaystyle 3(1) + 4 = A((1) - 1) + B((1) -5)`

Simplify:

`\displaystyle 7 = A(0) + B(-4) \Rightarrow -4B = 7`

Hence:

`\displaystyle B = -(7)/(4)`

To find A, let x = 5 (we choose this value because it allows us to cancel B):

`\displaystyle 3(5) + 4 = A((5)-1) + B((5) - 5)`

Simplify:

`\displaystyle 19 = A(4) + B(0) \Rightarrow A = (19)/(4)`

We had:

`\displaystyle (3x+4)/((x-5)(x-1)) = (A)/(x-5) + (B)/(x-1)`

Substitute:

`\displaystyle (3x+4)/((x-5)(x-1)) = ((19)/(4))/(x-5) +(-(7)/(4))/(x-1)`

In conclusion:

`\displaystyle (3x+4)/((x-5)(x-1)) = (19)/(4(x-5)) +(-7)/(4(x-1))`