Problem 46 - 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. \(\frac{5}{x+2}+\frac{3}{x-2}=\frac{12}{(x+2)(x-2)}\)

Step-by-Step Solution

Answer

The given equation has no solution since the variable \(x\) is equal to a value that is restricted.

Step 1: Identifying the Restrictions

Let's first identify the values of x that would make the denominator of each term zero. These are found by finding the roots of each denominator.\n1. For \(x+2=0\), the root is \(x=-2\).\n2. For \(x-2=0\), the root is \(x=2\).\nSo the restrictions to the values of x are \(x≠-2\) and \(x≠2\).

Step 2: Simplifying the Equation

To simplify the equation, it is best to perform cross-multiplication to get rid of the fraction part. We start by multiplying each term on both sides by \((x+2)(x-2)\), which results in: \(5(x-2) + 3(x+2) = 12\).

Step 3: Solving for x

Distribute and simplify the equation to find the value of x.\n1. Multiply through to get: \(5x-10+3x+6=12\).\n2. Combine like terms to get: \(8x-4=12\).\n3. Add 4 to both sides to get: \(8x=16\).\n4. Divide by 8 on both sides to get: \(x=2\).

Step 4: Check for Restrictions

Finally, we recheck the restrictions. The solution \(x=2\) is one of the restrictions, so it is not a valid solution. Thus, the given equation has no solution.

Key Concepts

Rational Equations with Variables in Denominators

When working with rational equations with variables in denominators, it is essential to understand the nature of rational expressions. A rational expression is a fraction in which both the numerator and the denominator are polynomials. The presence of variables in the denominators of these fractions introduces special considerations for solving the equations they form.

Firstly, a rational equation may not be valid for certain values of the variable that would cause any denominator to become zero because division by zero is undefined in mathematics. Therefore, solving these equations requires careful identification and exclusion of these values, which are termed as restrictions or excluded values. Here's an example to illustrate:

• Consider the rational equation \(\frac{5}{x+2} + \frac{3}{x-2} = \frac{12}{(x+2)(x-2)}\).
• To solve it, we must first exclude any values of \(x\) that would make either \(x+2\) or \(x-2\) equal to zero. These values are \(x = -2\) and \(x = 2\), respectively.

With this knowledge in place, we can comfortably proceed to solve the equation while keeping in mind that the solutions cannot include the restricted values.

Identifying Restrictions on Variables

Identifying restrictions on variables is a critical step in solving rational equations. As mentioned earlier, certain values for the variables can make the denominators in a rational expression zero, leading to an undefined expression. Here's how to identify these restrictions:

• Examine each denominator in the equation separately.
• Set each denominator equal to zero and solve for the variable.
• List these solutions as restrictions, as they are the values that cannot be used for the variable in the equation.

In our example problem, the equation \(\frac{5}{x+2} + \frac{3}{x-2} = \frac{12}{(x+2)(x-2)}\) presented two denominators: \(x+2\) and \(x-2\). The restrictions are found by setting these equal to zero and solving for \(x\).

Putting It Into Practice

For \(x+2=0\), the solution is \(x=-2\). Likewise, \(x-2=0\) yields \(x=2\). These values are restrictions since they make the denominators zero. Failure to identify and exclude these restricted values from our solutions could lead to nonsensical or incorrect results.

Cross-Multiplication Method

The cross-multiplication method is an effective technique for solving rational equations, especially when dealing with variables in the denominators. This method involves multiplying each side of the equation by a common denominator to eliminate the fractions, allowing us to work with a simpler equation. Here is a step-by-step breakdown of how the method is applied to our example equation:

• Identify a common denominator. For our equation \(\frac{5}{x+2} + \frac{3}{x-2} = \frac{12}{(x+2)(x-2)}\), the common denominator is \( (x+2)(x-2) \).
• Multiply each term in the equation by the common denominator.
• This step eliminates the denominators, leading to a polynomial equation.

By cross-multiplying, we transform the original equation into \(5(x-2) + 3(x+2) = 12\), which simplifies to \(8x-4=12\). Solving for \(x\) then gives us \(x=2\), but remember, this is one of our restrictions! So in this case, despite solving the equation successfully, we find that the solution is not valid, leading to the conclusion that the original rational equation has no solution. When using cross-multiplication, one must always verify that the resulting solution does not include any of the previously identified restrictions.