Solve Rational Equations A Comprehensive Step-by-Step Guide

In the realm of mathematics, solving equations is a fundamental skill. Among the various types of equations, rational equations often pose a challenge due to the presence of fractions with variables in the denominator. This article provides a comprehensive guide on how to solve rational equations, with a focus on a specific example:

$$\frac{s}{2s+6} - \frac{2}{5s+5} = \frac{5s^2 - 3s - 7}{10s^2 + 40s + 30}$$

We will delve into the step-by-step process of solving this equation, highlighting key concepts and techniques along the way. By understanding the methodology presented here, you will be well-equipped to tackle a wide range of rational equations.

1. Factoring the Denominators

The first crucial step in solving rational equations involves factoring the denominators of all fractions. This factorization helps identify common factors and simplifies the process of finding the least common denominator (LCD). In our example, we have the following denominators:

• $2s + 6 = 2(s + 3)$
• $5s + 5 = 5(s + 1)$
• $10s^2 + 40s + 30 = 10(s^2 + 4s + 3) = 10(s + 1)(s + 3)$

By factoring, we can clearly see the common factors of $(s + 1)$ and $(s + 3)$, which will be essential in determining the LCD.

Factoring denominators is a critical initial step in solving rational equations. This process allows us to identify common factors and simplify the expressions, ultimately making it easier to find the least common denominator (LCD). By carefully factoring each denominator, we can transform complex rational expressions into more manageable forms. In our specific example, we began by factoring the denominators as follows:

• $2s + 6$ factors into $2(s + 3)$. This reveals a simple linear factor, which is crucial for finding the LCD.
• $5s + 5$ factors into $5(s + 1)$. Another linear factor is identified, further aiding in the LCD determination.
• $10s^2 + 40s + 30$ factors into $10(s^2 + 4s + 3)$, which further factors into $10(s + 1)(s + 3)$. This quadratic expression is now fully factored, showcasing the common factors $(s + 1)$ and $(s + 3)$.

This meticulous factoring process not only simplifies the denominators but also highlights the common factors that will be essential when we combine the fractions. Recognizing these shared factors is a key element in efficiently solving rational equations. The factored forms allow us to clearly see the structure of each denominator, making it easier to identify the LCD and proceed with the subsequent steps in the solution process. Factoring is more than just a preliminary step; it's a fundamental technique that sets the stage for simplifying and solving rational equations effectively.

2. Finding the Least Common Denominator (LCD)

The next step is to determine the LCD of the fractions. The LCD is the smallest expression that is divisible by all the denominators. In this case, the LCD is $10(s + 1)(s + 3)$. This is because it includes all the factors from the denominators: 2, 5, $(s + 1)$, and $(s + 3)$.

Finding the least common denominator (LCD) is a pivotal step in solving rational equations. The LCD serves as a common ground, allowing us to combine fractions and eliminate denominators, thereby simplifying the equation. To find the LCD, we must consider all the factors present in the denominators. In our example, after factoring, we identified the denominators as $2(s + 3)$, $5(s + 1)$, and $10(s + 1)(s + 3)$. The LCD must include each of these factors to the highest power they appear in any denominator.

• The numerical coefficients are 2, 5, and 10. The least common multiple of these numbers is 10.
• The factors involving $s$ are $(s + 3)$ and $(s + 1)$. Both factors appear to the first power, so they both must be included in the LCD.

Therefore, the LCD is the product of these elements: $10(s + 1)(s + 3)$. This LCD is crucial because it allows us to rewrite each fraction with a common denominator, making it possible to add and subtract them. By multiplying both sides of the equation by the LCD, we will effectively eliminate the fractions, transforming the rational equation into a more manageable polynomial equation. This process is a cornerstone of solving rational equations, making the identification of the LCD a vital skill.

