Solving (x+10)/(x-15) ≤ 1 Algebraically A Step-by-Step Guide

#h1 Solving Rational Inequalities A Comprehensive Guide

In the realm of mathematics, particularly in algebra, inequalities play a crucial role. Inequalities help us define ranges and boundaries for solutions, rather than just pinpointing exact values as equations do. When dealing with rational expressions within inequalities, the complexity increases, demanding a strategic approach to arrive at the correct solution. This article delves into the step-by-step process of solving the rational inequality (x+10)/(x-15) ≤ 1, providing a detailed explanation to ensure clarity and understanding. Rational inequalities, which involve fractions with polynomials in the numerator and denominator, require careful consideration of critical points and intervals. These critical points, derived from both the numerator and the denominator, partition the number line into intervals. The inequality's solution is then found by testing values within these intervals. This method ensures that we account for all possible scenarios where the inequality holds true. The process begins with transforming the inequality into a form where one side is zero, which simplifies the subsequent analysis. This transformation involves algebraic manipulations that maintain the inequality's integrity while making it easier to identify the critical points. Once the inequality is in the standard form, the critical points are determined by setting both the numerator and the denominator to zero. These points are crucial because they are where the expression can change its sign. The next step involves creating a sign chart, which is a visual tool used to organize and analyze the sign of the expression in each interval defined by the critical points. The sign chart helps in identifying the intervals where the inequality is satisfied. Finally, the solution is expressed in interval notation, which provides a concise and clear representation of the range of values that satisfy the inequality. This comprehensive approach ensures that all aspects of the inequality are considered, leading to an accurate and complete solution. Understanding the nuances of rational inequalities is essential for various applications in mathematics and related fields. From optimization problems to modeling real-world scenarios, the ability to solve these inequalities is a valuable skill. This article aims to equip readers with the necessary knowledge and techniques to confidently tackle rational inequalities.

Step 1: Transform the Inequality

To effectively solve the rational inequality, our primary goal is to manipulate the given expression into a form that is easier to analyze. The standard approach is to get zero on one side of the inequality. This allows us to compare the expression to zero, making it simpler to determine when the inequality holds true. In this case, we start with the inequality: (x+10)/(x-15) ≤ 1. The first step involves subtracting 1 from both sides of the inequality. This maintains the balance of the inequality while moving all terms to one side. By subtracting 1, we obtain a new inequality: (x+10)/(x-15) - 1 ≤ 0. This form is more conducive to further algebraic manipulation. The next step is to combine the terms on the left side into a single fraction. To do this, we need to find a common denominator. In this case, the common denominator is (x-15). We rewrite 1 as (x-15)/(x-15) and subtract it from the original fraction. This gives us: (x+10)/(x-15) - (x-15)/(x-15) ≤ 0. Now that we have a common denominator, we can combine the numerators. This involves subtracting the second numerator from the first: (x+10 - (x-15))/(x-15) ≤ 0. Simplifying the numerator, we distribute the negative sign and combine like terms: (x+10 - x + 15)/(x-15) ≤ 0. This simplifies to: 25/(x-15) ≤ 0. This transformed inequality is now in a much simpler form, with a single fraction on one side and zero on the other. This makes it easier to identify the critical points and analyze the intervals where the inequality holds true. The transformation process is a crucial step in solving rational inequalities. It allows us to manipulate the expression into a form that is more manageable and easier to analyze. By getting zero on one side, we set the stage for identifying the critical points and determining the intervals where the inequality is satisfied. This initial step is essential for a successful solution. The importance of this step cannot be overstated, as it lays the foundation for the subsequent steps in the solution process. By transforming the inequality, we make it possible to apply the standard techniques for solving rational inequalities, ensuring an accurate and complete solution. This careful manipulation is a hallmark of effective problem-solving in algebra.

