- The first step correctly factors the quadratic expression, yielding \(\frac{x}{(x-2)(x-3)} + \frac{3}{x+3}\).
- To add these fractions, the Least Common Denominator (LCD) must be \((x-2)(x-3)(x+3)\).
- Riley's mistake occurs in Step 2, where she uses an incorrect common denominator of \((x-2)(x+3)\) and fails to include the \((x-3)\) factor.
- This results in the first fraction being changed from \(\frac{x}{(x-2)(x-3)}\) to \(\frac{x}{(x-2)(x+3)}\), which is not mathematically equivalent.
Explanation
- Analyzing the Expression and Step 1
The given expression to simplify is \(\frac{x}{x^2-5x+6} + \frac{3}{x+3}\). First, we factor the quadratic expression in the denominator of the first fraction: \(x^2 - 5x + 6 = (x-2)(x-3)\). Thus, the expression becomes \(\frac{x}{(x-2)(x-3)} + \frac{3}{x+3}\). This confirms that Step 1 is correct.
- Determining the Correct Common Denominator
To add two fractions with different denominators, we must find the Least Common Denominator (LCD). The denominators are \((x-2)(x-3)\) and \((x+3)\). The LCD is the product of these factors, which is \((x-2)(x-3)(x+3)\). To correctly combine these fractions, we must multiply the numerator and denominator of each fraction by the missing factors from the LCD. The first fraction \(\frac{x}{(x-2)(x-3)}\) should be multiplied by \(\frac{x+3}{x+3}\) to get \(\frac{x(x+3)}{(x-2)(x-3)(x+3)}\). The second fraction \(\frac{3}{x+3}\) should be multiplied by \(\frac{(x-2)(x-3)}{(x-2)(x-3)}\) to get \(\frac{3(x-2)(x-3)}{(x-2)(x-3)(x+3)}\).
- Identifying the Mistake
In Step 2, Riley wrote: \(\frac{x}{(x-2)(x+3)} + \frac{3(x-2)}{(x-2)(x+3)}\). Comparing this to our analysis, Riley incorrectly changed the denominator of the first fraction from \((x-2)(x-3)\) to \((x-2)(x+3)\). By doing this, she essentially replaced the factor \((x-3)\) with \((x+3)\) and failed to include the necessary factor \((x-3)\) in her common denominator. Consequently, the first term in Step 2 is not equivalent to the first term in Step 1. Her chosen common denominator \((x-2)(x+3)\) is missing the factor \((x-3)\), which led to an incorrect simplification.
Examples
When managing a warehouse, imagine you have two different types of shipping containers that need to be categorized into uniform batches. If you are grouping items of different sizes (represented by different denominators), you must find a common way to label them so that you can add or track them together. Failing to include all the necessary components of each group when creating a 'common' label leads to an incorrect inventory count, just as Riley’s algebraic error resulted in an incorrect sum of fractions. Precision in identifying common factors is essential for accurate resource management and problem-solving.