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

Answer:

$\displaystyle \frac{3x+4}{(x-5)(x-1)} = \frac{19}{4(x-5)} +\frac{-7}{4(x-1)}$

Step-by-step explanation:

We are given the rational expression:

$\displaystyle \frac{3x+4}{x^2 -6x +5}$

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:

$\displaystyle = \frac{3x+4}{(x-5)(x-1)}$

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

$\displaystyle \frac{3x+4}{(x-5)(x-1)} = \frac{A}{x-5} + \frac{B}{x-1}$

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

$\displaystyle \displaystyle \frac{3x+4}{(x-5)(x-1)}\left((x-5)(x-1)\right) = \left(\frac{A}{x-5} + \frac{B}{x-1}\right)\left( (x-5)(x-1)\right)$

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:

$\displaystyle B = -\frac{7}{4}$

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:

$\displaystyle 19 = A(4) + B(0) \Rightarrow A = \frac{19}{4}$

We had:

$\displaystyle \frac{3x+4}{(x-5)(x-1)} = \frac{A}{x-5} + \frac{B}{x-1}$

Substitute:

$\displaystyle \frac{3x+4}{(x-5)(x-1)} = \frac{\dfrac{19}{4}}{x-5} +\frac{-\dfrac{7}{4}}{x-1}$

In conclusion:

$\displaystyle \frac{3x+4}{(x-5)(x-1)} = \frac{19}{4(x-5)} +\frac{-7}{4(x-1)}$