Problem 39 - Algebra and Trigonometry with Analytic Geometry

Question

Solve the equation. $$\frac{4}{x+2}+\frac{1}{x-2}=\frac{5 x-6}{x^{2}-4}$$

Step-by-Step Solution

Answer

The solution is all real numbers except \(x = 2\) and \(x = -2\).

Step 1: Identify the Restrictions

The denominators in the equation are \(x+2\), \(x-2\), and \(x^2-4 = (x+2)(x-2)\). These are zero at \(x = -2\) and \(x = 2\). Therefore, these values are not allowed in the solution.

Step 2: Simplify the Right Side

Notice that \(x^2 - 4\) can be factored as \((x+2)(x-2)\). The equation can be rewritten as:\[ \frac{4}{x+2} + \frac{1}{x-2} = \frac{5x-6}{(x+2)(x-2)} \]The right-hand side is already in its simplest form based on the given expression.

Step 3: Create a Common Denominator

The denominators on the left side, \(x+2\) and \(x-2\), can be combined using the common denominator \((x+2)(x-2)\). Re-express the fractions:\[ \frac{4(x-2)}{(x+2)(x-2)} + \frac{1(x+2)}{(x-2)(x+2)} = \frac{5x-6}{(x+2)(x-2)} \]

Step 4: Combine the Fractions on the Left

Combine the fractions on the left to create a single fraction with the common denominator:\[ \frac{4(x-2) + 1(x+2)}{(x+2)(x-2)} \]Simplify the numerator:\[ 4(x-2) + 1(x+2) = 4x - 8 + x + 2 = 5x - 6 \]Thus, the equation becomes:\[ \frac{5x - 6}{(x+2)(x-2)} = \frac{5x - 6}{(x+2)(x-2)} \]

Step 5: Solve the Resulting Equation

Both sides of the equation are identical. Therefore, the equation holds for all values of \(x\) except where the expression is undefined. Based on the restrictions from Step 1, the solution is all real numbers except \(x = 2\) and \(x = -2\).

Key Concepts

Factoring

Factoring is one of the fundamental tools in algebra. It helps to break down complex expressions into simpler ones, often making them easier to solve or manipulate. In the context of rational equations, factoring allows us to simplify both the numerators and denominators.

When you look at the expression \( x^2 - 4 \), this can be factored into \((x+2)(x-2)\). This particular type of factoring is called the "difference of squares". It's a special pattern that appears in polynomials and is given by:\[ a^2 - b^2 = (a + b)(a - b) \]

This factoring is essential for solving rational equations because it helps reveal common denominators and simplify expressions to more easily manipulate and solve equations. Always remember to factor expressions completely to ensure you don't miss any simplifications.

Common Denominator

A common denominator is a shared multiple of the denominators of all the fractions in the equation. Using a common denominator is crucial when adding or subtracting rational expressions, just as it is when dealing with simple fractions.

In our rational equation, we had denominators like \(x+2\) and \(x-2\). By identifying the least common denominator, \((x+2)(x-2)\), we streamline the process of addition:

• Re-write each fraction with the common denominator.
• Ensure all terms are expressed with this denominator to ease calculation.

Using a common denominator allows for the combination of fractions and simplifies many algebraic manipulations. This approach creates a cohesive single fraction that can easily be used in solving the equation. Remember, finding and using the least common denominator makes solving these kinds of problems much more efficient.

Restrictions in algebra

When solving rational equations, it's important to identify restrictions. These are values for the variable that cause the denominators to be zero, making the expression undefined.

For example, in the equation given, denominators \(x+2\), \(x-2\), and \(x^2-4\) are present. Each of these expressions can result in a zero denominator:

• For \(x+2\), x must not be \(-2\).
• For \(x-2\), x must not be \(2\).
• Since \(x^2-4\) factors to \((x+2)(x-2)\), x cannot be \(-2\) or \(2\).

Identifying these restrictions before solving ensures you don't mistakenly conclude these values are solutions. Always check for these values first, as they are crucial for avoiding errors in rational equations.

Simplifying expressions

Simplifying expressions is about making the algebraic expressions as straightforward as possible. For rational equations, this often involves finding a common denominator, combining terms, and reducing the expressions to their simplest form.

For instance, in our example, we combine the fractions on the left after rewriting them with the common denominator. This creates a cleaner equation:

• Combine like terms in the numerators.
• Cancel out similar numerical terms if possible.
• Compare both sides to find simplifications.

In this case, the equation simplifies beautifully with \(5x - 6\) on both sides. This gives us an easily solvable equation, leading to the conclusion that it holds for all allowed values of \(x\). Simplifying expressions efficiently can reveal simple solutions for seemingly complex equations.