We are tasked with solving the partial fraction decomposition of:

$\frac{-3x + 1}{(4x - 1)(2x - 1)} = \frac{A}{4x - 1} + \frac{B}{2x - 1}$

Step 1: Combine the right-hand side

To express the right-hand side as a single fraction, we'll find a common denominator:

$\frac{A}{4x - 1} + \frac{B}{2x - 1} = \frac{A(2x - 1) + B(4x - 1)}{(4x - 1)(2x - 1)}$

Thus, the equation becomes:

$\frac{-3x + 1}{(4x - 1)(2x - 1)} = \frac{A(2x - 1) + B(4x - 1)}{(4x - 1)(2x - 1)}$

Step 2: Equate the numerators

Since the denominators are the same, we can equate the numerators:

$-3x + 1 = A(2x - 1) + B(4x - 1)$

Step 3: Expand and simplify

Expand the terms on the right-hand side:

$A(2x - 1) = 2Ax - A$ $B(4x - 1) = 4Bx - B$

So, the equation becomes:

$-3x + 1 = (2Ax - A) + (4Bx - B)$

Combine like terms:

$-3x + 1 = (2A + 4B)x - (A + B)$

Step 4: Set up a system of equations

Now, we equate the coefficients of $x$ and the constant terms on both sides:

1. Coefficient of $x$: $-3 = 2A + 4B$
2. Constant term: $1 = -A - B$

Step 5: Solve the system of equations

From the second equation:

$1 = -A - B \quad \Rightarrow \quad A = -1 - B$

Substitute $A = -1 - B$ into the first equation:

$-3 = 2(-1 - B) + 4B$ $-3 = -2 - 2B + 4B$ $-3 = -2 + 2B$ $-1 = 2B \quad \Rightarrow \quad B = -\frac{1}{2}$

Now substitute $B = -\frac{1}{2}$ into $A = -1 - B$:

$A = -1 - \left(-\frac{1}{2}\right) = -1 + \frac{1}{2} = -\frac{1}{2}$

Step 6: Write the final partial fraction decomposition

Thus, the partial fraction decomposition is:

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

Or, equivalently:

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

Would you like more details, or do you have any questions?

Here are 5 questions to consider:

1. How do you solve a partial fraction with repeated linear factors?
2. Can you apply partial fractions to improper fractions?
3. What happens if the degree of the numerator is greater than the denominator?
4. How do you handle partial fractions with quadratic factors in the denominator?
5. Can partial fraction decomposition be used to integrate rational functions?

Tip: Always check that the degree of the numerator is less than the degree of the denominator before applying partial fraction decomposition.