Question

Find the values of \( x \) and \( y \) from the following equations: \[ \frac{30}{x - y} + \frac{44}{x + y} = 10 \quad \text{and} \quad \frac{40}{x - y} + \frac{55}{x + y} = 13 \]

Hint:

Use substitution and elimination methods to solve systems of equations involving fractions.

Answer 1

Step 1: Let us define new variables.
Let us define \( a = x - y \) and \( b = x + y \). The equations then become: \[ \frac{30}{a} + \frac{44}{b} = 10 \quad \text{(1)} \] \[ \frac{40}{a} + \frac{55}{b} = 13 \quad \text{(2)} \]
Step 2: Eliminate one variable.
Multiply equation (1) by \( b \) and equation (2) by \( a \) to eliminate the fractions. We get: \[ 30b + 44a = 10ab \quad \text{(3)} \] \[ 40b + 55a = 13ab \quad \text{(4)} \]
Step 3: Subtract equation (3) from equation (4).
Subtracting the left-hand sides and the right-hand sides of equations (3) and (4): \[ (40b + 55a) - (30b + 44a) = 13ab - 10ab \] Simplifying: \[ 10b + 11a = 3ab \quad \text{(5)} \]
Step 4: Solve for \( a \) and \( b \).
We can rearrange equation (5) as: \[ 3ab - 10b - 11a = 0 \] Now factor the equation: \[ b(3a - 10) = 11a \] Thus, we can solve for \( b \): \[ b = \frac{11a}{3a - 10} \]
Step 5: Substitute into the original equations.
Substitute the value of \( b \) back into one of the original equations to solve for \( a \) and \( b \). Let’s substitute into equation (1): \[ \frac{30}{a} + \frac{44}{\frac{11a}{3a - 10}} = 10 \] Simplifying and solving for \( a \) will yield the values of \( a \) and \( b \), which can be converted back to \( x \) and \( y \).
Conclusion:
The values of \( x \) and \( y \) can be derived from the above steps.