Problem 54 - Elementary Algebra

Question

For Problems \(52-54\), solve each equation. $$ \frac{7 x-12}{x^{2}-16}-\frac{5}{x+4}=\frac{2}{x-4} $$

Step-by-Step Solution

Answer

All real numbers except \(x = 4\) and \(x = -4\).

Step 1: Identify the Least Common Denominator

First, identify the least common denominator (LCD) for the fractions in the equation. The denominators here are \((x^2 - 16)\), \((x+4)\), and \((x-4)\). Notice that \(x^2 - 16\) can be factored into \((x+4)(x-4)\). The LCD is then \((x+4)(x-4)\).

Step 2: Eliminate the Fractions

Multiply each term in the equation by the LCD to eliminate the fractions. \[ \frac{7x-12}{x^2-16} \times (x+4)(x-4) - \frac{5}{x+4} \times (x+4)(x-4) = \frac{2}{x-4} \times (x+4)(x-4) \]This simplifies to:\[ (7x-12) - 5(x-4) = 2(x+4) \]

Step 3: Simplify the Equation

Distribute and simplify the equation:Expand the terms:\[ 7x - 12 - 5x + 20 = 2x + 8 \]Combine like terms:\[ 2x + 8 = 2x + 8 \]

Step 4: Analyze the Simplified Equation

Recognize that the simplified equation is always true:Since both sides of the equation are equal, \(2x + 8 = 2x + 8\) holds for all values of \(x\) except those that make the original fractions undefined (\(x = 4\) and \(x = -4\)).

Step 5: State the Solution Set

The solution set includes all real numbers except the values that make the original denominators zero. Therefore, the solution is all real numbers except \(x = 4\) and \(x = -4\).

Key Concepts

Fractions

Fractions can initially appear complicated, but they hold the key to many algebraic solutions. At their core, a fraction consists of a numerator (the top part) and a denominator (the bottom part). In algebraic equations, fractions help express ratios or divisions between expressions.
Understanding fractions in algebra is crucial since many equations involve them.
When dealing with algebraic fractions, you often need to manipulate them to solve equations.
Key actions include:

• Adding or subtracting fractions: To do this, you need a common denominator.
• Simplifying fractions: This often involves factoring both numerator and denominator to reduce the fraction.

In the given equation, you notice the fractions: \( \frac{7x - 12}{x^2 - 16} \), \( \frac{5}{x + 4} \), and \( \frac{2}{x - 4} \). Solving equations with these fractions requires understanding how to manipulate them effectively. Each fraction represents a part of the expression that needs careful handling to maintain the equation's balance.

Least Common Denominator

Finding the least common denominator (LCD) is crucial when working with fractions, especially in algebraic expressions. The LCD is the smallest expression that includes all the denominators of the fractions involved.
Identifying the LCD allows you to combine or eliminate fractions more efficiently.
In the original exercise, the denominators were \( (x^2 - 16) \), \( (x+4) \), and \( (x-4) \).
Here’s how to find the LCD:

• First, factor any complex denominators. For instance, \( x^2 - 16 \) can be factored into \((x+4)(x-4)\).
• Identify the unique factors from all denominators. In this case, they are \((x+4)\) and \((x-4)\).
• Combine these factors to form the LCD: \((x+4)(x-4)\).

This LCD can now be used to multiply each term in the equation to eliminate the fractions, which simplifies the solving process significantly.

Simplification of Expressions

Simplification is a vital step in solving algebraic equations. It involves making the expression as straightforward as possible without changing its value.
In algebra, simplification may involve:

• Distributing factors across terms.
• Combining like terms.
• Cancelling out terms that appear on both sides of an equation.

In the exercise, after eliminating the fractions by multiplying through with the LCD, the equation becomes: \( (7x-12) - 5(x-4) = 2(x+4) \).
Next steps include expanding each part of the equation to combine like terms:

• Distribute the \(-5\) across \( (x-4) \) and \(2\) across \( (x+4) \).
• Combine the like terms resulting in the simplified equation: \( 2x + 8 = 2x + 8 \).

Ultimately, simplification reveals whether an equation holds true under certain conditions.
In this case, the equation confirms that it is valid for all \(x\) values except those that make the original denominators zero (i.e., \(x = 4\) and \(x = -4\)).