Problem 21 - College Algebra

Question

For the following exercises, solve each rational equation for \(x\). State all \(x\) -values that are excluded from the solution set. $$ \frac{1}{x}=\frac{1}{5}+\frac{3}{2 x} $$

Step-by-Step Solution

Answer

The solution is \( x = -\frac{5}{2} \), with \( x = 0 \) as the excluded value.

Step 1: Identify the Common Denominator

The equation to solve is \( \frac{1}{x} = \frac{1}{5} + \frac{3}{2x} \). To solve this equation, we must find a common denominator. The terms have denominators \( x \), \( 5 \), and \( 2x \). The least common denominator (LCD) is \( 10x \).

Step 2: Clear the Fractions

Multiply every term in the equation by the least common denominator (LCD), which is \( 10x \), to eliminate the fractions:\[10x \cdot \frac{1}{x} = 10x \cdot \frac{1}{5} + 10x \cdot \frac{3}{2x}\]After multiplying, simplify each term:- \( 10 \)- \( 2x \)- \( 15 \)So the equation becomes \( 10 = 2x + 15 \).

Step 3: Solve the Simplified Equation

Now, solve for \( x \) in the equation \( 10 = 2x + 15 \). Subtract 15 from both sides:\[10 - 15 = 2x\]This simplifies to:\[-5 = 2x\]

Step 4: Isolate \( x \)

Divide both sides by 2 to solve for \( x \):\[x = -\frac{5}{2}\]

Step 5: Identify Excluded Values

Look at the original equation \( \frac{1}{x} = \frac{1}{5} + \frac{3}{2x} \). Note that \( x \) cannot equal 0 because it would make the denominators zero, which is undefined. Hence, \( x = 0 \) is an excluded value.

Key Concepts

Common Denominator

Finding a common denominator is crucial when solving rational equations. In our problem, the equation is \( \frac{1}{x} = \frac{1}{5} + \frac{3}{2x} \). It involves three different denominators: \( x \), \( 5 \), and \( 2x \). To add or compare these fractions, they must share the same denominator. By identifying the least common denominator (LCD), we ensure that all fractions are compatible for further operations.

• The denominators \( x \) and \( 2x \) suggest a multiple of \( x \).
• Including the factor of \( 5 \) in \( \frac{1}{5} \), the least common multiple is \( 10x \).

The LCD, \( 10x \), becomes the building block to clear out the fractions by unifying all terms under a single denominator, thus simplifying the equation solving process.

Excluded Values

While simplifying rational equations, some \( x \)-values might lead to division by zero in the original equation. Such values are termed 'excluded values'. These values must be carefully identified as they cannot form part of the solution set for the equation.

• Whenever the variable appears in a denominator, equate that denominator to zero and solve for \( x \).
• In our exercise, the terms \( \frac{1}{x} \) and \( \frac{3}{2x} \) indicate that \( x = 0 \) would result in undefined values.

Therefore, \( x = 0 \) is specifically excluded even if it doesn't appear in the solution of the evaluated equation.

Fraction Elimination

To simplify equations, it is often helpful to eliminate fractions altogether. This is done by multiplying each term of the equation by the common denominator, which, in our case, is \( 10x \).

• Multiply each term by \( 10x \): \[10x \cdot \frac{1}{x} = 10x \cdot \frac{1}{5} + 10x \cdot \frac{3}{2x}\]
• After simplifying, the fractions disappear, leaving you with a straightforward equation: \( 10 = 2x + 15 \).

This converts the original rational equation into an easy linear equation. Solving becomes straightforward, significantly easing the process by allowing clear isolation of \( x \). Rational equation solving involves simplifying and clearing the fractions to proceed to standard algebraic methods.