Question

If \[ \frac{1}{(3x+1)(x-2)} = \frac{A}{3x+1} + \frac{B}{x-2} \quad {and} \quad \frac{x+1}{(3x+1)(x-2)} = \frac{C}{3x+1} + \frac{D}{x-2}, \] then \[ \frac{1}{(3x+1)(x-2)} = \frac{A}{3x+1} + \frac{B}{x-2}, { find } A + 3B = 0, A:C = 1:3, B:D = 2:3. \]

• A. \( A + 3B = 0, A:C = 1:3, B:D = 2:3 \)
• B. \( A + 3B = 0, A:C = 3:1, B:D = 3:2 \)
• C. \( A + 3B = 0, A:C = 3:2, B:D = 1:3 \)
• D. \( A + 3B = 0, A:C = 3:2, B:D = 1:3 \)

Hint:

When solving for coefficients in rational equations, equate the numerators and compare coefficients of like terms to find the values of the constants.

Answer 1

The Correct Option is D

We are given two rational expressions and asked to find the values of \( A \), \( B \), \( C \), and \( D \) such that the two equations hold true. 
Step 1: Consider the first equation: \[ \frac{1}{(3x+1)(x-2)} = \frac{A}{3x+1} + \frac{B}{x-2}. \] To combine the right-hand side into a single fraction, we use the common denominator \( (3x + 1)(x - 2) \): \[ \frac{A}{3x+1} + \frac{B}{x-2} = \frac{A(x-2) + B(3x+1)}{(3x+1)(x-2)}. \] Since the denominators on both sides are the same, equate the numerators: \[ 1 = A(x - 2) + B(3x + 1). \] Expanding both terms: \[ 1 = A(x - 2) + B(3x + 1) = A x - 2A + 3B x + B. \] Now, collect like terms: \[ 1 = (A + 3B)x + (-2A + B). \] For the equation to hold true for all values of \( x \), the coefficients of \( x \) and the constant terms must be equal on both sides:
- Coefficient of \( x \): \( A + 3B = 0 \),
- Constant term: \( -2A + B = 1 \).
Step 2: Now, consider the second equation: \[ \frac{x+1}{(3x+1)(x-2)} = \frac{C}{3x+1} + \frac{D}{x-2}. \] Following similar steps as before, combine the right-hand side into a single fraction: \[ \frac{C}{3x+1} + \frac{D}{x-2} = \frac{C(x-2) + D(3x+1)}{(3x+1)(x-2)}. \] Equating the numerators: \[ x + 1 = C(x - 2) + D(3x + 1). \] Expanding both terms: \[ x + 1 = Cx - 2C + 3Dx + D. \] Collect like terms: \[ x + 1 = (C + 3D)x + (-2C + D). \] For the equation to hold true for all values of \( x \), the coefficients of \( x \) and the constant terms must be equal:
- Coefficient of \( x \): \( C + 3D = 1 \),
- Constant term: \( -2C + D = 1 \).
Step 3: Now we have the system of equations: 1. \( A + 3B = 0 \),
2. \( -2A + B = 1 \),
3. \( C + 3D = 1 \),
4. \( -2C + D = 1 \).
Solving these, we get: \[ A:C = 3:2, \quad B:D = 1:3. \] Thus, the correct answer is \( A + 3B = 0, A:C = 3:2, B:D = 1:3 \).