Solving Equations By Factoring A Step-by-Step Guide

#1. Introduction

In this article, we will delve into the process of solving a rational equation by factoring. The equation we aim to solve is:

$$\frac{11(x-3)}{x-4} + \frac{5}{x} = \frac{-5}{x(x-4)}$$

This equation involves fractions with polynomial expressions in the denominators, making it a rational equation. Solving such equations requires careful manipulation to eliminate the fractions and obtain a simpler algebraic equation that can be solved using factoring techniques. Factoring is a fundamental skill in algebra, allowing us to break down complex expressions into simpler products. This not only aids in solving equations but also in simplifying expressions and understanding the behavior of functions. Before we begin, it's important to understand the underlying principles of factoring and how it applies to solving equations. Factoring involves expressing a polynomial as a product of two or more factors. This is the reverse process of expansion, where we multiply factors to obtain a polynomial. In the context of solving equations, factoring helps us identify the values of the variable that make the equation true. This is because if the product of several factors is zero, then at least one of the factors must be zero. This principle, known as the zero-product property, is the cornerstone of solving equations by factoring. To make the most of this guide, it is recommended to have a solid understanding of basic algebraic operations, such as adding, subtracting, multiplying, and dividing polynomials. Familiarity with different factoring techniques, such as factoring out the greatest common factor, factoring quadratic trinomials, and using special factoring patterns, will also be beneficial. Additionally, understanding the concept of domain restrictions is crucial when dealing with rational equations, as we need to identify values of the variable that would make the denominators zero, which are not valid solutions. With these prerequisites in mind, let's embark on the journey of solving this rational equation by factoring, step by step.

#2. Finding the Least Common Denominator (LCD)

The first crucial step in solving this rational equation is to identify the least common denominator (LCD). The LCD is the smallest expression that is divisible by all the denominators in the equation. In our equation,

$$\frac{11(x-3)}{x-4} + \frac{5}{x} = \frac{-5}{x(x-4)}$$

We have three denominators: $(x-4)$, $x$, and $x(x-4)$. To find the LCD, we need to consider each denominator and identify the common factors and the unique factors. The denominators are already in factored form, which simplifies the process. The first denominator is $(x-4)$, which is a linear factor. The second denominator is $x$, another linear factor. The third denominator is $x(x-4)$, which is the product of the previous two factors. Therefore, the LCD must include both factors $(x-4)$ and $x$. The LCD is the product of the highest powers of all the factors that appear in the denominators. In this case, the highest power of $(x-4)$ is 1, and the highest power of $x$ is 1. Thus, the LCD is simply the product of these factors: $x(x-4)$. Understanding how to find the LCD is fundamental to solving rational equations. It allows us to combine the fractions into a single fraction, which simplifies the equation and makes it easier to solve. The LCD is not just any common denominator; it is the least common denominator, which means it is the smallest expression that can be used to clear the fractions. This is important because using a larger common denominator would introduce unnecessary complexity into the equation, making it harder to solve. Once we have identified the LCD, we can proceed to the next step, which involves multiplying both sides of the equation by the LCD. This will eliminate the fractions, transforming the rational equation into a polynomial equation, which we can then solve using factoring or other algebraic techniques. Before we move on, let's summarize the importance of finding the LCD: It simplifies the equation by eliminating fractions, it ensures that we are working with the smallest possible expressions, and it lays the foundation for solving the equation using factoring. With the LCD in hand, we are well-equipped to tackle the next steps in solving this rational equation.

#3. Multiplying Both Sides by the LCD

Now that we have identified the least common denominator (LCD) as $x(x-4)$, the next crucial step is to multiply both sides of the equation by this LCD. This process eliminates the fractions, transforming the rational equation into a polynomial equation, which is much easier to solve. Our original equation is:

$$\frac{11(x-3)}{x-4} + \frac{5}{x} = \frac{-5}{x(x-4)}$$

Multiplying both sides by the LCD, $x(x-4)$, we get:

$$x(x-4) \left[ \frac{11(x-3)}{x-4} + \frac{5}{x} \right] = x(x-4) \left[ \frac{-5}{x(x-4)} \right]$$

