Question

Solve: \[ \frac{4}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 3 \] \[ \frac{8}{\sqrt{x}} - \frac{9}{\sqrt{y}} = 1 \]

Hint:

When solving equations involving square roots, substitution is an effective method to simplify the expressions.

Answer 1

Let \( \frac{1}{\sqrt{x}} = a \) and \( \frac{1}{\sqrt{y}} = b \). Then, the given equations become: \[ 4a + 3b = 3 \quad \text{(1)} \] \[ 8a - 9b = 1 \quad \text{(2)}. \] We solve this system of linear equations. Multiply equation (1) by 3 and equation (2) by 1 to eliminate \( b \): \[ 12a + 9b = 9 \quad \text{(3)} \] \[ 8a - 9b = 1 \quad \text{(4)}. \] Now, add equations (3) and (4): \[ (12a + 9b) + (8a - 9b) = 9 + 1 \] \[ 20a = 10 \quad \Rightarrow \quad a = \frac{10}{20} = \frac{1}{2}. \] Substitute \( a = \frac{1}{2} \) into equation (1): \[ 4 \times \frac{1}{2} + 3b = 3 \quad \Rightarrow \quad 2 + 3b = 3 \quad \Rightarrow \quad 3b = 1 \quad \Rightarrow \quad b = \frac{1}{3}. \] Now, recall that \( a = \frac{1}{\sqrt{x}} \) and \( b = \frac{1}{\sqrt{y}} \). So: \[ \frac{1}{\sqrt{x}} = \frac{1}{2} \quad \Rightarrow \quad \sqrt{x} = 2 \quad \Rightarrow \quad x = 4, \] \[ \frac{1}{\sqrt{y}} = \frac{1}{3} \quad \Rightarrow \quad \sqrt{y} = 3 \quad \Rightarrow \quad y = 9. \]
Conclusion: The values of \( x \) and \( y \) are \( x = 4 \) and \( y = 9 \).