Solving Equations With Fractions Eliminating Denominators By Multiplying By The LCD

When dealing with equations involving fractions, a powerful technique to simplify the process is to eliminate the denominators. This is achieved by multiplying both sides of the equation by the Least Common Denominator (LCD). The LCD is the smallest multiple that all the denominators in the equation share. By strategically multiplying by the LCD, we can transform a complex fractional equation into a more manageable equation involving only integers.

Understanding the LCD and Its Role

The Least Common Denominator (LCD) is a crucial concept when working with fractions, especially in equations. To grasp its significance, we must first understand the Least Common Multiple (LCM). The LCM of two or more numbers is the smallest number that is a multiple of each of the given numbers. For example, the LCM of 3 and 4 is 12, as 12 is the smallest number divisible by both 3 and 4. The LCD is essentially the LCM of the denominators in a set of fractions.

When solving equations with fractions, the LCD becomes our tool for clearing denominators. Multiplying both sides of the equation by the LCD effectively cancels out the denominators, transforming the equation into a simpler form without fractions. This makes the equation much easier to solve using standard algebraic techniques.

The process of finding the LCD involves identifying the denominators in the equation and then determining their LCM. If the denominators are simple integers, finding the LCM is straightforward. However, when dealing with algebraic expressions in the denominators, we need to factor each denominator and identify the unique factors. The LCD is then constructed by taking the highest power of each unique factor present in the denominators.

For instance, consider the equation ${\frac{x+2}{3x} - \frac{1}{x-2} = \frac{x-3}{3x}}$. The denominators are ${3x}$ and ${x-2}$. To find the LCD, we observe that the factors are ${3}$, ${x}$, and ${x-2}$. Therefore, the LCD is ${3x(x-2)}$. Multiplying both sides of the equation by this LCD will eliminate the fractions, simplifying the equation and making it easier to solve for ${x}$.

Step-by-Step Multiplication by the LCD

Let's illustrate the process of eliminating denominators by multiplying by the LCD with a specific example. Consider the equation:

$$\frac{x+2}{3x} - \frac{1}{x-2} = \frac{x-3}{3x}$$

Our goal is to eliminate the fractions by multiplying both sides of the equation by the Least Common Denominator (LCD).

1. Identify the Denominators

The denominators in this equation are ${3x}$ and ${x-2}$.

2. Determine the LCD

To find the LCD, we need to find the Least Common Multiple (LCM) of the denominators. In this case, the denominators are ${3x}$ and ${x-2}$. Since these expressions have no common factors, the LCD is simply their product:

$$LCD = 3x(x-2)$$

3. Multiply Both Sides of the Equation by the LCD

We multiply both sides of the equation by the LCD, ${3x(x-2)}$:

$$3x(x-2) \left[ \frac{x+2}{3x} - \frac{1}{x-2} \right] = 3x(x-2) \left[ \frac{x-3}{3x} \right]$$

4. Distribute the LCD on the Left Side

Distribute the LCD to each term on the left side of the equation:

$$3x(x-2) \cdot \frac{x+2}{3x} - 3x(x-2) \cdot \frac{1}{x-2} = 3x(x-2) \cdot \frac{x-3}{3x}$$

5. Simplify by Cancelling Common Factors

Now, we simplify each term by cancelling out the common factors:

• In the first term, ${3x}$ cancels out:

  $$(x-2)(x+2)$$

• In the second term, ${(x-2)}$ cancels out:

  $$-3x$$

• In the third term, ${3x}$ cancels out:

  $$(x-2)(x-3)$$

6. Write the Simplified Equation

After cancelling out the common factors, we are left with the following equation:

$$(x-2)(x+2) - 3x = (x-2)(x-3)$$

7. Expand and Simplify Further

Expand the products:

$$(x^2 - 4) - 3x = x^2 - 5x + 6$$

Combine like terms:

$$x^2 - 3x - 4 = x^2 - 5x + 6$$

8. Solve for x

Subtract ${x^2}$ from both sides:

$$-3x - 4 = -5x + 6$$

Add ${5x}$ to both sides:

$$2x - 4 = 6$$

Add 4 to both sides:

$$2x = 10$$

Divide by 2:

$$x = 5$$

Thus, by multiplying the original equation by the LCD, we successfully eliminated the denominators and simplified the equation, making it easier to solve for ${x}$.

