Problem 32 - Algebra for College Students

Question

Simplify each complex fraction. $$\frac{\frac{3}{2+x}-\frac{4}{2-x}}{\frac{1}{x+2}-\frac{3}{x-2}}$$

Step-by-Step Solution

Answer

\frac{-(7x+2)}{2(x+4)}

Step 1: Simplify the Numerator

First, simplify the numerator of the complex fraction, which is \frac{3}{2+x} - \frac{4}{2-x}. Find a common denominator for the two fractions. Notice that (2-x) can be written as -(x-2). So, the common denominator is (2+x)(x-2).\[ \frac{3}{2+x} - \frac{4}{2-x} = \frac{3(x-2) - 4(-1)(2+x)}{(2+x)(x-2)} = \frac{3(x-2) + 4(2+x)}{(2+x)(x-2)} \]Simplify the numerator:\[ 3(x-2) + 4(2+x) = 3x - 6 + 8 + 4x = 7x + 2 \]Now the numerator is:\[ \frac{7x+2}{(2+x)(x-2)} \]

Step 2: Simplify the Denominator

Next, simplify the denominator of the complex fraction, which is \frac{1}{x+2} - \frac{3}{x-2}. The common denominator here is also (x+2)(x-2).\[ \frac{1}{x+2} - \frac{3}{x-2} = \frac{1(x-2) - 3(x+2)}{(x+2)(x-2)} = \frac{x-2 - 3x - 6}{(x+2)(x-2)} \]Simplify the numerator:\[ x - 2 - 3x - 6 = -2x - 8 \]Now the denominator is:\[ \frac{-2x-8}{(x+2)(x-2)} \]

Step 3: Divide the Numerator by the Denominator

Now, divide the simplified numerator by the simplified denominator:\[ \frac{\frac{7x+2}{(2+x)(x-2)}}{\frac{-2x-8}{(x+2)(x-2)}} \]This can be written as:\[ \frac{7x+2}{(2+x)(x-2)} \times \frac{(x+2)(x-2)}{-2x-8} \]Notice that the (x+2)(x-2) terms cancel out, leaving us with:\[ \frac{7x+2}{-2(x+4)} \]

Step 4: Simplify the Remaining Expression

Simplify the remaining expression:\[ \frac{7x+2}{-2(x+4)} \]to reveal the final simplified expression:\[ = \frac{-(7x+2)}{2(x+4)} \]or\[ = \frac{-(7x+2)}{2(x+4)} \]

Key Concepts

Finding Common Denominators

When simplifying complex fractions, the first thing you often need to do is find a common denominator. This is crucial because it allows you to combine fractions by providing a unified base. Consider this fraction: \[ \frac{3}{2+x} - \frac{4}{2-x} \]. At a glance, the denominators look different, but they have a relationship. Notice that \(2-x\) can be rewritten as \(-(x-2)\). Therefore, the common denominator is \((2 + x)(x - 2)\).
By aligning the denominators, we can rewrite the original expression as:

• \[ \frac{3(x-2) - 4(-1)(2+x)}{(2+x)(x-2)} = \frac{3(x-2) + 4(2+x)}{(2+x)(x-2)}\]

This way, we convert the fractions into a single fraction, making manipulation easier in subsequent steps.

Algebraic Manipulation

Once you've found the common denominators, the next step involves manipulating the algebraic expressions. This often means expanding and simplifying terms. Continuing from the common denominator, you get:

• \[3(x-2) + 4(2+x) = 3x - 6 + 8 + 4x = 7x + 2\]

By combining like terms, you attain a cleaner, simplified numerator or denominator. Algebraic manipulation is essentially about organizing and simplifying expressions to make them more manageable. In this case, the numerator becomes \[ \frac{7x+2}{(2+x)(x-2)} \]. Performing similar steps on the denominator helps to balance the entire fraction for the next stages.

Canceling Common Terms

After simplifying the numerator and the denominator separately, the next step is to divide these expressions. This involves flipping the denominator and multiplying. For our example, we have:

• \[ \frac{\frac{7x+2}{(2+x)(x-2)}}{\frac{-2x-8}{(x+2)(x-2)}} = \frac{7x+2}{(2+x)(x-2)} \times \frac{(x+2)(x-2)}{-2x-8} \]

Notice that \((x+2)(x-2) \) appears in both the numerator and the denominator, meaning these terms can be canceled out. This simplification is crucial because it reduces the fraction to its irreducible form, leaving us with:

• \[ \frac{7x+2}{-2(x+4)} \]

Simplifying Expressions

The final step in solving our complex fraction involves simplifying the remaining expression. Here, you manipulate the remaining terms to get the simplest form possible. For the given problem, we need to work with\(* \frac{7x + 2}{-2(x+4)}*\):

• To make the expression more standard, we can factor out the negative sign from the denominator: \[ \frac{7x+2}{-2(x+4)} \rightarrow \frac{-(7x+2)}{2(x+4)} \]

This last bit of simplification ensures that the fraction is in its most reduced form. It's important always to look for these opportunities to reduce complexity further, making the final answer clean and easy to interpret.