Problem 48 - Algebra for College Students

Question

Set up an equation and solve each problem. The difference between two whole numbers is 8 , and the difference between their reciprocals is \(\frac{1}{6}\). Find the two numbers.

Step-by-Step Solution

Answer

The two numbers are 12 and 4.

Step 1: Define Variables

Let's define the two whole numbers as \(x\) and \(y\). Without loss of generality, we assume \(x > y\). Therefore, the difference between the two numbers is given by \(x - y = 8\).

Step 2: Set Up Equation for Reciprocals

According to the problem, the difference between the reciprocals of \(x\) and \(y\) is \(\frac{1}{6}\). This gives us the equation \(\frac{1}{y} - \frac{1}{x} = \frac{1}{6}\).

Step 3: Solve the Reciprocals Equation

Let's solve \(\frac{1}{y} - \frac{1}{x} = \frac{1}{6}\). Rewrite this as \(\frac{x-y}{xy} = \frac{1}{6}\). Using the first equation \(x-y=8\), substitute to obtain \(\frac{8}{xy} = \frac{1}{6}\). Hence, \(xy = 48\).

Step 4: Solve the System of Equations

We now have two equations: \(x-y=8\) and \(xy=48\). From the first equation, express \(x\) as \(x = y + 8\). Substitute \(x = y + 8\) into the second equation to get \((y + 8)y = 48\). This simplifies to \(y^2 + 8y - 48 = 0\). Solve this quadratic equation.

Step 5: Solve the Quadratic Equation

To solve \(y^2 + 8y - 48 = 0\), apply the quadratic formula \(y = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\) with \(a=1\), \(b=8\), and \(c=-48\). Calculate the discriminant: \(b^2 - 4ac\) gives \(64 + 192 = 256\). Thus, \(y = \frac{{-8 \pm \sqrt{256}}}{2}\).

Step 6: Calculate Values of y and x

The solutions for \(y\) are \(y = \frac{-8 + 16}{2} = 4\) and \(y = \frac{-8 - 16}{2} = -12\). Since \(y\) must be a whole number, choose \(y=4\). Substitute \(y=4\) into \(x = y + 8\) to get \(x = 12\).

Step 7: Verify the Solution

Check if the numbers \(x = 12\) and \(y = 4\) satisfy the original conditions: \(x-y=8\) holds since \(12-4=8\), and the difference between the reciprocals \(\frac{1}{4} - \frac{1}{12} = \frac{1}{6}\) is also correct.

Key Concepts

Reciprocal

In mathematics, the reciprocal of a number is simply the "flip" of the number when it's in fraction form. For example, the reciprocal of a whole number \(a\) is \(\frac{1}{a}\). This concept is especially important because multiplying a number by its reciprocal gives the result of 1. Let's consider the numbers in the exercise: You have two numbers, \(x\) and \(y\), with given reciprocals that differ by \(\frac{1}{6}\). This reciprocal can be seen as an expression of balance – denoting how two values relate when one is inversely proportional to another.

In solving problems involving reciprocals:

• First, set up an equation that represents the reciprocal relationship, like \(\frac{1}{y} - \frac{1}{x} = \frac{1}{6}\).
• Simplify the equation using algebraic manipulations, such as finding a common denominator or factoring.

Reciprocals are essential in solving equations that involve fractions, especially when those fractions are set in conditions involving differences or sums.

Quadratic Equation

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. In this exercise, once you transform the system of equations, you reach a quadratic equation: \(y^2 + 8y - 48 = 0\). Solving this equation is fundamental for finding the values of \(y\).

When solving quadratic equations, the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) is often used:

• Start by identifying the coefficients \(a\), \(b\), and \(c\).
• Calculate the discriminant \(b^2 - 4ac\). This helps determine the nature of the roots.
• Substitute the values into the formula to find \(y\).

The solution provides all potential roots for \(y\), but only roots that satisfy the initial conditions of the problem, such as positivity or integer value, are considered valid.

System of Equations

A system of equations is a collection of two or more equations with the same set of variables. The goal is to find a common solution that satisfies all equations concurrently. In this problem, the system includes \(x - y = 8\) and \(xy = 48\). These represent the conditions defined by the differences of numbers and their product.

When solving systems of equations, it is often useful to:

• Solve one equation for one variable, such as \(x = y + 8\).
• Substitute this expression into the other equation.
• Solve the resulting equation to find one of the variables. Use this found value to back-substitute and find the other variable.

This method ensures accuracy and leverages the relationships between the equations to quickly find solutions that satisfy all the given conditions. Solving systems of equations forms the basis for many mathematical applications in science, engineering, and real-world problem-solving.