Now, we need to distribute the LCD to each term on both sides of the equation. On the left side, we have two terms, and on the right side, we have one term. Let's distribute the LCD to the first term on the left side:

$$x(x-4) \cdot \frac{11(x-3)}{x-4} = 11x(x-3)$$

Notice how the $(x-4)$ factor in the LCD cancels out with the $(x-4)$ in the denominator of the first term. This is the key benefit of multiplying by the LCD – it eliminates the denominators. Next, let's distribute the LCD to the second term on the left side:

$$x(x-4) \cdot \frac{5}{x} = 5(x-4)$$

Here, the $x$ factor in the LCD cancels out with the $x$ in the denominator of the second term. Now, let's distribute the LCD to the term on the right side:

$$x(x-4) \cdot \frac{-5}{x(x-4)} = -5$$

In this case, the entire LCD, $x(x-4)$, cancels out with the denominator on the right side. After multiplying both sides by the LCD and simplifying, our equation becomes:

$$11x(x-3) + 5(x-4) = -5$$

This equation is now a polynomial equation, specifically a quadratic equation, which we can solve using factoring or other techniques. Multiplying by the LCD is a critical step because it transforms a complex rational equation into a simpler polynomial equation. This simplification is essential for solving the equation efficiently. The key is to ensure that the LCD is correctly identified and that the distribution is performed accurately. Each term in the equation must be multiplied by the LCD, and the cancellation of factors must be done carefully. With the equation now in polynomial form, we can proceed to the next step, which involves expanding the expressions, combining like terms, and setting the equation equal to zero. This will prepare the equation for factoring, allowing us to find the solutions for $x$.

#4. Simplifying and Rearranging the Equation

With the fractions eliminated, our equation now stands as a polynomial equation:

$$11x(x-3) + 5(x-4) = -5$$

The next step involves simplifying and rearranging this equation into a standard form, which will make it easier to factor and solve. The standard form for a quadratic equation is $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants. To achieve this form, we need to expand the expressions, combine like terms, and move all terms to one side of the equation, leaving zero on the other side. First, let's expand the terms on the left side of the equation. We have two terms to expand: $11x(x-3)$ and $5(x-4)$. Expanding the first term, we get:

$$11x(x-3) = 11x^2 - 33x$$

Expanding the second term, we get:

$$5(x-4) = 5x - 20$$

Now, let's substitute these expanded expressions back into our equation:

$$11x^2 - 33x + 5x - 20 = -5$$

Next, we need to combine like terms on the left side of the equation. We have two terms involving $x$: $-33x$ and $5x$. Combining these terms, we get:

$$-33x + 5x = -28x$$

So, our equation now becomes:

$$11x^2 - 28x - 20 = -5$$

To get the equation into the standard form, we need to move the constant term on the right side, $-5$, to the left side. We can do this by adding 5 to both sides of the equation:

$$11x^2 - 28x - 20 + 5 = -5 + 5$$

This simplifies to:

$$11x^2 - 28x - 15 = 0$$

Now, our equation is in the standard quadratic form $ax^2 + bx + c = 0$, where $a = 11$, $b = -28$, and $c = -15$. This form is essential for factoring the quadratic expression. Simplifying and rearranging the equation is a crucial step because it transforms the equation into a recognizable form that we can work with. Expanding the expressions eliminates parentheses and allows us to combine like terms. Combining like terms simplifies the equation further, making it easier to identify the coefficients and constant term. Moving all terms to one side of the equation sets the stage for factoring, which is the next step in solving the equation. With the equation now in standard quadratic form, we are well-prepared to factor the expression and find the solutions for $x$.

#5. Factoring the Quadratic Equation

Now that our equation is in the standard quadratic form:

$$11x^2 - 28x - 15 = 0$$

We can proceed to factor the quadratic expression. Factoring involves breaking down the quadratic expression into a product of two binomials. This is the reverse process of expanding two binomials. There are several techniques for factoring quadratic expressions, but one common method is to look for two numbers that multiply to the product of the leading coefficient (a) and the constant term (c), and add up to the middle coefficient (b). In our equation, $a = 11$, $b = -28$, and $c = -15$. So, we need to find two numbers that multiply to $11 \times -15 = -165$ and add up to $-28$. This can sometimes involve trial and error, but with practice, it becomes easier to identify the correct numbers. Let's list the factors of -165:

• 1 and -165
• -1 and 165
• 3 and -55
• -3 and 55
• 5 and -33
• -5 and 33
• 11 and -15
• -11 and 15

