Problem 43 - Algebra 2

Question

Solve each equation. Check each solution. $$ \frac{2}{x-3}-\frac{4}{x+3}=\frac{8}{x^{2}-9} $$

Step-by-Step Solution

Answer

The solution to the equation is \(x=5\).

Step 1: Simplify the Equation

To make a rational equation easier to solve, combine terms to a single fraction. Here, find a common denominator and rewrite the fractions. The denominator \(x^{2}-9\) is equal to \((x-3)(x+3)\), which is the product of the denominators of the first and second fraction. The whole expression then simplifies to: \[\frac{2(x+3)}{(x-3)(x+3)} - \frac{4(x-3)}{(x-3)(x+3)} = \frac{8}{(x-3)(x+3)}.\]

Step 2: Simplify Further

Now, further simplification can be achieved by getting rid of the common denominator. This can be achieved by multiplying the entire equation by \((x-3)(x+3)\) which results in: \[2(x+3) - 4(x-3) = 8.\]

Step 3: Solve for x

By simplifying the above formula by distribution and collecting like terms, the expression becomes: \[2x + 6 - 4x +12 = 8.\] Which simplifies to \[-2x + 18 = 8.\] Finally, solving for x gives: \[x=5.\]

Step 4: Check the Solution

The solution is checked by substituting it back into the original equation and verifying whether it holds true: \[\frac{2}{5-3} - \frac{4}{5+3} = \frac{8}{5^{2}-9}.\] The equation balances out, hence \(x=5\) is a valid solution.

Key Concepts

Common Denominator

When solving rational equations, identifying a common denominator is a crucial first step. It helps transform multiple fractions into a single unified expression. In this exercise, we deal with fractions like \(\frac{2}{x-3}\) and \(\frac{4}{x+3}\). To solve them efficiently, we recognize that the expression \(x^2-9\) is a result of multiplying \((x-3)\) and \((x+3)\).
\((x-3)(x+3)\) is known as the least common denominator (LCD) in this scenario. Using it allows us to combine these fractions effectively. Here's why it matters:

• The LCD simplifies the process of combining fractions by creating a common ground for numerical operations.
• It facilitates easier calculations as it transforms the entire rational equation into a workable format.
• This step eliminates the separate denominators, causing our equation to be expressed in a single fraction.

In this scenario, once the fractions have the common denominator, you can rewrite the expression as a single fraction so that the equation is solvable.

Solving Equations

Once a common denominator has unified different fractions, we can proceed to solve the equation. This is an essential phase, as it involves simplifying the equation and finding values that satisfy it.
The process involves eliminating the denominator by multiplying each term by the least common denominator, \((x-3)(x+3)\), effectively removing it and simplifying the equation to:\[2(x+3) - 4(x-3) = 8.\]
This reduction creates a linear equation, which simplifies further by applying basic algebraic operations.

• Using distribution, expand each term: \(2(x+3)\) becomes \(2x + 6\) and \(-4(x-3)\) becomes \(-4x + 12\).
• Combine like terms to streamline the equation, leaving you with \(-2x + 18 = 8\).
• To isolate \(x\), solve the equation, leading to our result: \(x = 5\).

This solution step is crucial because it represents how to concretely apply algebra to reach an exact numerical result.

Checking Solutions

Checking your solution is a vital step to verifying the accuracy of the solution. After arriving at \(x=5\), it's important to substitute this value back into the original equation to ensure it satisfies the equation.
The original equation is:\[\frac{2}{x-3} - \frac{4}{x+3} = \frac{8}{x^2-9}.\]
Upon substituting \(x=5\), verify:
\[\frac{2}{5-3} - \frac{4}{5+3} = \frac{8}{5^2-9}.\]
Which simplifies to:

• \(\frac{2}{2} = 1\)
• \(\frac{4}{8} = 0.5\)
• \(5^2=25\), so \(\frac{8}{16} = 0.5\)
• Check whether \(1 - 0.5 = 0.5\)

Because both sides of the equation are equal, this confirms that \(x=5\) is a valid and correct solution. This cross-verification is crucial because it ensures that we didn't make any errors in our calculations. Ensuring solutions meet the original condition authenticates our mathematical process.