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

Step-by-Step Solution

Answer

The solution to the equation is that it has no solution as the result \(x = 2\) is a restriction for x.

Step 1: Identify Restrictions

The denominators are \(x+2\), \(x-2\), and \((x+2)(x-2)\). The restrictions on x, thus, are the solutions to the equations \(x+2 = 0\), \(x-2 = 0\), and \((x+2)(x-2) = 0\). Solving these, we find the restrictions are \(x = -2\) and \(x = 2\).

Step 2: Get a Common Denominator

To combine these fractions, they all need to share the same denominator. In this case, the common denominator is \( (x+2)(x-2) \). So we can rewrite the equation as \[\frac{5(x-2)}{(x+2)(x-2)}+\frac{3(x+2)}{(x+2)(x-2)}-\frac{12}{(x+2)(x-2)}\].

Step 3: Simplify and Solve

Now we can combine these fractions and solve for x: \[\frac{5x-10+3x+6-12}{(x+2)(x-2)} = 0\]This simplifies to: \[\frac{8x-16}{(x+2)(x-2)} = 0\]If the fraction equals zero, that means the numerator must be zero. So we can set 8x-16 = 0 and solve for x, which gives us \(x = 2\). However, this was one of our restrictions, thus the equation has no solution.

Key Concepts

Denominator Restrictions

Rational equations involve expressions where the variable appears in the denominator. This is important because dividing by zero is undefined in mathematics. When dealing with rational equations, the first step is to identify any restrictions on the variable—the values that make any denominator zero. These are called denominator restrictions.

For example, in the equation \( \frac{5}{x+2} + \frac{3}{x-2} - \frac{12}{(x+2)(x-2)} \), we need to ensure that none of the denominators equal zero. This requires us to solve \( x+2=0 \) and \( x-2=0 \).

• Solving \( x+2=0 \) gives us \( x = -2 \).
• Solving \( x-2=0 \) gives us \( x = 2 \).

These are our restrictions, meaning \( x \) cannot equal \( -2 \) or \( 2 \). By identifying these, we prevent our equation from becoming undefined.

Common Denominator

Once denominator restrictions are identified, the next step is often to combine the fractions by finding a common denominator. A common denominator is a shared multiple of all the denominators in the equation, allowing them to be combined into a single fraction.

In the equation \( \frac{5}{x+2} + \frac{3}{x-2} - \frac{12}{(x+2)(x-2)} \), the common denominator is \((x+2)(x-2)\). Each term must be rewritten over this common denominator so they can be easily combined.

• Rewrite \( \frac{5}{x+2} \) as \( \frac{5(x-2)}{(x+2)(x-2)} \).
• Rewrite \( \frac{3}{x-2} \) as \( \frac{3(x+2)}{(x+2)(x-2)} \).
• The last term \( \frac{12}{(x+2)(x-2)} \) is already over the common denominator.

After rewriting, you can now proceed to combine these into one fraction.

Solving Rational Equations

Solving rational equations involves simplifying the combined fraction and solving for the variable, keeping denominator restrictions in mind. A rational equation equals zero when its numerator equals zero.

Combining the fractions in \( \frac{5(x-2) + 3(x+2) - 12}{(x+2)(x-2)} = 0 \) results in the expression \( \frac{8x - 16}{(x+2)(x-2)} = 0 \). For the rational expression to equal zero, the numerator must be zero, as the denominator is non-zero due to restrictions.

Set \( 8x - 16 = 0 \) and solve for \( x \):

• Add 16 to both sides: \( 8x = 16 \)
• Divide by 8: \( x = 2 \)

However, \( x = 2 \) is one of the restrictions we identified earlier. Thus, it cannot be a solution, and the equation has no valid solution. Always ensure that potential solutions do not violate the initial restrictions outlined.