Question

If \( \frac{2x + 3}{(x + 1)(x - 3)} = \frac{a}{x + 1} + \frac{b}{x - 3} \), then \( a + b \) is equal to

• A. 1
• B. 2
• C. \( \frac{9}{4} \)
• D. \( -\frac{1}{4} \)

Hint:

When solving for coefficients, express the right-hand side of the equation with a common denominator and compare the numerators.

Answer 1

The Correct Option is B

Step 1: Set up the partial fractions.
We are given the equation: \[ \frac{2x + 3}{(x + 1)(x - 3)} = \frac{a}{x + 1} + \frac{b}{x - 3} \] To combine the fractions on the right-hand side, get a common denominator: \[ \frac{a}{x + 1} + \frac{b}{x - 3} = \frac{a(x - 3) + b(x + 1)}{(x + 1)(x - 3)} \]

Step 2: Set the numerators equal.
Since the denominators are the same on both sides, we can set the numerators equal to each other: \[ 2x + 3 = a(x - 3) + b(x + 1) \]

Step 3: Expand both sides.
Expand the right-hand side: \[ 2x + 3 = ax - 3a + bx + b \] Now combine like terms: \[ 2x + 3 = (a + b)x + (-3a + b) \]

Step 4: Compare coefficients.
Equate the coefficients of \( x \) and the constant terms on both sides:
- From the \( x \)-terms: \( a + b = 2 \)
- From the constant terms: \( -3a + b = 3 \)

Step 5: Solve the system of equations.
We have the system: \[ a + b = 2 \quad \text{and} \quad -3a + b = 3 \] Solve for \( a \) and \( b \): - From \( a + b = 2 \), we get \( b = 2 - a \). - Substitute into \( -3a + b = 3 \): \[ -3a + (2 - a) = 3 \quad \implies \quad -3a + 2 - a = 3 \quad \implies \quad -4a = 1 \quad \implies \quad a = -\frac{1}{4} \] Now substitute \( a = -\frac{1}{4} \) into \( b = 2 - a \): \[ b = 2 - \left( -\frac{1}{4} \right) = 2 + \frac{1}{4} = \frac{9}{4} \]

Step 6: Find \( a + b \).
Thus, \( a + b = -\frac{1}{4} + \frac{9}{4} = 2 \), corresponding to option (B).