Question

If \(\frac{48}{x+y}-\frac{6}{x-y}=10\) and \(\frac{15}{x+y}+\frac{4}{x-y}=9\), then \(x^{2}+y^{2}=\)

• A. 3
• B. 4
• C. 5
• D. 6

Hint:

Whenever reciprocals of linear expressions occur repeatedly, substitute them with new variables to convert the problem into a system of linear equations.

Answer 1

The Correct Option is C

Step 1: Convert the equations into linear form.
Let \[ u=\frac1{x+y},\qquad v=\frac1{x-y} \] Then the given equations become \[ 48u-6v=10 \] \[ 15u+4v=9 \] Dividing the first equation by 2, \[ 24u-3v=5 \qquad ...(1) \] \[ 15u+4v=9 \qquad ...(2) \]

Step 2: Solve for \(u\) and \(v\).
Multiply equation (1) by 4: \[ 96u-12v=20 \] Multiply equation (2) by 3: \[ 45u+12v=27 \] Adding, \[ 141u=47 \] \[ u=\frac13 \] Therefore, \[ x+y=3 \] Substituting \(u=\frac13\) into equation (2): \[ 15\left(\frac13\right)+4v=9 \] \[ 5+4v=9 \] \[ v=1 \] Hence, \[ x-y=1 \]

Step 3: Find \(x\) and \(y\).
\[ x+y=3,\qquad x-y=1 \] Adding, \[ 2x=4 \] \[ x=2 \] Therefore, \[ y=1 \]

Step 4: Calculate \(x^2+y^2\).
\[ x^2+y^2=2^2+1^2 \] \[ =4+1 \] \[ =5 \] {5}