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

Let's simplify the system of equations by making substitutions.
Let $a = \frac{1}{x+y}$ and $b = \frac{1}{2x-y}$.
The equations become:
1) $3a + 12b = \frac{7}{3}$
2) $6a + 18b = \frac{11}{3}$
Multiply equation (1) by 2 to make the coefficient of 'a' the same in both equations.
$2 \times (3a + 12b) = 2 \times \frac{7}{3} \implies 6a + 24b = \frac{14}{3}$ (Equation 3)
Now, subtract equation (2) from equation (3):
$(6a + 24b) - (6a + 18b) = \frac{14}{3} - \frac{11}{3}$.
$6b = \frac{3}{3} = 1$.
$b = \frac{1}{6}$.
Substitute $b = \frac{1}{6}$ into equation (1):
$3a + 12\left(\frac{1}{6}\right) = \frac{7}{3}$.
$3a + 2 = \frac{7}{3}$.
$3a = \frac{7}{3} - 2 = \frac{7-6}{3} = \frac{1}{3}$.
$a = \frac{1}{9}$.
Now, substitute back the original expressions for $a$ and $b$:
$a = \frac{1}{x+y} = \frac{1}{9} \implies x+y = 9$ (Equation I)
$b = \frac{1}{2x-y} = \frac{1}{6} \implies 2x-y = 6$ (Equation II)
Now we have a simple system of linear equations. Add Equation I and Equation II:
$(x+y) + (2x-y) = 9 + 6$.
$3x = 15 \implies x = 5$.
Substitute $x=5$ into Equation I:
$5 + y = 9 \implies y = 4$.
Finally, calculate the required value $x^2 + y^2$:
$x^2 + y^2 = 5^2 + 4^2 = 25 + 16 = 41$.