Problem 45 - College Algebra Essentials

Question

Contain rational equations with variables in denominators. For each equation, a. write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind , solve the equation. \(\frac{3}{x+2}+\frac{2}{x-2}=\frac{8}{(x+2)(x-2)}\)

Step-by-Step Solution

Answer

The given equation has no solution.

Step 1: Identify the Restrictions

To find the values of the variable that make a denominator zero, simply solve each denominator from the equation for zero. The denominators are \(x+2\), \(x-2\), and \((x+2)(x-2)\). Solving these for zero, we get \(x+2=0\) provides \(x=-2\) and \(x-2=0\) provides \(x=2\). Therefore, the values of \(x\) that make the denominator zero are x=-2 and x=2.

Step 2: Simplify the Given Equation

Given the equation \(\frac{3}{x+2}+\frac{2}{x-2}=\frac{8}{(x+2)(x-2)}\). The LCD for this equation is \((x+2)(x-2)\). Multiply each term in the equation by the LCD to clear the denominators: \[3(x-2)+2(x+2)=8\]. Simplifying the equation gives \(3x-6+2x+4=8\), or \(5x-2=8\).

Step 3: Solve for the Variable

Now, solve the simplified equation \(5x-2=8\) for \(x\). Add 2 to both sides of the equation to get \(5x = 10\). Then, divide both sides by 5 to solve for \(x\), which gives \(x=2\). However, x=2 is one of our restrictions and hence it is an extraneous solution. Therefore, the given equation has no solution.

Key Concepts

Variables in Denominators

In rational equations, variables often appear in denominators. This means they are located at the bottom part of a fraction. For example, in the rational equation \(\frac{3}{x+2}+\frac{2}{x-2}=\frac{8}{(x+2)(x-2)}\), the variables \(x+2\) and \(x-2\) are our denominators. This placement can complicate things because when a denominator equals zero, the entire fraction becomes undefined. To avoid such issues, it's crucial to identify these variables and consider their roles within the equation.
By managing the variables in the denominators carefully, you can establish a pathway to effectively solve the equation, ensuring you do not trip over undefined operations. Handling these correctly is the first step in rational equation problems.

Restrictions on Variables

When working with rational equations, it is vital to find the restrictions on the variables. Restrictions are values that make the denominator zero, thus making the equation undefined. In the example \(\frac{3}{x+2}+\frac{2}{x-2}=\frac{8}{(x+2)(x-2)}\), you determine restrictions by setting each denominator equal to zero and solving for \(x\).

• \(x+2=0\) resolves to \(x=-2\)
• \(x-2=0\) resolves to \(x=2\)

These solutions are not permissible values for \(x\) since they would make the equation undefined. Recognizing these values early on will help you avoid missteps when solving the problem and can also keep you mindful of potential extraneous solutions.

Extraneous Solutions

Extraneous solutions arise when solving rational equations, especially after multiplying to clear denominators. These solutions stem from values that solve the simplified form of an equation but do not satisfy the original equation.
It's essential to compare any solution back to your list of restrictions. If any solution matches a restricted value, it is deemed extraneous. In our exercise, \(x=2\) was obtained as a solution, yet \(x=2\) was also a value that made a denominator zero, hence it was not valid. Extraneous solutions like this highlight why double-checking your results against restrictions is key in rational equations.

Solving Equations with Fractions

Handling equations with fractions, especially rational ones, requires a clear strategy. One effective approach is to eliminate fractions by multiplying each term by the least common denominator (LCD). This technique clears out the denominators, simplifying the equation.
In our example, we find the LCD is \((x+2)(x-2)\). By multiplying each term by this LCD, the equation transforms into \(3(x-2)+2(x+2)=8\). This clears the fractions, allowing you to work with a simpler, linear equation \(5x-2=8\). Solving this gives \(x=2\), but remember to cross-reference with your restriction notes to identify any extraneous solutions. By following structured steps, you can solve these equations efficiently without overlooking critical details.