Problem 85 - Algebra and Trigonometry Real Mathematics, Real People

Question

Consider the equation $$\frac{6}{(x-3)(x-1)}=\frac{3}{x-3}+\frac{4}{x-1}$$ Without performing any calculations, explain how to clear this equation of fractions. Is it possible that this process will introduce an extraneous solution? If so, describe two ways to determine whether a solution is extraneous.

Step-by-Step Solution

Answer

To clear the equation of fractions, multiply each term by the common denominator \((x-3)(x-1)\). Yes, this process can potentially introduce extraneous solutions. Two ways to determine an extraneous solution include: 1) substituting the found solutions back into the original equation to see if they hold true, 2) checking if the solutions derived after clearing the equation of fractions are among the excluded values from the common denominator.

Step 1: Identify Common Denominator

The first step is to identify a common denominator for the fractions. In the given equation, the common denominator would be \((x-3)(x-1)\).

Step 2: Clear the Fractions

Once the common denominator has been identified, you can now multiply each term of the equation by the common denominator. This process of multiplying allows clearing out the fractions from the equation.

Step 3: Explain Extraneous Solutions

Extraneous solutions are solutions that we find through the process of solving an equation, but are not valid when substituted back into the original equation. The process of clearing fractions by multiplying each term of the equation can introduce such solutions.

Step 4: Check for Extraneous Solutions

One way to check for extraneous solutions is to substitute the solutions back into the original equation. If they make the equation true, they are valid solutions; otherwise, they are extraneous. Another way to avoid extraneous solutions is to make a note of the excluded values from the common denominator which make it zero, prior to multiplying through. This should indicate which solution values will not satisfy the original equation.

Key Concepts

Finding a Common Denominator

When solving rational equations, identifying a common denominator is a crucial initial step. To explain it straightforwardly, think of it as the least common ground all fractions involved can share. For the equation \(\frac{6}{(x-3)(x-1)}=\frac{3}{x-3}+\frac{4}{x-1}\), the common denominator is \( (x-3)(x-1) \), which is the product of the individual denominators. By finding this shared denominator, we can more easily combine the fractions on one side and simplify the equation. It's key for students to understand that finding a common denominator does not change the values of the fractions; it simply provides a consistent reference point for combining them.

Why do we seek a common denominator? This approach allows us to compare and combine fractions without losing their inherent value. By using the common denominator, we can eliminate the fractional components, which simplifies the solving process and helps students see the underlying structure of the equation more clearly.

Clearing Equations of Fractions

Once the common denominator is found, the next move in our mathematical choreography is to clear the equation of fractions. This step is known as 'clearing fractions' and means we're ridding the equation of its fractional form. We do this by multiplying each term by the common denominator. In our example, this would involve multiplying each term by \( (x-3)(x-1) \). The result is an equation void of fractions, which typically makes it straightforward to solve.

Clearing fractions simplifies the equation into a format that's often easier for students to manipulate, laying down the groundwork for finding the solutions. As fractions sometimes intimidate students, this process can alleviate some of that mathematical anxiety. It's important to emphasize that when clearing fractions, we are not altering the equation's solutions—just its appearance.

Extraneous Solutions

An intriguing aspect of solving rational equations is the emergence of extraneous solutions. These are 'solutions' that appear during the solving process but do not actually satisfy the original equation. A classic example is when we multiply an equation by an expression that could be zero; this action can introduce solutions that aren't valid when plugged back into the original equation.

To illustrate, after clearing fractions in our example, we might derive a solution that would make the original denominators zero—these are automatically extraneous. It's critical for students to grasp that while clearing fractions is a powerful tool, it also carries the responsibility of vigilance against these false solutions. Recognizing and excluding extraneous solutions is as important as finding the actual solutions because it reinforces the understanding of the equation's constraints.

Checking Solutions

The final act in our mathematical performance is checking the solutions. This is the quality control phase, where we verify that the solutions are true to the original equation. After finding potential solutions, we plug them back into the equation to ensure they work. This stage helps to weed out any extraneous solutions encountered along the way.

To properly check solutions, we substitute them individually into the original equation and look for consistency. Does the equation hold true, or does it fall apart? If consistent, we have a true solution. If not, we likely have an extraneous solution. Documenting the excluded values—that is, values that would make any denominator zero—before we begin solving can save us time in this stage. Students should understand that this process is not just a formality but a fundamental part of solving rational equations, reflecting the importance of accuracy and verification in mathematics.