Clearing Fractions in Equations

Introduction

Key Concepts

Understanding Fractions in Equations

In algebra, fractions often appear in equations that involve variables. A fraction in an equation represents the division of two expressions, where the numerator is divided by the denominator. For example, in the equation $\frac{2x + 3}{4} = 5$, the entire expression $2x + 3$ is divided by 4. Solving such equations requires eliminating the fraction to simplify the equation and isolate the variable.

The Importance of Clearing Fractions

Clearing fractions is essential because it transforms the equation into a simpler form, making it easier to solve. Fractions can complicate calculations, especially when multiple fractions are involved. By eliminating the denominators, students can work with whole numbers or simpler expressions, reducing the potential for errors and simplifying the solving process.

Steps to Clear Fractions

1. Identify the Least Common Denominator (LCD): Determine the least common multiple of all denominators in the equation. For example, in the equation $\frac{2x}{3} + \frac{4}{5} = 6$, the LCD of 3 and 5 is 15.
2. Multiply Both Sides by the LCD: Multiply every term in the equation by the LCD to eliminate the fractions. Using the previous example, multiplying both sides by 15 gives:

Simplifying each term:

Examples of Clearing Fractions

Example 1:

Solve for $x$ in the equation: $$\frac{3x - 2}{5} = 4$$

Solution:

1. Identify the LCD, which is 5.
2. Multiply both sides by 5:
$$5 \cdot \frac{3x - 2}{5} = 5 \cdot 4$$ $$3x - 2 = 20$$
3. Add 2 to both sides:
$$3x = 22$$
4. Divide by 3:
$$x = \frac{22}{3}$$

Therefore, $x = \frac{22}{3}$.

Example 2:

Solve for $y$ in the equation: $$\frac{2y + 5}{3} + \frac{y - 1}{4} = 7$$

Solution:

1. Identify the LCD of 3 and 4, which is 12.
2. Multiply both sides by 12:
$$12 \cdot \frac{2y + 5}{3} + 12 \cdot \frac{y - 1}{4} = 12 \cdot 7$$ $$4(2y + 5) + 3(y - 1) = 84$$
3. Expand the terms:
$$8y + 20 + 3y - 3 = 84$$
4. Combine like terms:
$$11y + 17 = 84$$
5. Subtract 17 from both sides:
$$11y = 67$$
6. Divide by 11:
$$y = \frac{67}{11}$$

Therefore, $y = \frac{67}{11}$.

Handling Multiple Fractions

When an equation contains multiple fractions with different denominators, identifying the LCD is crucial. Suppose you have the equation: $$\frac{x}{2} + \frac{3x}{4} = 5$$

The denominators are 2 and 4, with an LCD of 4. Multiply both sides by 4 to eliminate the fractions:

Thus, $x = 4$.

Equations with Variables in the Denominator

Equations where variables appear in the denominator, such as: $$\frac{1}{x} + 2 = 5$$ require careful manipulation. To clear the fraction:

1. Subtract 2 from both sides:
$$\frac{1}{x} = 3$$
2. Take the reciprocal of both sides:
$$x = \frac{1}{3}$$

Therefore, $x = \frac{1}{3}$.

Checking Solutions

It’s essential to verify solutions by substituting them back into the original equation to ensure they are correct and do not result in undefined expressions, especially when variables were present in denominators.

Example:

Given $x = \frac{22}{3}$ from the first example, substitute back: $$\frac{3\left(\frac{22}{3}\right) - 2}{5} = \frac{22 - 2}{5} = \frac{20}{5} = 4$$ Which matches the original equation, confirming the solution is correct.

Special Cases

When dealing with special cases, such as equations that result in undefined expressions or no solution, it's important to recognize these scenarios.

Example:

Solve for $x$: $$\frac{2}{x} = \frac{2}{x-1}$$

Solution:

1. Multiply both sides by $x(x-1)$ to eliminate fractions:
$$2(x-1) = 2x$$
2. Expand and simplify:
$$2x - 2 = 2x$$
3. Subtract $2x$ from both sides:
$$-2 = 0$$

This is a contradiction, indicating there is no solution to the equation.

Application in Real-World Problems

Clearing fractions is not only a theoretical exercise but also has practical applications. For instance, in financial calculations involving rates, such as interest rates or speed, fractions frequently appear. Simplifying these equations allows for easier computation and better understanding of the underlying relationships.

Real-World Example:

Suppose you are calculating the speed of two cars traveling towards each other. Car A travels at $\frac{3}{4}$ of the speed of Car B. If the combined speed is 75 mph, find the speed of each car.

Solution:

1. Let the speed of Car B be $x$ mph. Then, the speed of Car A is $\frac{3}{4}x$ mph.
2. The combined speed is: $$\frac{3}{4}x + x = 75$$
3. Multiply both sides by 4 to eliminate fractions: $$4 \cdot \frac{3}{4}x + 4x = 300$$ $$3x + 4x = 300$$ $$7x = 300$$ $$x = \frac{300}{7} \approx 42.86 \text{ mph}$$
4. Therefore, Car B travels at approximately 42.86 mph, and Car A travels at: $$\frac{3}{4} \times 42.86 \approx 32.14 \text{ mph}$$

Thus, Car A travels at approximately 32.14 mph, and Car B at approximately 42.86 mph.

Common Mistakes to Avoid

• Incorrect LCD: Failing to identify the correct least common denominator can lead to errors in clearing fractions.
• Misapplying Multiplication: Forgetting to multiply every term by the LCD can leave fractions in the equation.
• Ignoring Variable Restrictions: Not considering values that would make denominators zero, leading to undefined expressions.
• Calculation Errors: Simple arithmetic mistakes can result in incorrect solutions.

Advanced Techniques

For more complex equations involving fractions, such as those with variables in denominators or multiple fractions, advanced techniques may be required. These include factoring, using the quadratic formula, or applying substitution methods.

Example:

Solve for $x$: $$\frac{1}{x} + \frac{1}{x+2} = \frac{5}{6}$$

Solution:

1. Find the LCD, which is $x(x+2) \cdot 6$.
2. Multiply both sides by $6x(x+2)$: $$6(x+2) + 6x = 5x(x+2)$$
3. Expand and simplify: $$6x + 12 + 6x = 5x^2 + 10x$$ $$12x + 12 = 5x^2 + 10x$$
4. Rearrange the equation: $$5x^2 - 2x - 12 = 0$$
5. Apply the quadratic formula: $$x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 5 \cdot (-12)}}{2 \cdot 5}$$ $$x = \frac{2 \pm \sqrt{4 + 240}}{10}$$ $$x = \frac{2 \pm \sqrt{244}}{10}$$ $$x = \frac{2 \pm 2\sqrt{61}}{10}$$ $$x = \frac{1 \pm \sqrt{61}}{5}$$

Thus, the solutions are: $$x = \frac{1 + \sqrt{61}}{5} \quad \text{and} \quad x = \frac{1 - \sqrt{61}}{5}$$

Since $x$ cannot be negative in this context, only the positive solution is considered valid.

Comparison Table

Summary and Key Takeaways