Problem 55 - College Algebra and Calculus: An Applied Approach

Question

A student states that the solution to the equation \(\frac{2}{x(x-2)}+\frac{5}{x}=\frac{1}{x-2}\) is \(x=2\). Describe and correct the student's error.

Step-by-Step Solution

Answer

The student is incorrect; the equation does not have a solution. The value \(x = 2\) makes the denominator of the first and the third term in the original equation undefined, which means \(x = 2\) cannot be a solution.

Step 1: Identify the invalid values

First and foremost, figure out the values of \(x\) that lead to a denominator of zero. These are \(x=0\) and \(x=2\). In mathematics, dividing by zero is illegal, so we know these cannot be valid solutions.

Step 2: Eliminate fractions

To simplify the calculation, multiply each term by \(x(x-2)\), the least common denominator (LCD) for the fractions in the equation. This gives us \(2 + 5(x - 2) = x\).

Step 3: Simplify and rearrange the equation

Simplify the equation to have it in its standard form, by distributing the 5 in the second term on the left hand side. This gives \(2 + 5x - 10 = x\), which simplifies to \(4x = 8\).

Step 4: Solve for \(x\)

Dividing by 4 on both sides, we get \(x = 2\). However, as noted earlier, \(x = 2\) is not a valid solution as it leads to division by zero in the original equation.

Step 5: Verify other solutions

The great part about algebra is that you can always check your solutions. Substitute \(x = 2\) back into the initial equation to check. You will see that it does not hold since the denominator of one of the terms would become zero. Hence, the given equation has no solution.

Key Concepts

Least Common Denominator

When solving equations with algebraic fractions, it is essential to find the least common denominator (LCD) to combine the terms. The LCD is the smallest number that each denominator can divide into evenly. It's like a common ground for all fractions involved, allowing us to eliminate the fractions and solve a simpler equation.

In our exercise, the fractions \( \frac{2}{x(x-2)} \) and \( \frac{5}{x} \) have denominators that can be combined into the LCD \( x(x-2) \). By multiplying every term in the equation by the LCD, we can then work with whole numbers instead of fractions, making the steps to find the solution much clearer and less error-prone. It's important to remember that finding the LCD is a key tool in solving rational equations effectively.

Division by Zero

One of the most crucial rules in algebra—and indeed, in all of mathematics—is that division by zero is undefined. This is because dividing a number by zero does not produce a sensible result. In the context of our exercise, the values of \( x \) that make any denominator zero are not permissible as solutions to the equation.

For instance, if a student suggests \( x=2 \) as a solution, but this value turns the denominator of \( \frac{2}{x(x-2)} \) into zero, then it's an invalid solution. Division by zero is not just illegal in mathematics; it fundamentally breaches the concept that for any number \( a \) divided by another number \( b \) (\( \frac{a}{b} \)), the result should be a number that, when multiplied by \( b \) gives \( a \) again. With \( b \) as zero, there is no such number, as anything multiplied by zero is always zero.

Algebraic Fractions

Algebraic fractions are just like regular fractions, but instead of numbers, they have variables (like \( x \) or \( y \) in their numerators or denominators. Solving equations with algebraic fractions often involves finding a common denominator and sometimes factoring polynomials.

In our problem, both \( \frac{2}{x(x-2)} \) and \( \frac{1}{x-2} \) are algebraic fractions. It is necessary to handle these expressions with care, ensuring that when we perform operations like addition, subtraction, or finding equivalent fractions, we consider the behavior of the variable and avoid any undefined states like division by zero.

Equation Simplification

The beauty of algebra is in simplifying complex expressions into more manageable forms. Equation simplification might involve expanding products, combining like terms, or canceling common factors. The goal is to isolate the variable on one side of the equation to find its value.

In the step-by-step solution, after multiplying through by the LCD, the equation \(2 + 5(x - 2) = x\) is simplified by distributing and rearranging terms, eventually giving us \(4x = 8\). Simplifying further, we get \(x = 2\), which seems like a solution but is not valid due to the restriction involving division by zero. Thus, understanding and applying simplification techniques is vital to solving algebraic equations successfully while keeping an eye out for any restrictions.