3. Multiplying by the LCD

To eliminate the fractions, we multiply both sides of the equation by the LCD, $10(s + 1)(s + 3)$. This step is crucial because it transforms the rational equation into a simpler polynomial equation.

$$10(s + 1)(s + 3) \left[\frac{s}{2(s + 3)} - \frac{2}{5(s + 1)}\right] = 10(s + 1)(s + 3) \left[\frac{5s^2 - 3s - 7}{10(s + 1)(s + 3)}\right]$$

Distributing the LCD on the left side, we get:

$$10(s + 1)(s + 3) \cdot \frac{s}{2(s + 3)} - 10(s + 1)(s + 3) \cdot \frac{2}{5(s + 1)} = 5s(s + 1) - 4(s + 3)$$

On the right side, the LCD cancels out the denominator, leaving us with:

$$5s^2 - 3s - 7$$

Multiplying by the LCD is a transformative step in solving rational equations. This process effectively clears the fractions, converting the equation into a more manageable polynomial form. By multiplying both sides of the equation by the LCD, we ensure that all denominators are canceled out, simplifying the equation significantly. In our example, we identified the LCD as $10(s + 1)(s + 3)$. Multiplying both sides of the equation by this LCD yields:

$$10(s + 1)(s + 3) \left[\frac{s}{2(s + 3)} - \frac{2}{5(s + 1)}\right] = 10(s + 1)(s + 3) \left[\frac{5s^2 - 3s - 7}{10(s + 1)(s + 3)}\right]$$

Distributing the LCD on the left side involves multiplying each term inside the brackets by the LCD. This allows us to cancel out the denominators:

• For the first term, $10(s + 1)(s + 3) \cdot \frac{s}{2(s + 3)}$, the $2(s + 3)$ in the denominator cancels with the corresponding factor in the LCD, leaving $5s(s + 1)$.
• For the second term, $10(s + 1)(s + 3) \cdot \frac{2}{5(s + 1)}$, the $5(s + 1)$ in the denominator cancels with the corresponding factor in the LCD, leaving $4(s + 3)$.

On the right side of the equation, the entire denominator $10(s + 1)(s + 3)$ cancels out with the LCD, leaving just the numerator $5s^2 - 3s - 7$. This step effectively transforms the rational equation into a polynomial equation, which is much easier to solve. The resulting equation is a crucial stepping stone towards finding the solution(s) for the original rational equation. Multiplying by the LCD is a powerful technique that simplifies the equation and sets the stage for the subsequent algebraic manipulations.

4. Simplifying and Solving the Equation

After multiplying by the LCD, we have the equation:

$$5s(s + 1) - 4(s + 3) = 5s^2 - 3s - 7$$

Expanding and simplifying, we get:

$$5s^2 + 5s - 4s - 12 = 5s^2 - 3s - 7$$

$$5s^2 + s - 12 = 5s^2 - 3s - 7$$

Subtracting $5s^2$ from both sides:

$$s - 12 = -3s - 7$$

Adding $3s$ to both sides:

$$4s - 12 = -7$$

Adding 12 to both sides:

$$4s = 5$$

Dividing by 4:

$$s = \frac{5}{4}$$

Simplifying and solving the resulting equation after multiplying by the LCD is a critical phase in finding the solution to the rational equation. This step involves expanding, combining like terms, and isolating the variable. In our example, after multiplying by the LCD, we arrived at the equation:

$$5s(s + 1) - 4(s + 3) = 5s^2 - 3s - 7$$

The first step in simplifying this equation is to expand the terms on the left side:

• $5s(s + 1)$ expands to $5s^2 + 5s$.
• $-4(s + 3)$ expands to $-4s - 12$.

So the equation becomes:

$$5s^2 + 5s - 4s - 12 = 5s^2 - 3s - 7$$

Next, we combine like terms on the left side:

$$5s^2 + s - 12 = 5s^2 - 3s - 7$$

Now, we begin to isolate the variable $s$. First, we subtract $5s^2$ from both sides of the equation to simplify it:

$$s - 12 = -3s - 7$$

Next, we add $3s$ to both sides to get all terms involving $s$ on one side:

$$4s - 12 = -7$$

Then, we add 12 to both sides to isolate the term with $s$:

$$4s = 5$$

Finally, we divide both sides by 4 to solve for $s$:

$$s = \frac{5}{4}$$

This algebraic manipulation is a systematic process of reducing the equation to its simplest form and isolating the variable. Each step builds upon the previous one, leading us closer to the solution. The careful execution of these steps is essential to ensure accuracy and arrive at the correct solution. Simplifying the equation allows us to see the underlying relationship between the terms and effectively solve for the unknown variable.

