Problem 71 - Elementary Algebra

Question

For exercises \(35-86\), simplify. $$ \frac{4}{v^{2}-1}-\frac{2}{v-1} $$

Step-by-Step Solution

Answer

\( \frac{-2}{v+1} \)

Step 1: Identify and Factor the Denominators

First, factor the denominator of the first fraction: \[ v^2 - 1 = (v - 1)(v + 1)\] Now the problem looks like this: \[ \frac{4}{(v-1)(v+1)} - \frac{2}{v-1} \]

Step 2: Rewrite the Second Fraction with a Common Denominator

To combine the fractions, they need a common denominator. The common denominator is \((v-1)(v+1)\). Rewrite the second fraction with this common denominator: \[ \frac{2}{v-1} = \frac{2(v+1)}{(v-1)(v+1)} \] Now rewrite the expression as: \[ \frac{4}{(v-1)(v+1)} - \frac{2(v+1)}{(v-1)(v+1)} \]

Step 3: Combine the Fractions

Now that both fractions have the same denominator, subtract the numerators: \[ \frac{4 - 2(v+1)}{(v-1)(v+1)} = \frac{4 - 2v - 2}{(v-1)(v+1)} \] Simplify the numerator: \[ 4 - 2v - 2 = 2 - 2v \] Rewrite the expression: \[ \frac{2 - 2v}{(v-1)(v+1)} \]

Step 4: Simplify the Result

Factor out \(-2\) from the numerator: \[ \frac{-2(v-1)}{(v-1)(v+1)} \] Cancel \((v-1)\) from the numerator and the denominator: \[ \frac{-2}{v+1} \]

Key Concepts

Factoring Polynomials

Factoring polynomials means expressing a polynomial as a product of simpler polynomials. For instance, in our exercise, we need to factor the denominator of the first fraction:

• The denominator is \[ v^2 - 1 \]. Here, we recognize this as a difference of squares, which can be factored as \[ (v-1)(v+1) \].

By factoring, we transform expressions into more manageable forms.
This is crucial in simplifying algebraic fractions, making further steps more straightforward.

Common Denominator

When working with multiple fractions, finding a common denominator is key. It allows you to combine the fractions easily.

• Let's consider our second fraction from the exercise: \[ \frac{2}{v-1} \].
• To combine it with the first fraction, \[ \frac{4}{(v-1)(v+1)} \], we need to have the same denominator across both fractions.
• So, we rewrite \[ \frac{2}{v-1} \] as \[ \frac{2(v+1)}{(v-1)(v+1)} \].

Now, both fractions share the denominator \[ (v-1)(v+1) \].
This common ground is essential for the next steps.

Fraction Simplification

Simplifying fractions involves reducing expressions to their simplest form. Here’s how we did it in our problem:

• First, we expressed both fractions with a common denominator: \[ \frac{4}{(v-1)(v+1)} - \frac{2(v+1)}{(v-1)(v+1)} \].
• Second, we combined them by subtracting the numerators: \[ \frac{4 - 2(v+1)}{(v-1)(v+1)} \].

This resulted in \[ \frac{4 - 2v - 2}{(v-1)(v+1)} \], which simplifies to \[ \frac{2 - 2v}{(v-1)(v+1)} \].
Always look to simplify further if possible, like factoring out common terms or canceling out common factors.

Combining Fractions

Combining fractions is often necessary to simplify complex algebraic expressions. This involves:

• Ensuring all fractions have a common denominator.
• Combining the numerators according to the operation (addition or subtraction).

In our exercise, both fractions were rewritten with the common denominator \[ (v-1)(v+1) \].
We then combined them by subtracting the numerator terms: \[ 4 - 2(v+1) \] to get \[ 2 - 2v \].
Finally, we simplified by factoring and canceling out \[ (v-1) \], resulting in \[ \frac{-2}{v+1} \].
Follow these steps to consistently simplify algebraic fractions effectively.