Problem 43 - Prealgebra

Question

Solve each equation by first finding the LCD for the fractions in the equation and then multiplying both sides of the equation by it.(Assume \(x\) is not 0 in Problems \(39-46\).) $$\frac{3}{x}-\frac{2}{x}=\frac{1}{5}$$

Step-by-Step Solution

Answer

The solution is \(x = 5\).

Step 1: Identify the Fractions

In the equation \(\frac{3}{x} - \frac{2}{x} = \frac{1}{5}\), identify the fractions involved: \(\frac{3}{x}\), \(\frac{2}{x}\), and \(\frac{1}{5}\). The denominators are \(x\) and \(5\).

Step 2: Find the Least Common Denominator (LCD)

Determine the least common denominator for the fractions. The denominators \(x\) and \(5\) are distinct. The least common denominator (LCD) will be the product of these distinct denominators: \(x \cdot 5 = 5x\).

Step 3: Multiply the Entire Equation by the LCD

Multiply every term in the equation by the LCD \(5x\) to eliminate the fractions: \[5x \left(\frac{3}{x}\right) - 5x \left(\frac{2}{x}\right) = 5x \left(\frac{1}{5}\right)\] This simplifies to: \[5 \cdot 3 - 5 \cdot 2 = x\] Resulting in: \(15 - 10 = x\)

Step 4: Solve the Simplified Equation

Simplify the equation from the previous step: \[15 - 10 = x\] Subtracting gives: \[5 = x\]

Step 5: Verify the Solution

Substitute \(x = 5\) back into the original equation to verify correctness: \[\frac{3}{5} - \frac{2}{5} = \frac{1}{5}\] Simplify the left side: \[\frac{3 - 2}{5} = \frac{1}{5}\] Since both sides of the equation are equal, \(x = 5\) is verified as the correct solution.

Key Concepts

Least Common Denominator

The Least Common Denominator (LCD) is a key aspect of working with fraction equations. When you encounter an equation involving fractions, the denominators tell you how many equal parts something is divided into. To solve such equations, it's crucial to have a common ground for comparison, which is where the LCD comes in.
Finding the LCD involves identifying the smallest number that can evenly divide all denominators in the equation.

• For example, in the equation \(\frac{3}{x} - \frac{2}{x} = \frac{1}{5}\), the denominators are \(x\) and \(5\).
• Since they are different, we take the product of these distinct values to find the LCD: \(x \cdot 5 = 5x\).

The LCD allows us to clear the fractions by multiplying the entire equation through by this value. This simplification is crucial for solving the fractions step, turning the equation into a more straightforward problem that can be easily solved.

Solving Fractions

Once you find the Least Common Denominator, the next step is to use it to solve the equation. Here's why this step is so important: multiplying by the LCD helps to remove the fractions from the equation, making it simpler to handle.
Let's look at our example again. By multiplying every term in the equation \(\frac{3}{x} - \frac{2}{x} = \frac{1}{5}\) by the LCD \(5x\), we eliminate the fractions. So:

• Multiply: \(5x \times \frac{3}{x} - 5x \times \frac{2}{x} = 5x \times \frac{1}{5}\).
• This simplifies the fractions: \(5 \cdot 3 - 5 \cdot 2 = x\).

The above simplification reduces the equation to \(15 - 10 = x\).
Now, the fractions are gone, and you can easily solve for \(x\). This process shows how multiplying by the LCD turns a complicated equation into something manageable, by focusing on solving basic arithmetic where fractions no longer exist.

Prealgebra

Prealgebra provides the foundation for understanding various mathematical concepts like fraction equations. It helps you get familiar with operations involving fractions, multiplication, and equations—all important skills for advanced math.
When dealing with fraction problems in prealgebra:

• You learn to identify common factors and denominators to simplify expressions and equations.
• Understanding the interplay between numerators and denominators builds the skills necessary for algebra and beyond.
• Prealgebra also introduces concepts such as solving for unknowns, an essential step for algebraic problem-solving.

For instance, solving the fraction equation \(\frac{3}{x} - \frac{2}{x} = \frac{1}{5}\) using the LCD method is a perfect example of how prealgebra helps in approaching the problem systematically. This practical application of prealgebra is crucial because, by the time you progress in mathematics, these fundamental skills become essential tools.