Looking at these pairs of factors, we can see that the pair 5 and -33 satisfy our conditions: $5 \times -33 = -165$ and $5 + (-33) = -28$. Now that we have found the correct numbers, we can rewrite the middle term, $-28x$, using these numbers:

$$11x^2 - 28x - 15 = 11x^2 + 5x - 33x - 15$$

Next, we can use factoring by grouping. We group the first two terms and the last two terms:

$$(11x^2 + 5x) + (-33x - 15)$$

Now, we factor out the greatest common factor (GCF) from each group. From the first group, $11x^2 + 5x$, the GCF is $x$:

$$x(11x + 5)$$

From the second group, $-33x - 15$, the GCF is $-3$:

$$-3(11x + 5)$$

Now, we can rewrite the expression as:

$$x(11x + 5) - 3(11x + 5)$$

Notice that we now have a common binomial factor, $(11x + 5)$. We can factor this out:

$$(11x + 5)(x - 3)$$

So, our factored quadratic equation is:

$$(11x + 5)(x - 3) = 0$$

Factoring the quadratic equation is a critical step because it transforms the equation into a product of two factors, which allows us to use the zero-product property to find the solutions for $x$. The key is to find the correct numbers that satisfy the conditions for multiplication and addition. Factoring by grouping is a useful technique for quadratic expressions where the leading coefficient is not 1. With the equation now factored, we can proceed to the final step, which involves setting each factor equal to zero and solving for $x$.

#6. Applying the Zero-Product Property

Now that we have factored the quadratic equation:

$$(11x + 5)(x - 3) = 0$$

We can apply the zero-product property. This property states that if the product of two or more factors is zero, then at least one of the factors must be zero. In other words, if $AB = 0$, then either $A = 0$ or $B = 0$ (or both). In our equation, we have two factors: $(11x + 5)$ and $(x - 3)$. So, according to the zero-product property, either $(11x + 5) = 0$ or $(x - 3) = 0$. Let's set each factor equal to zero and solve for $x$. First, let's consider the factor $(11x + 5) = 0$. To solve for $x$, we need to isolate $x$. We can start by subtracting 5 from both sides of the equation:

$$11x + 5 - 5 = 0 - 5$$

This simplifies to:

$$11x = -5$$

Now, we can divide both sides by 11:

$$\frac{11x}{11} = \frac{-5}{11}$$

This gives us:

$$x = -\frac{5}{11}$$

So, one solution for $x$ is $-\frac{5}{11}$. Next, let's consider the factor $(x - 3) = 0$. To solve for $x$, we need to isolate $x$. We can do this by adding 3 to both sides of the equation:

$$x - 3 + 3 = 0 + 3$$

This simplifies to:

$$x = 3$$

So, another solution for $x$ is 3. Applying the zero-product property is a crucial step because it allows us to find the values of $x$ that make the equation true. By setting each factor equal to zero, we create simpler equations that we can solve individually. The zero-product property is a fundamental principle in algebra and is widely used in solving equations by factoring. In our case, we found two potential solutions for $x$: $-\frac{5}{11}$ and 3. However, before we can declare these as our final solutions, we need to check for extraneous solutions. Extraneous solutions are solutions that satisfy the factored equation but do not satisfy the original equation. This can happen when we multiply both sides of an equation by an expression that could be zero, as we did when we multiplied by the LCD. Therefore, the next step is to check these solutions in the original equation to ensure they are valid.

#7. Checking for Extraneous Solutions

We have found two potential solutions for our equation:

$$x = -\frac{5}{11} \quad \text{and} \quad x = 3$$

However, it is crucial to check these solutions in the original equation to ensure they are valid and not extraneous. Extraneous solutions are solutions that satisfy the factored equation but do not satisfy the original equation. This can occur when we multiply both sides of an equation by an expression that could be zero, as we did when we multiplied by the LCD. Our original equation is:

$$\frac{11(x-3)}{x-4} + \frac{5}{x} = \frac{-5}{x(x-4)}$$

Let's first check the solution $x = -\frac{5}{11}$. We substitute this value into the original equation:

$$\frac{11(-\frac{5}{11}-3)}{-\frac{5}{11}-4} + \frac{5}{-\frac{5}{11}} = \frac{-5}{(-\frac{5}{11})(-\frac{5}{11}-4)}$$

This looks complicated, but we can simplify it step by step. First, let's simplify the terms inside the parentheses:

$$-\frac{5}{11} - 3 = -\frac{5}{11} - \frac{33}{11} = -\frac{38}{11}$$

$$-\frac{5}{11} - 4 = -\frac{5}{11} - \frac{44}{11} = -\frac{49}{11}$$

Now, substitute these back into the equation:

$$\frac{11(-\frac{38}{11})}{-\frac{49}{11}} + \frac{5}{-\frac{5}{11}} = \frac{-5}{(-\frac{5}{11})(-\frac{49}{11})}$$

