1. **Problem statement:** The sum of two numbers is 18 and the sum of their reciprocals is \(\frac{9}{40}\). Find the numbers. 2. **Formulate the equations:** Let the two numbers be \(x\) and \(y\). Given: \[ x + y = 18 \] \[ \frac{1}{x} + \frac{1}{y} = \frac{9}{40} \] 3. **Use the sum and product of the numbers:** From the reciprocal sum, \[ \frac{1}{x} + \frac{1}{y} = \frac{x + y}{xy} = \frac{9}{40} \] Substitute \(x + y = 18\): \[ \frac{18}{xy} = \frac{9}{40} \] 4. **Solve for the product \(xy\):** \[ 18 \times 40 = 9 \times xy \] \[ 720 = 9xy \] \[ xy = \frac{720}{9} = 80 \] 5. **Form the quadratic equation:** The numbers \(x\) and \(y\) satisfy \[ x + y = 18, \quad xy = 80 \] The quadratic equation with roots \(x\) and \(y\) is \[ x^2 - (x+y)x + xy = 0 \implies x^2 - 18x + 80 = 0 \] 6. **Solve the quadratic equation:** \[ \Delta = b^2 - 4ac = (-18)^2 - 4 \times 1 \times 80 = 324 - 320 = 4 \] \[ x = \frac{18 \pm \sqrt{4}}{2} = \frac{18 \pm 2}{2} \] \[ x = 10 \quad \text{or} \quad x = 8 \] 7. **Find the numbers:** The two numbers are \(10\) and \(8\). **Final answer:** The numbers are \(10\) and \(8\).