5. Checking for Extraneous Solutions

It is essential to check for extraneous solutions, which are solutions that satisfy the simplified equation but not the original rational equation. This typically occurs when a solution makes one of the denominators in the original equation equal to zero.

Our potential solution is $s = \frac{5}{4}$. Let's check if this value makes any of the original denominators zero:

• $2s + 6 = 2(\frac{5}{4}) + 6 = \frac{5}{2} + 6 = \frac{17}{2} \neq 0$
• $5s + 5 = 5(\frac{5}{4}) + 5 = \frac{25}{4} + 5 = \frac{45}{4} \neq 0$
• $10s^2 + 40s + 30 = 10(\frac{5}{4})^2 + 40(\frac{5}{4}) + 30 = 10(\frac{25}{16}) + 50 + 30 = \frac{125}{8} + 80 \neq 0$

Since $s = \frac{5}{4}$ does not make any of the denominators zero, it is a valid solution.

Checking for extraneous solutions is a crucial step in solving rational equations. Extraneous solutions are values that satisfy the simplified equation but do not satisfy the original equation because they make one or more of the denominators equal to zero. These solutions must be discarded to ensure the accuracy of our final answer. To check for extraneous solutions, we substitute our potential solution back into the original equation's denominators.

In our example, we found a potential solution of $s = \frac{5}{4}$. The original equation's denominators were $2s + 6$, $5s + 5$, and $10s^2 + 40s + 30$. We need to ensure that substituting $s = \frac{5}{4}$ into each of these denominators does not result in zero.

1. Checking $2s + 6$:

   $$2(\frac{5}{4}) + 6 = \frac{5}{2} + 6 = \frac{5}{2} + \frac{12}{2} = \frac{17}{2}$$

   Since $\frac{17}{2}$ is not zero, this denominator is valid.

2. Checking $5s + 5$:

   $$5(\frac{5}{4}) + 5 = \frac{25}{4} + 5 = \frac{25}{4} + \frac{20}{4} = \frac{45}{4}$$

   Since $\frac{45}{4}$ is not zero, this denominator is also valid.

3. Checking $10s^2 + 40s + 30$:

   $$10(\frac{5}{4})^2 + 40(\frac{5}{4}) + 30 = 10(\frac{25}{16}) + 50 + 30 = \frac{250}{16} + 80 = \frac{125}{8} + 80$$

   To simplify, we find a common denominator:

   $$\frac{125}{8} + \frac{640}{8} = \frac{765}{8}$$

   Since $\frac{765}{8}$ is not zero, this denominator is also valid.

Since substituting $s = \frac{5}{4}$ into all the original denominators does not result in zero, we can confirm that $s = \frac{5}{4}$ is a valid solution and not an extraneous one. This step underscores the importance of verifying solutions in the context of rational equations, where the presence of variables in the denominators can lead to values that are algebraically correct but mathematically invalid in the original equation. Rigorous checking ensures that our solutions are accurate and reliable.

6. The Solution

Therefore, the solution to the equation is $s = \frac{5}{4}$.

In conclusion, solving rational equations involves a series of steps, including factoring denominators, finding the LCD, multiplying by the LCD, simplifying the equation, solving for the variable, and checking for extraneous solutions. By following these steps carefully, you can confidently solve a wide range of rational equations.

Rational equations can often seem daunting, but by following a systematic approach, they can be solved with confidence. Our journey through solving the equation $\frac{s}{2s+6} - \frac{2}{5s+5} = \frac{5s^2 - 3s - 7}{10s^2 + 40s + 30}$ has highlighted the key steps involved in this process. From factoring the denominators to checking for extraneous solutions, each step plays a crucial role in arriving at the correct answer. The solution, $s = \frac{5}{4}$, was obtained by meticulously applying these steps, demonstrating the effectiveness of a structured approach. This method not only helps in solving this specific equation but also provides a blueprint for tackling other rational equations. The ability to confidently solve rational equations is a valuable skill in mathematics, enabling us to address problems in various fields that involve ratios and proportions.

Solve the equation: $\frac{s}{2s+6} - \frac{2}{5s+5} = \frac{5s^2 - 3s - 7}{10s^2 + 40s + 30}$. If there is no solution, enter $\varnothing$.

Solving Rational Equations Step-by-Step Guide