Question

The sum of two numbers is 18, and the sum of their reciprocals is $\frac{1}{4}$. Find the numbers.

Answer 1

Step 1: Understand the problem:
We are given two numbers. Let the two numbers be \( x \) and \( y \). The following conditions are given:
- The sum of the numbers is 18, so:
\[ x + y = 18 \] - The sum of their reciprocals is \( \frac{1}{4} \), so:
\[ \frac{1}{x} + \frac{1}{y} = \frac{1}{4} \]

Step 2: Manipulate the second equation:
We can combine the two fractions in the second equation by finding a common denominator. The common denominator of \( x \) and \( y \) is \( xy \), so:
\[ \frac{1}{x} + \frac{1}{y} = \frac{x + y}{xy} \] Substitute \( x + y = 18 \) from the first condition:
\[ \frac{18}{xy} = \frac{1}{4} \] Now, solve for \( xy \):
\[ 18 = \frac{xy}{4} \quad \Rightarrow \quad xy = 72 \]

Step 3: Solve the system of equations:
We now have the following system of equations:
1. \( x + y = 18 \)
2. \( xy = 72 \)
These are the sum and product of the two numbers \( x \) and \( y \). We can solve this system by using the quadratic equation.

Step 4: Form a quadratic equation:
From the sum and product of the roots, we can form a quadratic equation whose roots are \( x \) and \( y \). The equation is:
\[ t^2 - (x + y)t + xy = 0 \] Substitute \( x + y = 18 \) and \( xy = 72 \) into this equation:
\[ t^2 - 18t + 72 = 0 \]

Step 5: Solve the quadratic equation:
Solve the quadratic equation \( t^2 - 18t + 72 = 0 \) using the quadratic formula:
\[ t = \frac{-(-18) \pm \sqrt{(-18)^2 - 4(1)(72)}}{2(1)} \] Simplifying the discriminant:
\[ t = \frac{18 \pm \sqrt{324 - 288}}{2} = \frac{18 \pm \sqrt{36}}{2} = \frac{18 \pm 6}{2} \] Thus, the two possible values for \( t \) are:
\[ t = \frac{18 + 6}{2} = \frac{24}{2} = 12 \quad \text{or} \quad t = \frac{18 - 6}{2} = \frac{12}{2} = 6 \]

Step 6: Conclusion:
The two numbers are \( x = 12 \) and \( y = 6 \), or vice versa.
Thus, the numbers are 12 and 6.