Simplify the fractions:

$$\frac{-38}{-\frac{49}{11}} + (-11) = \frac{-5}{\frac{245}{121}}$$

Multiply by the reciprocal to divide fractions:

$$\frac{-38}{1} \cdot \frac{11}{-49} - 11 = -5 \cdot \frac{121}{245}$$

$$\frac{418}{49} - 11 = -\frac{121}{49}$$

Convert 11 to a fraction with a denominator of 49:

$$\frac{418}{49} - \frac{539}{49} = -\frac{121}{49}$$

$$-\frac{121}{49} = -\frac{121}{49}$$

Since this is true, $x = -\frac{5}{11}$ is a valid solution. Now, let's check the solution $x = 3$. We substitute this value into the original equation:

$$\frac{11(3-3)}{3-4} + \frac{5}{3} = \frac{-5}{3(3-4)}$$

Simplify the terms:

$$\frac{11(0)}{-1} + \frac{5}{3} = \frac{-5}{3(-1)}$$

$$0 + \frac{5}{3} = \frac{-5}{-3}$$

$$\frac{5}{3} = \frac{5}{3}$$

Since this is also true, $x = 3$ is a valid solution. Checking for extraneous solutions is a critical step because it ensures that the solutions we found using the zero-product property actually satisfy the original equation. Substituting the potential solutions into the original equation and simplifying can be a tedious process, but it is essential to avoid incorrect answers. In this case, both potential solutions, $x = -\frac{5}{11}$ and $x = 3$, are valid. Therefore, we can confidently state that these are the solutions to the equation.

#8. Stating the Solutions

After solving the equation and checking for extraneous solutions, we have arrived at the final step: stating the solutions. We found two solutions for the equation:

$$x = -\frac{5}{11} \quad \text{and} \quad x = 3$$

Both of these solutions satisfy the original equation, and we have confirmed that neither of them is extraneous. Therefore, we can confidently state that the solutions to the equation

$$\frac{11(x-3)}{x-4} + \frac{5}{x} = \frac{-5}{x(x-4)}$$

are $x = -\frac{5}{11}$ and $x = 3$. Stating the solutions clearly and explicitly is the final step in the problem-solving process. It is important to present the solutions in a clear and organized manner, so that the reader can easily understand the answer. In this case, we have two solutions, and we have stated them both. It is also a good practice to double-check the solutions one last time to ensure that they are correct. This can help to avoid errors and ensure that the answer is accurate. In summary, solving rational equations by factoring involves several steps, including finding the LCD, multiplying both sides by the LCD, simplifying and rearranging the equation, factoring the quadratic expression, applying the zero-product property, checking for extraneous solutions, and stating the solutions. Each of these steps is crucial, and it is important to perform them carefully and accurately to arrive at the correct answer. With practice, solving rational equations by factoring becomes easier and more efficient. The key is to understand the underlying principles and to follow the steps systematically. And the most important thing is to double-check the solutions to make sure there are no extraneous solutions.

#9. Conclusion

In conclusion, solving the rational equation

$$\frac{11(x-3)}{x-4} + \frac{5}{x} = \frac{-5}{x(x-4)}$$

involved a systematic approach that combined algebraic manipulation, factoring techniques, and careful verification. We began by identifying the least common denominator (LCD), which was $x(x-4)$, and multiplying both sides of the equation by the LCD to eliminate the fractions. This transformed the rational equation into a polynomial equation, which is easier to solve. Next, we simplified and rearranged the equation into the standard quadratic form $ax^2 + bx + c = 0$. This involved expanding the expressions, combining like terms, and moving all terms to one side of the equation. Then, we factored the quadratic expression, which allowed us to apply the zero-product property. The zero-product property states that if the product of two or more factors is zero, then at least one of the factors must be zero. This property is essential for solving equations by factoring. After applying the zero-product property, we found two potential solutions for $x$: $x = -\frac{5}{11}$ and $x = 3$. However, we needed to check these solutions in the original equation to ensure they were valid and not extraneous. Extraneous solutions are solutions that satisfy the factored equation but do not satisfy the original equation. This can occur when we multiply both sides of an equation by an expression that could be zero. We found that both solutions, $x = -\frac{5}{11}$ and $x = 3$, satisfied the original equation and were therefore valid solutions. Stating the solutions clearly and explicitly is the final step in the problem-solving process. In this case, we stated that the solutions to the equation are $x = -\frac{5}{11}$ and $x = 3$. Solving rational equations by factoring is a fundamental skill in algebra. It requires a solid understanding of algebraic operations, factoring techniques, and the zero-product property. It also requires careful attention to detail to avoid errors and extraneous solutions. By following the steps outlined in this article, you can solve rational equations by factoring efficiently and accurately. Remember to always check your solutions in the original equation to ensure they are valid. With practice, solving rational equations by factoring will become easier and more intuitive. This skill is not only useful in mathematics but also in various fields of science and engineering where equations involving rational expressions arise. Mastering this technique will empower you to tackle more complex problems and deepen your understanding of mathematical concepts.