Problem 61 - Elementary and Intermediate Algebra

Question

Perform the operations. Simplify, if possible. $$ \frac{b}{b+1}-\frac{b-1}{b+2} $$

Step-by-Step Solution

Answer

\(\frac{b}{b+1}-\frac{b-1}{b+2} = \frac{2b+1}{(b+1)(b+2)}\).

Step 1: Identify the Least Common Denominator (LCD)

To combine the fractions, identify a common denominator. The denominators are \(b+1\) and \(b+2\). The LCD will be the product of these two expressions: \((b+1)(b+2)\).

Step 2: Rewrite each Fraction with the LCD

Rewrite each fraction using the LCD. For the first fraction, multiply both the numerator and denominator by \(b+2\):\[\frac{b}{b+1} = \frac{b(b+2)}{(b+1)(b+2)} = \frac{b^2 + 2b}{(b+1)(b+2)}\]For the second fraction, multiply both the numerator and denominator by \(b+1\):\[\frac{b-1}{b+2} = \frac{(b-1)(b+1)}{(b+2)(b+1)} = \frac{b^2 - 1}{(b+1)(b+2)}\]

Step 3: Subtract the Numerators

Now that the fractions have a common denominator, subtract the numerators:\[\frac{b^2 + 2b}{(b+1)(b+2)} - \frac{b^2 - 1}{(b+1)(b+2)} = \frac{(b^2 + 2b) - (b^2 - 1)}{(b+1)(b+2)}\]Simplify the numerator:\[b^2 + 2b - b^2 + 1 = 2b + 1\]

Step 4: Write the Final Expression

Build the final simplified fraction using the LCD:\[\frac{2b + 1}{(b+1)(b+2)}\]This is the final expression as no further simplification is possible.

Key Concepts

Understanding the Least Common Denominator

When working with algebraic fractions, much like numerical fractions, we need to have a common denominator in order to add or subtract them. This common ground is called the Least Common Denominator (LCD).
For simple numbers, this can be something like finding a common multiple. In algebraic terms, however, the LCD is the smallest expression that can completely divide each of the original denominators.In our exercise, we're working with the fractions \(\frac{b}{b+1}\) and \(\frac{b-1}{b+2}\). Each fraction has a different denominator, those being \(b+1\) and \(b+2\).
To find the LCD, we multiply these denominators together:

• Step 1: The first denominator is \(b+1\).
• Step 2: The second denominator is \(b+2\).
• Step 3: Multiply the denominators: \((b+1)(b+2)\). This gives us our LCD.

By using the LCD \((b+1)(b+2)\), we ensure that both fractions can be transformed to have the same bottom value, allowing them to be subtracted directly.

Steps in Numerator Subtraction

Once both fractions have the same denominator, the next task is to focus on the numerators. Only at this point can we perform addition or subtraction between the fractions. With our newly acquired LCD, we can rewrite each fraction. This step is crucial to successfully performing the operation.Here is what happens next:

• First fraction: \(\frac{b \cdot (b+2)}{(b+1)(b+2)} = \frac{b^2 + 2b}{(b+1)(b+2)}\)
• Second fraction: \(\frac{(b-1) \cdot (b+1)}{(b+2)(b+1)} = \frac{b^2 - 1}{(b+1)(b+2)}\)

You'll then subtract the second numerator from the first. The key task here is correctly distributing the negative sign. Doing this, the expression inside the numerator becomes:\[(b^2 + 2b) - (b^2 - 1) = b^2 + 2b - b^2 + 1 = 2b + 1\].Accurate subtraction here transitions us smoothly to the next stage: fraction simplification.

Getting to Fraction Simplification

Simplification is the final process in solving algebraic fraction problems. After combining the numerators, we typically look for any further simplifications. Simplifying fractions means reducing them to their simplest form or finding common factors that can cancel out.In the exercise expression, once we've combined the numerators to get \(2b + 1\), the fraction looks like this:\[\frac{2b + 1}{(b+1)(b+2)}\].
The next step is to check if there is any factor that is common in the numerator and the denominator.Inspect:

• Numerator: \(2b + 1\)
• Denominator expansion: \((b+1)(b+2) = b^2 + 3b + 2\)

Since there are no common factors in both terms, the fraction is already in its simplest form.
Reducing fractions is important in ensuring the expression is both concise and comprehensive, making it easier to understand and interpret!