Problem 21 - Algebra 2 and Trigonometry

Question

Simplify each expression. In each case, list any values of the variables for which the fractions are not defined. \(\frac{3}{2 x}-\frac{1+\frac{1}{x}}{x+1}\)

Step-by-Step Solution

Answer

The simplified expression is \( \frac{1}{2x} \) and it is undefined for \(x = 0\) and \(x = -1\).

Step 1: Identify the Denominators

In the expression \(\frac{3}{2x}-\frac{1+\frac{1}{x}}{x+1}\), identify the denominators. The first term has a denominator of \(2x\) and the second term has a denominator of \(x+1\). Additionally, the term \(\frac{1}{x}\) inside the second fraction introduces another denominator which is \(x\).

Step 2: Determine Values that Make Denominators Zero

We determine the values of \(x\) that will make any of the denominators zero, which makes the expression undefined. For \(2x\), if \(x = 0\), the denominator is zero. For \(x+1\), if \(x = -1\), the denominator is zero. Hence, the expression is undefined for \(x = 0\) and \(x = -1\).

Step 3: Convert the Compound Fraction

The second term \(-\frac{1+\frac{1}{x}}{x+1}\) is a compound fraction. To simplify, find a common denominator within \(1 + \frac{1}{x}\), which is \(x\). Express it as \(\frac{x+1}{x}\) and then substitute back into the expression: \(-\frac{\frac{x+1}{x}}{x+1}\).

Step 4: Simplify the Expression

Simplify the expression \(-\frac{\frac{x+1}{x}}{x+1}\). Notice that \(x+1\) is both the numerator and the denominator in the compound fraction itself, thus they cancel each other out: \(-\frac{1}{x}\).

Step 5: Combine the Simplified Fractions

The expression is now \(\frac{3}{2x} - \frac{1}{x}\). To subtract these fractions, find a common denominator, which is \(2x\). Rewrite \(-\frac{1}{x}\) as \(-\frac{2}{2x}\) so they have a common denominator.

Step 6: Subtract the Fractions

With a common denominator of \(2x\), subtract the numerators: \(\frac{3}{2x} - \frac{2}{2x} = \frac{3 - 2}{2x} = \frac{1}{2x}\). This is the simplified expression.

Key Concepts

Undefined Fractions

In algebra, fractions are undefined when their denominators are zero. This is important because division by zero is not possible, leading to undefined expressions.
Imagine you have some pencils, and you're trying to divide them among no boxes. The division doesn't make sense; there's no way to distribute the pencils.
For the given expression, we identified several points where the fractions become undefined:

• In the denominator of the first term, which is \(2x\), the fraction is undefined if \(x = 0\).
• The second fraction has the denominator \(x + 1\), making it undefined if \(x = -1\).
• Finally, within the compound fraction, we found a denominator of \(x\), again undefined for \(x = 0\).

In summary, for the original expression, the variable \(x\) must not be equal to 0 or -1 to prevent undefined fractions.

Common Denominator

Finding a common denominator is crucial when adding or subtracting fractions, especially in algebra.
A common denominator is like a common ground. It lets fractions "speak the same language" so you can work with them as a single entity.
In the initial expression, we simplified \(\frac{3}{2x} - \frac{1}{x}\) by finding a common denominator.

• The common denominator between \(2x\) and \(x\) is \(2x\).
• We then rewrote \(-\frac{1}{x}\) as \(-\frac{2}{2x}\) to ensure both fractions were expressed over \(2x\).

This technique allows the fractions to be subtracted smoothly, leading to a simplified expression. With a common denominator in place, subtracting their numerators gives you the clean result.

Compound Fraction

A compound fraction looks complex at first because it has a fraction within a fraction.
Imagine a burger within a bun; layering can make things seem complicated, but breaking it down reveals familiar components.
The compound fraction in our problem was \(-\frac{1+\frac{1}{x}}{x+1}\). To simplify, we first focused on the inner fraction \(\frac{1}{x}\).

• By finding a common denominator within "\(1+\frac{1}{x}\)," we expressed it as \(\frac{x+1}{x}\).
• This turned the compound fraction into something simpler: \(-\frac{\frac{x+1}{x}}{x+1}\).

At this point, the expression becomes easier to handle—like removing the layers of a complicated dish to focus just on the main ingredients.

Simplification Steps

Simplifying algebraic expressions involves breaking down complex parts into simpler ones. It's like untangling a knot until you have straight sections of string.
Follow these steps for smoother algebraic simplification:
First, identify denominators and check for undefined variables.
Then, simplify any compound fractions by removing the layers inside them.

• For \(-\frac{\frac{x+1}{x}}{x+1}\), noticed how \(x+1\) appears in both the numerator and denominator of the outer fraction. They cancel each other out.
• After canceling, you're left with \(-\frac{1}{x}\).
• Now, rewrite any remaining fractions with a common denominator for straightforward addition or subtraction.
• In this context, \(\frac{3}{2x} - \frac{2}{2x} = \frac{1}{2x}\).

Finally, neatly subtract the numerators to arrive at the simplified expression. Each step untangles a part of the problem, leaving you with an elegant final answer.