Common Mistakes to Avoid

When multiplying equations by the LCD to eliminate denominators, several common mistakes can occur. Recognizing and avoiding these pitfalls is crucial for accurate problem-solving. One frequent error is failing to distribute the LCD to every term in the equation. The LCD must be multiplied by each term on both sides of the equation to ensure that the equality remains valid. For instance, in the equation ${\frac{x+2}{3x} - \frac{1}{x-2} = \frac{x-3}{3x}}$, the LCD, ${3x(x-2)}$, must be multiplied by ${\frac{x+2}{3x}}$, ${-\frac{1}{x-2}}$, and ${\frac{x-3}{3x}}$ individually.

Another common mistake is incorrectly identifying the LCD. The LCD is the Least Common Multiple (LCM) of the denominators, not just any common multiple. To find the LCD, one must factor each denominator completely and then take the highest power of each unique factor. For example, if the denominators are ${x^2 - 4}$ and ${x + 2}$, factoring ${x^2 - 4}$ into ${(x + 2)(x - 2)}$ reveals that the LCD is ${(x + 2)(x - 2)}$, not just ${(x^2 - 4)(x + 2)}$.

Careless cancellation of terms is another source of error. When multiplying by the LCD, terms can only be cancelled if they are common factors in both the numerator and the denominator. For example, when multiplying ${3x(x-2)}$ by ${\frac{x+2}{3x}}$, the ${3x}$ terms can be cancelled, but one cannot cancel terms that are part of a sum or difference. Finally, after solving for ${x}$, it's essential to check for extraneous solutions. These are solutions that satisfy the transformed equation but not the original equation. Extraneous solutions often arise when the original equation contains rational expressions, and a solution makes one of the denominators zero. Therefore, it is always necessary to substitute the solutions back into the original equation to ensure they are valid.

Practical Examples and Exercises

To solidify your understanding of multiplying by the LCD, let's explore some practical examples and exercises. These examples will illustrate the step-by-step process and highlight key considerations for different types of equations.

Example 1: Solve the equation:

$$\frac{2}{x} + \frac{3}{2x} = 7$$

1. Identify the denominators: The denominators are ${x}$ and ${2x}$.

2. Determine the LCD: The LCD of ${x}$ and ${2x}$ is ${2x}$.

3. Multiply both sides by the LCD:

   $$2x \left( \frac{2}{x} + \frac{3}{2x} \right) = 2x(7)$$

4. Distribute and simplify:

   $$2x \cdot \frac{2}{x} + 2x \cdot \frac{3}{2x} = 14x$$

   $$4 + 3 = 14x$$

5. Solve for ${x}$:

   $$7 = 14x$$

   $$x = \frac{1}{2}$$

Example 2: Solve the equation:

$$\frac{1}{x-1} + \frac{2}{x+1} = \frac{3}{x^2-1}$$

1. Identify the denominators: The denominators are ${x-1}$, ${x+1}$, and ${x^2-1}$.

2. Determine the LCD: Note that ${x^2-1}$ can be factored as ${(x-1)(x+1)}$. Thus, the LCD is ${(x-1)(x+1)}$.

3. Multiply both sides by the LCD:

   $$(x-1)(x+1) \left( \frac{1}{x-1} + \frac{2}{x+1} \right) = (x-1)(x+1) \left( \frac{3}{x^2-1} \right)$$

4. Distribute and simplify:

   $$(x-1)(x+1) \cdot \frac{1}{x-1} + (x-1)(x+1) \cdot \frac{2}{x+1} = (x-1)(x+1) \cdot \frac{3}{(x-1)(x+1)}$$

   $$(x+1) + 2(x-1) = 3$$

5. Solve for ${x}$:

   $$x + 1 + 2x - 2 = 3$$

   $$3x - 1 = 3$$

   $$3x = 4$$

   $$x = \frac{4}{3}$$

Exercises:

1. Solve: ${\frac{4}{x} - \frac{2}{3x} = 5}$
2. Solve: ${\frac{1}{x+2} - \frac{3}{x-2} = \frac{2}{x^2-4}}$

By working through these examples and exercises, you will gain confidence in your ability to multiply equations by the LCD and eliminate denominators effectively.

Conclusion

In conclusion, multiplying by the Least Common Denominator (LCD) is a fundamental technique for solving equations involving fractions. This method simplifies equations by eliminating denominators, transforming them into more manageable forms. By understanding the concept of the LCD, following the step-by-step multiplication process, avoiding common mistakes, and practicing with examples and exercises, you can master this technique. Mastering this skill not only simplifies the process of solving equations but also builds a solid foundation for more advanced algebraic concepts. Remember, the key to success lies in careful attention to detail and consistent practice.