Problem 60 - Algebra 2

Question

Simplify each expression. $$ \frac{4}{2 x-6}+\frac{x+1}{x-3} $$

Step-by-Step Solution

Answer

The result of simplifying the given expression is \( \frac{x+3}{x-3}\).

Step 1: Understand What Common Denominator is Required

The first step in this process is understanding what the common denominator must be. Since the two denominators in the fraction are \(2x-6\) and \(x-3\), and because \(2x-6\) can be simplified to \(2(x-3)\), it is clear that \(2(x-3)\) is the least common denominator.

Step 2: Equivalent Fractions

In the second step, we change each fraction to an equivalent one that has a denominator of \(2(x-3)\). The first fraction already has this denominator, so leave this as is. To change the second fraction, multiply both the numerator and denominator by 2. This gives us the expression \(\frac{4}{2x-6} + \frac{2(x+1)}{2(x-3)}\).

Step 3: Adding the Fractions

Now that both fractions have a common denominator, they can be added together. This gives us \(\frac{4 + 2(x+1)}{2(x-3)}\).

Step 4: Simplify the Numerator

Distribute the 2 in the numerator of the new fraction and then combine like terms. This simplifies the fraction to \(\frac{4+2x+2}{2(x-3)} = \frac{2x+6}{2(x-3)}\).

Step 5: Further Reduce the Fraction

The fraction \(\frac{2x+6}{2(x-3)}\), can be further reduced by factoring out a 2 from the numerator which yields the expression \(\frac{2(x+3)}{2(x-3)}\). After which the 2's can be cancelled to obtain the simplified result \( \frac{x+3}{x-3}\).

Key Concepts

Common Denominator

When working with algebraic fractions, finding a common denominator is essential to perform operations like addition or subtraction. The common denominator is the least common multiple (LCM) of the denominators from the fractions you want to combine. In our original problem:

• We have two fractions: \( \frac{4}{2x-6} \) and \( \frac{x+1}{x-3} \).
• Notice that \( 2x-6 \) can be rewritten as \( 2(x-3) \).
• This means one fraction already has a denominator in terms of \( (x-3) \).

Therefore, the common denominator for both fractions is \( 2(x-3) \). Utilizing the common denominator allows us to rewrite both fractions with the same bottom term, paving the way for seamless addition or subtraction later. Ensuring both fractions have the same denominator is critical, as it aligns their scale, making further operations logical and error-free.

Simplifying Fractions

Once you've aligned fractions with a common denominator, the next goal is to simplify them. Simplifying fractions involves reducing them to their most compact form without changing their value. This often means factoring expressions to find and cancel out common terms.Consider our expression:

• After converting to a common denominator, we have \( \frac{4 + 2(x+1)}{2(x-3)} \).
• Distribute the 2 within the numerator to get \( \frac{4 + 2x + 2}{2(x-3)} \).
• Simplify by combining like terms to obtain \( \frac{2x + 6}{2(x-3)} \).

This expression can be further simplified:

• Notice that the numerator, \( 2x + 6 \), can be factored as \( 2(x+3) \).
• Now, \( \frac{2(x+3)}{2(x-3)} \) allows us to cancel the common factor of 2 from the numerator and denominator.

This cancelation results in the simplified expression \( \frac{x+3}{x-3} \). By simplifying, calculations are easier, and your results are cleaner and more interpretable.

Equivalent Fractions

Equivalent fractions are different representations of the same value. When dealing with algebraic fractions, equivalent fractions help to match denominators and compute results like sums or differences seamlessly.In our exercise:

• We start with fractions \( \frac{4}{2x-6} \) and \( \frac{x+1}{x-3} \).
• To form equivalent fractions with a shared denominator, note that the first fraction already is \( \frac{4}{2(x-3)} \).
• To make the second fraction equivalent, multiply both its top and bottom by 2, getting \( \frac{2(x+1)}{2(x-3)} \).

Through these transformations:

• We've created equivalent fractions with the desired common denominator.
• This careful multiplication and transformation processes allow for seamless addition, evidenced by resulting expressions such as \( \frac{4+2(x+1)}{2(x-3)} \).

Equivalent fractions maintain the same value before and after transformation, which is crucial for accurate calculations in algebraic manipulations.