Solving for Two Integers Based on Their Reciprocal Sums

One common problem in algebra involves finding two positive integers based on the sum of their reciprocals. Such a problem can be approached through algebraic equations and simple arithmetic. Let's break down the steps involved in solving such a problem.

Problem Formulation

Let the two positive integers be denoted by x and y where x is the smaller integer and y is twice x. Thus, the relationship between the two integers is given by:

y 2x

The problem states that the sum of the reciprocals of the two positive integers is equal to 3/10. This can be formulated as:

1/x 1/y 3/10

Solution Steps

Substituting y 2x into the equation, we get:

1/x 1/(2x) 3/10

Step 1: Combine the Fractions

To combine the fractions on the left side of the equation, we find a common denominator:

(2/x) (1/x) 3/10

This simplifies to:

3/x 3/10

Step 2: Cross-Multiplication

Next, we cross-multiply to solve for x:

3 * 10 3 * x

This simplifies to:

30 3x

Dividing both sides by 3, we get:

x 10

Step 3: Determine y

Since y 2x, we substitute the value of x to find y:

y 2 * 10 20

Therefore, the two integers are:

boxed{10} and boxed{20}

Additional Examples

Here are a few more examples to further illustrate the process:

Example 1

Let the integers be a and b such that 1/a 1/b 3/14. Assume b a/2:

1/a 1/(2a) 3/14

This simplifies to:

3/(2a) 3/14

By cross-multiplication:

6a 42

Solving for a gives:

a 7 b 14

Example 2

Consider the case where b 2a and 1/a 1/b 3/8:

1/a 1/(2a) 3/8

Multiplying through by 2a:

3 6a / 8

Solving for a gives:

a 4 b 8

Example 3

Let one number be x and the other be 2x. The sum of their reciprocals is 3/8:

1/x 1/(2x) 3/8

Multiplying through by 2x gives:

3 6x / 8

Solving for x results in:

x 4 2x 8

Conclusion

By solving these algebraic equations, we can find two positive integers based on their reciprocal sums. The examples illustrate the step-by-step process of combining fractions, cross-multiplying, and solving for the unknown variables. This approach can be applied to similar problems involving the sum of reciprocals.