Problem 68 - Introductory Algebra for College Students

Question

perform the indicated operation or operations. Simplify the result, if possible. $$\frac{3 x}{(x+1)^{2}}-\left[\frac{5 x+1}{(x+1)^{2}}-\frac{3 x+2}{(x+1)^{2}}\right]$$

Step-by-Step Solution

Answer

After following these steps, the simplified result of the operation given is \(\frac{1}{x+1}\)

Step 1: Evaluate the Expression Inside the Square Brackets

To begin with, solve the operation inside the square brackets, which is subtraction of two fractions: \(\frac{5x+1}{(x+1)^2} - \frac{3x+2}{(x+1)^2}\). Both fractions have the same denominator, so subtract directly, remembering that subtracting a fraction is equivalent to adding a fraction with a negative numerator. Thus, the expression becomes \(\frac{(5x+1)-(3x+2)}{(x+1)^2} = \frac{2x-1}{(x+1)^2}\)

Step 2: Evaluate the Entire Expression

Now, the entire expression simplifies to \(\frac{3x}{(x+1)^2} - \frac{2x-1}{(x+1)^2}\). The two terms have common denominators, so we can combine them into a single fraction. Hence it simplifies to \(\frac{3x-(2x-1)}{(x+1)^2}\)

Step 3: Simplify the Numerator

It remains to simplify the numerator of the resulting fraction from Step 2. By order of operations, first subtract the 2x, then subtract the -1, hence it simplifies to \(\frac{3x-2x+1}{(x+1)^2} = \frac{x+1}{(x+1)^2}\)

Step 4: Simplify the Entire Fraction

Finally, simplify the fraction to obtain the result. \(x+1\) is a common factor in the numerator and denominator, so we cancel them out. This simplifies the expression to \(\frac{1}{x+1}\)

Key Concepts

Subtraction of Fractions

Subtraction of fractions might seem tricky at first, but once you get the hang of it, it's pretty straightforward. When subtracting fractions, the key step involves making sure that the denominators, which are the bottom parts of the fractions, are the same. If the denominators are different, you'll need to find a common one. Once you have identical denominators, you can subtract the numerators, like you would with simple numbers.
Why do we need a common denominator? Because fractions are parts of a whole, and the denominator tells us how many equal parts the whole is divided into. To subtract correctly, each fraction must refer to the same-sized parts.
For example, in the expression \( \frac{5x+1}{(x+1)^2} - \frac{3x+2}{(x+1)^2} \), the denominators are already the same \((x+1)^2\). This makes it easy to subtract the numerators directly, leading to \( \frac{(5x+1)-(3x+2)}{(x+1)^2} = \frac{2x-1}{(x+1)^2} \). Remember, subtraction is essentially adding the negative of the second term.

Common Denominators

When dealing with fractions, ensuring that the denominators are common—or the same—is crucial for both addition and subtraction. This allows us to perform operations directly on the numerators without altering the symmetrical balance of each fraction.
To find a common denominator, look for the least common multiple (LCM) of the denominators if they are different. In cases where the denominators are already identical, like \((x+1)^2\) in our example, there's nothing additional to do.

• If the denominators were different, finding a common denominator might involve multiplying each fraction by a suitable expression that makes all denominators the same.
• The goal is to transform each fraction into an equivalent fraction that has this shared denominator.

Working with a common denominator makes the process of subtraction straightforward, as demonstrated in the simplification from \( \frac{3x}{(x+1)^2} - \frac{2x-1}{(x+1)^2} \) to \( \frac{x+1}{(x+1)^2} \). The fractions combine easily into a single fraction.

Simplifying Algebraic Expressions

Simplifying is all about making an expression as simple as possible without changing its value. In algebra, this often involves reducing fractions and canceling common factors. By simplifying, complex fractions become easier to understand and work with.
In our problem, after subtracting the fractions, we reach \( \frac{x+1}{(x+1)^2} \). Now, look at both the numerator and the denominator. Since \(x+1\) appears in both, they can "cancel" each other out.

• The numerator \(x+1\) is a factor in the denominator as well, so removing it leaves you with \(\frac{1}{x+1}\).
• Remember that canceling common factors is a crucial step in simplifying algebraic expressions, as it reduces the expression to its simplest form.

The final result, \( \frac{1}{x+1} \), shows perfectly how simplification can transform a seemingly complex expression into a clean and manageable form. Practice recognizing and canceling common factors to make algebra more approachable.