Question

If \[ \frac{3}{x + y} + \frac{12}{2x - y} = \frac{7}{3} \text{ and } \frac{6}{x + y} + \frac{18}{2x - y} = \frac{11}{3}, \text{ then } x^2 + y^2 = \]

• A. 34
• B. 41
• C. 25
• D. 7

Hint:

When solving simultaneous equations, defining new variables can often simplify the process. Remember to substitute back carefully to find the final solution.

Answer 1

The Correct Option is B

Step 1: Defining variables.
Let \( a = x + y \) and \( b = 2x - y \). The given equations then become: \[ \frac{3}{a} + \frac{12}{b} = \frac{7}{3} \quad \text{(A)} \] \[ \frac{6}{a} + \frac{18}{b} = \frac{11}{3} \quad \text{(B)} \]

Step 2: Solving the system of equations.
We can multiply equation (A) by 2 and subtract from equation (B) to eliminate \( a \). Multiplying (A) by 2: \[ \frac{6}{a} + \frac{24}{b} = \frac{14}{3} \quad \text{(C)} \] Now subtract equation (B) from equation (C): \[ \left( \frac{6}{a} + \frac{24}{b} \right) - \left( \frac{6}{a} + \frac{18}{b} \right) = \frac{14}{3} - \frac{11}{3} \] \[ \frac{6}{b} = \frac{3}{3} = 1 \] Thus, \[ b = 6 \]

Step 3: Substituting \( b = 6 \) in equation (A).
Substitute \( b = 6 \) in equation (A): \[ \frac{3}{a} + \frac{12}{6} = \frac{7}{3} \] \[ \frac{3}{a} + 2 = \frac{7}{3} \] \[ \frac{3}{a} = \frac{7}{3} - 2 = \frac{1}{3} \] Thus, \[ a = 9 \]

Step 4: Finding \( x^2 + y^2 \).
We now have \( a = x + y = 9 \) and \( b = 2x - y = 6 \). Solve these two equations: 1) \( x + y = 9 \)
2) \( 2x - y = 6 \) Add these equations: \[ x + y + 2x - y = 9 + 6 \] \[ 3x = 15 \] \[ x = 5 \] Substitute \( x = 5 \) into \( x + y = 9 \): \[ 5 + y = 9 \] \[ y = 4 \] Finally, calculate \( x^2 + y^2 \): \[ x^2 + y^2 = 5^2 + 4^2 = 25 + 16 = 41 \]

Final Answer: 41.