Problem 6 - Elementary Algebra

Question

A positive integer is 2 more than another. If the sum of the reciprocal of the smaller and twice the reciprocal of the larger is \(17 / 35,\) then find the two integers.

Step-by-Step Solution

Answer

The two integers are 5 and 7.

Step 1: Set Up Variables

Let the smaller integer be \( x \). Then the larger integer is \( x + 2 \).

Step 2: Set Up Reciprocals

The reciprocal of the smaller integer is \( \frac{1}{x} \), and the reciprocal of the larger integer (which is twice) is \( \frac{2}{x+2} \).

Step 3: Write the Equation

According to the problem, \( \frac{1}{x} + \frac{2}{x + 2} = \frac{17}{35} \).

Step 4: Find a Common Denominator

The common denominator for the left side of the equation is \( x(x+2) \). Thus, the equation becomes \( \frac{x+2+2x}{x(x+2)} = \frac{17}{35} \).

Step 5: Simplify the Equation

Simplify the numerator: \( \frac{3x+2}{x(x+2)} = \frac{17}{35} \).

Step 6: Cross-Multiply to Solve for x

Cross-multiply to solve for \( x \): \( 35(3x + 2) = 17x(x+2) \). This becomes \( 105x + 70 = 17x^2 + 34x \).

Step 7: Rearrange into Quadratic Form

Rearrange the equation: \( 17x^2 - 71x - 70 = 0 \).

Step 8: Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 17 \), \( b = -71 \), and \( c = -70 \). Calculate the discriminant and solve for \( x \).

Step 9: Calculate Solutions

The positive integer solutions for \( x \) are found to be \( x = 5 \) and \( x = -\frac{14}{17} \). Since \( x \) must be a positive integer, \( x = 5 \).

Step 10: Determine the Two Integers

The smaller integer is \( x = 5 \). The larger integer is \( x + 2 = 7 \).

Key Concepts

Reciprocals

In algebra, a **reciprocal** is what you get when you switch the numerator and the denominator of a fraction. For example, the reciprocal of a number like 5 is simply \( \frac{1}{5} \). Reciprocals are essential because they show the inverse of operations, particularly in division.

• The reciprocal of \( x \) is \( \frac{1}{x} \).
• For a number to have a reciprocal, it cannot be zero.

In the context of the given exercise, we need to find the reciprocals of the two integers. These reciprocals help form the equation that will allow us to find the solution. Defining these reciprocals correctly sets the stage to proceed with solving the problem by establishing foundational relationships between the numbers.

Positive Integers

**Positive integers** are numbers greater than zero without any fractions or decimals. They are part of the whole number family. In mathematics, these numbers are often used because they represent countable quantities.

• Examples include 1, 2, 3, and so on.
• In the context of our exercise, we are specifically looking at two such numbers where one is 2 more than the other.

Given the problem, we must identify two positive integers such that when studied using certain reciprocal relationships, they provide an exact sum. Crucially, when dealing with such problems, ensure all calculated values remain positive integers to conform to the problem's requirements.

Quadratic Equation

A **quadratic equation** is a polynomial equation of degree 2. It takes the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable.

• The general shape of its graph is a parabola.
• Solving involves methods like factoring, using the quadratic formula, or completing the square.

In our exercise, a quadratic equation is formed during the solution process after setting up our reciprocal relationships. By rearranging terms into the standard quadratic form, we can apply the quadratic formula to systematically find values of \( x \). Solving these equations is crucial in finding solutions that match the integer rules and relationships given.

Cross-Multiplication

One algebraic technique used in solving equations involving fractions is **cross-multiplication**. This method is useful when we have an equation with two fractions set equal to each other, \( \frac{a}{b} = \frac{c}{d} \). By cross-multiplying, we eliminate the fractions:

• It simplifies the equation to \( a \times d = b \times c \).
• Ensures easier manipulation and solution of the equation.

In our exercise, cross-multiplication is applied to the equation \( \frac{3x+2}{x(x+2)} = \frac{17}{35} \). By cross-multiplying, this helps in transforming the complex fraction problem into a standard form that is more manageable and allows us to progress to solving the quadratic equation.