Problem 56 - Intermediate Algebra

Question

Solve equation. If a solution is extraneous, so indicate. \(\frac{2}{x}+\frac{1}{2}=\frac{7}{2 x}\)

Step-by-Step Solution

Answer

The solution is \(x = 3\). It is not extraneous.

Step 1: Identify the Common Denominator

To eliminate the fractions in the equation, we first identify the least common denominator (LCD) among the fractions. The denominators are \(x\), \(2\), and \(2x\). Thus, the LCD is \(2x\).

Step 2: Rewrite Each Fraction with the LCD

We rewrite each fraction with a denominator of \(2x\). This means that each fraction will be converted as follows: \(\frac{2}{x} = \frac{2 \cdot 2}{2x} = \frac{4}{2x}\)\(\frac{1}{2} = \frac{1 \cdot x}{2x} = \frac{x}{2x}\)\(\frac{7}{2x}\) already has the desired denominator.

Step 3: Eliminate Fractions by Multiplying by the LCD

Multiply each term in the equation by the LCD, \(2x\), to eliminate the denominators:\[2x \left( \frac{4}{2x} + \frac{x}{2x} \right) = 2x \cdot \frac{7}{2x}\]This reduces to:\[4 + x = 7\]

Step 4: Solve for \(x\)

Solve the linear equation \(4 + x = 7\) by isolating \(x\):Subtract 4 from both sides:\[x = 7 - 4\]Which simplifies to:\[x = 3\]

Step 5: Check for Extraneous Solutions

Substitute \(x = 3\) back into the original equation to verify that it satisfies the equation:\[ \frac{2}{3} + \frac{1}{2} = \frac{7}{6} \]Convert fractions to have a common denominator of 6:\[ \frac{4}{6} + \frac{3}{6} = \frac{7}{6}\]Since both sides of the equation are equal, \(x = 3\) is a valid and non-extraneous solution.

Key Concepts

Least Common Denominator

When dealing with rational equations like \( \frac{2}{x} + \frac{1}{2} = \frac{7}{2x} \), the first step is to address the denominators involved. In this case, the fractions have different denominators. To simplify the process, we use the Least Common Denominator (LCD), which is the smallest number that all denominators can divide without leaving a remainder.
In our equation, the denominators are \( x \), \( 2 \), and \( 2x \). The LCD here is \( 2x \), because:

• \( 2x \) can be divided by \( x \) resulting in \( 2 \).
• \( 2x \) can be divided by \( 2 \) resulting in \( x \).
• \( 2x \) is itself one of the denominators.

By multiplying the entire equation by the LCD, we eliminate the fractions. This transformation simplifies our task and converts the rational equation into a form that is easier to resolve.

Linear Equations

After converting the rational equation with the least common denominator, we often end up with a simpler linear equation. In our exercise, multiplying through by \( 2x \) transforms the equation into \( 4 + x = 7 \). This is a linear equation because it can be expressed in the form \( ax + b = c \).
Linear equations usually have one variable and are straightforward to solve. The principal goal is to "isolate" the variable to determine its value. Here's how you can solve \( 4 + x = 7 \):

• Subtract 4 from both sides of the equation: \( x = 7 - 4 \).
• This simplifies to \( x = 3 \).

Linear equations usually entail a series of inverse operations—addition is undone by subtraction, and vice versa. The answer found is considered valid unless identified otherwise after verification.

Extraneous Solutions

When solving rational equations, there is a chance of arriving at extraneous solutions. These are results that algebraically satisfy the simplified version of the equation but do not hold when plugged into the original equation. This occurs when the operations used to simplify the equation inadvertently alter its original meaning.
To prevent this, you should always verify your solutions. Look at our solved example. After finding \( x = 3 \), we checked by substituting back into the original equation:

• Calculate the left side: \( \frac{2}{3} + \frac{1}{2} = \frac{4}{6} + \frac{3}{6} = \frac{7}{6} \).
• The right side of the equation simplifies directly to \( \frac{7}{6} \).

Both sides are equal, confirming that \( x = 3 \) is not an extraneous solution. Always test solutions for rational equations to ensure conclusions are both algebraically and contextually correct.