Problem 76 - Intermediate Algebra

Question

Solve equation. If a solution is extraneous, so indicate. \(x^{-1}+2=3 x^{-1}\) \(\left(\text {Hint: Use } x^{-n}=\frac{1}{x^{n}}\right)\)

Step-by-Step Solution

Answer

The solution is \(x = 1\). It is not extraneous.

Step 1: Interpret the Equation

The equation given is \(x^{-1} + 2 = 3x^{-1}\). Using the hint that \(x^{-n} = \frac{1}{x^n}\), rewrite the equation as \(\frac{1}{x} + 2 = 3 \cdot \frac{1}{x}\).

Step 2: Combine Terms

Move all terms involving \(\frac{1}{x}\) to one side: \(\frac{1}{x} - 3\cdot\frac{1}{x} = -2\). This simplifies to \(-2\cdot\frac{1}{x} = -2\).

Step 3: Solve for \(\frac{1}{x}\)

Isolate \(\frac{1}{x}\) by dividing both sides by \(-2\), leading to \(\frac{1}{x} = 1\).

Step 4: Find \(x\)

Since \(\frac{1}{x} = 1\), take the reciprocal of both sides to find \(x = 1\).

Step 5: Check for Extraneous Solutions

Substitute \(x = 1\) back into the original equation to test: \(1^{-1} + 2 = 3(1^{-1})\). This becomes \(1 + 2 = 3\), confirming that \(x = 1\) satisfies the equation and is not extraneous.

Key Concepts

Rational Equations

Rational equations are mathematical statements that involve ratios or fractions. They are often represented with variables in the denominators. Solving these types of equations requires finding a common denominator or manipulating the equation to eliminate the fractions.
To solve rational equations efficiently, the process often involves:

• Identifying the terms that contain fractions.
• Eliminating the fractions by multiplying through by a common denominator.
• Simplifying and solving the resulting polynomial equation.

In our given problem, the rational equation is initially written with negative exponents. We rewrite it as fractions using the rule: \(x^{-1} = \frac{1}{x}\). This step transforms the initial equation into a more recognizable rational form, making it easier to solve. Always remember to check for any restrictions on the variable, such as values that make a denominator zero, as these might not be valid solutions.

Extraneous Solutions

During the process of solving equations, especially rational ones, you may encounter solutions that seem valid algebraically but do not satisfy the original equation. These are known as extraneous solutions.
Extraneous solutions often arise when you perform operations that are not reversible, such as squaring both sides of an equation or multiplying by a variable expression. It's important to verify each solution by substituting back into the original equation to confirm its validity.
In our exercise, we solved for \(x\) and checked if the solution satisfies the original equation. This test helps avoid assumptions and ensures no extraneous solutions are mistakenly identified as valid. Here, when we tested \(x = 1\), it indeed satisfied the initial equation without introducing any mathematical inconsistency, confirming it is not extraneous.

Reciprocals

Reciprocals play a crucial role in solving equations involving fractions and rational expressions. The reciprocal of a number is essentially its 'flip'—if the original number is \(a\), the reciprocal is \(\frac{1}{a}\).
In equations, especially those with negative exponents, finding the reciprocal can be a vital step. For instance, when \(x^{-1}\) is present in an equation, it is effectively a shortcut for writing \(\frac{1}{x}\). Recognizing this allows you to reframe the problem in terms of fractions, which can simplify the solving process.
In our problem, we used reciprocals to transform the equation and isolate \(\frac{1}{x}\). By understanding reciprocals, you can easily switch between different forms of an equation and apply the appropriate algebraic techniques to find a solution.