Partial fraction decomposition of $\frac{x-4}{(x-2)(x-3)}$

Question

I'm trying to do the partial fraction decomposition of the following rational expression:$$\frac{x-4}{(x-2)(x-3)}$$

Here are the steps I preformed: \begin{align*} x-4 & = \frac{A}{x-3} + \frac{B}{x-2}\\ x-4 & = Ax - 3A + Bx - 2A\\ x-4 & = x(A+B) - (3A + 2B) \end{align*}

Form a system of equations by equating the coefficients of like powers of $x$:$$\begin{cases}A+B = 1\\ -3A - 2B = 0\end{cases}$$

Solving the system by substitution: \begin{align*} A & = -B+ 1\\ -3(-B+1) -2B & = 0\\ 3B -3 -2B & = 0\\ 3B - 2B & = 3\\ B & = 3\\ A + 3 & = 1\\ A & = - 2 \end{align*}

So, my final decomposition is$$\frac{-2}{x-2} + \frac{3}{x-3} .$$

However, the answer in the back of my packet is$$\frac{2}{x-2} + \frac{-1}{x-3} .$$

What have I done wrong?

Answer 1

The Answer in your book is correct and your calculation is wrong. I am writting correct one for you now.

$$\dfrac{x-4}{(x-2)(x-3)}=\dfrac{A}{x-2}+\dfrac{B}{x-3}=\dfrac{Ax-3A+Bx-2B}{(x-2)(x-3)}=\dfrac{(A+B)x+(-3A-2B)}{(x-2)(x-3)}$$

so we should have

$$A+B=1,-3A-2B=-4$$

By multiping first equation by 2 and adding result by second one you will have $-A=-2$ so $A=2$ and by $A+B=1$ we have $B=-1$

Answer 2

Your expression is $\dfrac{x-4}{(x-2)(x-3)}$

Now you have $A(x-3)+B(x-2)=x-4$

Plug in $x=3$ to cancel out $A$,

You get the value of $B$. Now you can calculate the value of $A$. You have to cancel out the multiple variables to make the calculation easier.

Answer 3

The partial fraction decomposition is given by:

$$\displaystyle \frac{x-4}{(x-3) (x-2)} = \frac{2}{x-2}-\frac{1}{x-3}$$

We would have written:

$$\displaystyle \frac{x-4}{(x-3) (x-2)} = \frac{A}{x-3}+ \frac{B}{x-2}$$

This yields:

$$x - 4 = A ( x- 2) + B(x-3)$$

Equating terms yields:

$$A+ B = 1$$

$$-2A -3B = -4$$

Solving for the two variables yields:

$$A = -1, B = 2$$