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: Let the two numbers be $x$ and $y$ x + y = 18 \tag{1}. \frac{1}{x} + \frac{1}{y} = \frac{1}{4} \tag{2}. Step 2: Rewrite equation (2) Using $\frac{1}{x} + \frac{1}{y} = \frac{x + y}{xy}$: $\frac{x + y}{xy} = \frac{1}{4}.$ Substitute $x + y = 18$ from (1): \frac{18}{xy} = \frac{1}{4} \implies xy = 72 \tag{3}. Step 3: Solve the quadratic equation From (1) and (3), $x$ and $y$ are roots of the quadratic equation: $t^2 - (x + y)t + xy = 0 \implies t^2 - 18t + 72 = 0.$ Factorize: $t^2 - 18t + 72 = (t - 12)(t - 6) = 0.$ $t = 12 \quad \text{or} \quad t = 6.$ Step 4: Find the numbers The two numbers are $12$ and $6$. Correct Answer: The numbers are $12$ and $6$.

Answer 2

Step 1: Let the two numbers be $x$ and $y$ x + y = 18 \tag{1}. \frac{1}{x} + \frac{1}{y} = \frac{1}{4} \tag{2}. Step 2: Rewrite equation (2) Using $\frac{1}{x} + \frac{1}{y} = \frac{x + y}{xy}$: $\frac{x + y}{xy} = \frac{1}{4}.$ Substitute $x + y = 18$ from (1): \frac{18}{xy} = \frac{1}{4} \implies xy = 72 \tag{3}. Step 3: Solve the quadratic equation From (1) and (3), $x$ and $y$ are roots of the quadratic equation: $t^2 - (x + y)t + xy = 0 \implies t^2 - 18t + 72 = 0.$ Factorize: $t^2 - 18t + 72 = (t - 12)(t - 6) = 0.$ $t = 12 \quad \text{or} \quad t = 6.$ Step 4: Find the numbers The two numbers are $12$ and $6$. Correct Answer: The numbers are $12$ and $6$.