Problem 23 - Algebra 1

Question

Simplify the expression. $$\frac{x}{x-10}+\frac{x+4}{x+6}$$

Step-by-Step Solution

Answer

The simplified form of the given expression is \( \frac{2(x^2+20)}{(x+6)(x-10)} \).

Step 1: Find the least common denominator

To add two fractions, their denominators must be the same. The least common denominator(LCD) is the smallest number that both denominators (x-10 and x+6) can divide into. Observe that no factors are common between the two denominators x-10 and x+6, thus the LCD is the product of these two denominators, \((x-10)(x+6)\).

Step 2: Express fractions with the least common denominator

Rewrite both fractions using the LCD. Multiply both the numerator and the denominator of the first fraction by \(x+6\) and the numerator and the denominator of the second fraction by \(x-10\). This results in \(\frac{x(x+6)}{(x-10)(x+6)} + \frac{(x+4)(x-10)}{(x+6)(x-10)}\).

Step 3: Simplify the fractions' numerators

Apply the distributive property to multiply the numerators out in both fractions. This results in \(\frac{x^2+6x}{(x-10)(x+6)} + \frac{x^2-6x+40}{(x+6)(x-10)}\).

Step 4: Add the fractions

Since both fractions have the same denominator now, they can be added. By doing so, the numerator becomes \(x^2+6x+x^2-6x+40\). Therefore, the fractions sum to \(\frac{2x^2+40}{(x+6)(x-10)}.\)

Step 5: Simplify the result

The numerator can be simplified by factoring out common terms. After factoring 2 from the numerator, the final simplified expression is \(\frac{2(x^2+20)}{(x+6)(x-10)}.\)

Key Concepts

Least Common Denominator

When adding algebraic fractions, one crucial step is to find the Least Common Denominator (LCD). The LCD is like a shared platform that allows different fractions to come together seamlessly. In this exercise, we have the fractions \(\frac{x}{x-10}\) and \(\frac{x+4}{x+6}\). Since their denominators are distinct expressions \(x-10\) and \(x+6\), we must find a common base for these denominators to make adding them possible.

• First, analyze the denominators, \(x-10\) and \(x+6\). They don't share any common factors apart from 1.
• This means that the simplest LCD is just the product of these two expressions, \((x-10)(x+6)\).
• Transform each fraction so that they both share this denominator.

This foundational step ensures that both fractions can be added easily, setting the stage for further simplification.

Distributive Property

The Distributive Property is a handy tool when dealing with algebraic expressions, especially when you need to multiply terms. It's like a magic wand that helps spread one factor across terms within parentheses. Let’s see how it works.In our exercise, after rewriting fractions with the LCD, we get \(\frac{x(x+6)}{(x-10)(x+6)}\) and \(\frac{(x+4)(x-10)}{(x+6)(x-10)}\). Here, we use the distributive property to expand the numerators:

• First, distribute \(x\) across \(x+6\) to get \(x^2 + 6x\).
• Then, distribute \(x + 4\) across \(x - 10\) to obtain \(x^2 - 10x + 4x - 40\), which further simplifies to \(x^2 - 6x - 40\).

By using the distributive property, the expressions inside the numerators are expanded, allowing for easier addition because now they're polynomial expressions that can be simplified later.

Factoring

Factoring is breaking down a complex expression into simpler parts that, when multiplied, give the original expression back. It's a reverse process of multiplication that simplifies expressions by identifying and extracting common factors.In the final step of our exercise, the expression after combining the fractions becomes \(\frac{2x^2 + 40}{(x+6)(x-10)}\). To factor the numerator:

• Look for a common factor in \(2x^2 + 40\). Notice that both terms share a factor of 2.
• Factor out the 2, leaving \(2(x^2 + 20)\).

Factoring made the expression tidier and easier to interpret, resulting in the final simplified expression: \(\frac{2(x^2 + 20)}{(x+6)(x-10)}\). This step is particularly useful in further algebraic operations or solving equations that involve such expressions.