Problem 45 - College Algebra

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 original equation simplifies to \(x = 2\), which is an identified restriction, making it an invalid solution. Thus, there is no valid solution to this equation.

Step 1: Identify Restrictions

The restrictions are the values of the variable \(x\) that make the denominator zero. If we set each denominator equal to zero, we get \(x+2=0\) and \(x-2=0\), which imply \(x=-2\) and \(x=2\) respectively.

Step 2: Simplify the Rational Equation

First, find a common denominator to add or subtract the fractions. The common denominator of \(x+2\), \(x-2\), and \((x+2)(x-2)\) is \((x+2)(x-2)\). Thus, multiply the numerators and denominators of all fractions by \((x+2)(x-2)\) to get: \[3(x-2)+2(x+2)-8=0\] Expand this to get: \[3x-6+2x+4-8=0\] Simplify to: \[5x -10 = 0\] Dividing by 5 gives: \[x = 2\]

Step 3: Check Solution Against Restrictions

The solution \(x=2\) is one of the identified restrictions, which indicates that \(x=2\) is not a valid solution for the original equation. The restrictions are \(x \neq -2\) and \(x \neq 2\).

Key Concepts

Denominator Restrictions

When working with rational equations, one important step is to determine the values that make the denominator zero. These are called denominator restrictions. Why? Because division by zero is undefined in mathematics and leads to errors in equations. To find these restrictions, take each denominator and set it equal to zero.In the original equation, we have den
ominator expressions of \(x+2\) and \(x-2\). Setting \(x+2=0\) gives us \(x=-2\). Similarly, \(x-2=0\) gives \(x=2\). Therefore, the restrictions are \(x eq -2\) and \(x eq 2\). These values must be excluded from the solution set since they would cause the equation to be undefined.

Common Denominator

Finding a common denominator is crucial when adding or subtracting fractions, as it allows us to combine the fractions into a single expression. For rational equations, this typically involves multiplying the denominators of all fractions involved. In the given exercise, the denominators are \(x+2\), \(x-2\), and their product, \((x+2)(x-2)\). Consequently, the common denominator is \((x+2)(x-2)\). To simplify the equation, we rewrite each term with this common denominator. Multiply each fraction's numerator by the denominator "missing" from that term. This results in a unified expression, making it easier to simplify and solve the equation.

Solving Rational Equations

Once the equation is simplified using a common denominator, the next step involves solving it just like any other equation. Start by distributing any constants across terms within the brackets. Using the simplified expression \(3(x-2)+2(x+2)-8=0\), distribute to get \(3x-6+2x+4-8=0\). Combine like terms to simplify further, resulting in \(5x - 10 = 0\). To isolate \(x\), add 10 to both sides to obtain \(5x = 10\), and then divide by 5. Hence, \(x = 2\). Remember, solving rational equations requires checking for restrictions to ensure the solution does not make any denominator zero.

Excluded Values

After solving a rational equation, it's paramount to remember the importance of excluded values. These are solutions that were identified as restrictions at the start of the problem.For this example, we found that \(x=2\) turns the denominator zero, making it an excluded value. Although obtaining \(x=2\) as a solution seems feasible mathematically, its exclusion stems from it making the original equation undefined. Such values should be omitted from the final set of solutions. Thus, always cross-check your solutions against any restrictions before concluding your work on rational equations.