Step 2: Find the Critical Points

Identifying the critical points is a pivotal step in solving rational inequalities. Critical points are the values of x that make either the numerator or the denominator of the rational expression equal to zero. These points are crucial because they divide the number line into intervals where the expression's sign remains constant. Understanding how to find and interpret these critical points is essential for determining the solution set of the inequality. In the transformed inequality 25/(x-15) ≤ 0, we need to consider both the numerator and the denominator. The numerator is 25, which is a constant. Since 25 can never be equal to zero, there is no critical point derived from the numerator in this specific case. However, it's important to remember that in other rational inequalities, the numerator often contains a variable expression that can be set to zero to find critical points. The denominator is (x-15). To find the critical point associated with the denominator, we set it equal to zero and solve for x: x - 15 = 0. Adding 15 to both sides of the equation, we get x = 15. This value, x = 15, is a critical point because it makes the denominator equal to zero. When the denominator of a fraction is zero, the expression is undefined. This is a crucial consideration because the inequality may change its behavior around this point. The critical point x = 15 divides the number line into two intervals: (-∞, 15) and (15, ∞). These intervals represent the ranges of x values that we will test to determine where the inequality 25/(x-15) ≤ 0 holds true. It is important to note that the critical point x = 15 is not included in the solution set because it makes the denominator zero, resulting in an undefined expression. This exclusion is often indicated by using an open circle on a number line or parentheses in interval notation. In summary, finding the critical points involves identifying the values of x that make the numerator or denominator equal to zero. These points are essential for partitioning the number line and determining the intervals where the inequality is satisfied. In this case, the critical point x = 15 is the only point we need to consider. Understanding the significance of critical points is fundamental to solving rational inequalities. These points serve as boundaries, separating regions where the expression's sign remains consistent. By identifying these points, we can systematically analyze the inequality and determine its solution set. This step is a cornerstone of the overall solution process.

Step 3: Create a Sign Chart

Creating a sign chart is a vital technique for solving inequalities, particularly rational inequalities. A sign chart provides a visual representation of how the sign of an expression changes across different intervals, which are delineated by the critical points. This tool helps in systematically determining where the inequality holds true. In our case, we have the transformed inequality 25/(x-15) ≤ 0 and the critical point x = 15. This critical point divides the number line into two intervals: (-∞, 15) and (15, ∞). The sign chart will help us analyze the sign of the expression 25/(x-15) in each of these intervals. To construct the sign chart, we first draw a number line and mark the critical point x = 15 on it. This point separates the number line into the two intervals we mentioned earlier. Next, we choose a test value within each interval. These test values will help us determine the sign of the expression in each interval. For the interval (-∞, 15), we can choose a test value such as x = 0. Substituting x = 0 into the expression 25/(x-15), we get 25/(0-15) = 25/(-15) = -5/3. Since the result is negative, the expression 25/(x-15) is negative in the interval (-∞, 15). We indicate this on the sign chart by placing a negative sign (-) above the interval. For the interval (15, ∞), we can choose a test value such as x = 20. Substituting x = 20 into the expression 25/(x-15), we get 25/(20-15) = 25/5 = 5. Since the result is positive, the expression 25/(x-15) is positive in the interval (15, ∞). We indicate this on the sign chart by placing a positive sign (+) above the interval. Now that we have completed the sign chart, we can analyze it to determine where the inequality 25/(x-15) ≤ 0 is satisfied. We are looking for intervals where the expression is either negative or equal to zero. From the sign chart, we see that the expression is negative in the interval (-∞, 15). Since the inequality includes “less than or equal to,” we also need to consider where the expression is equal to zero. However, in this case, the numerator 25 is never zero, so the expression can never be equal to zero. Therefore, the solution includes only the interval where the expression is negative. The sign chart is an invaluable tool for solving inequalities. It provides a clear and organized way to analyze the sign of an expression across different intervals. By using test values, we can efficiently determine the sign in each interval and identify the regions where the inequality is satisfied. This visual aid simplifies the process and reduces the likelihood of errors.

Step 4: Express the Solution

After constructing the sign chart and analyzing the intervals, the final step is to express the solution to the inequality. This involves translating the information from the sign chart into a clear and concise representation of the values that satisfy the original inequality. The solution is typically expressed in interval notation, which is a standard way of denoting intervals of real numbers. In our case, we have the inequality 25/(x-15) ≤ 0, and the sign chart indicates that the expression is negative in the interval (-∞, 15). Since the inequality includes “less than or equal to,” we also need to consider where the expression is equal to zero. However, the numerator 25 is never zero, so the expression can never be equal to zero. Therefore, we only consider the interval where the expression is negative. The interval (-∞, 15) represents all real numbers less than 15. The parenthesis around 15 indicates that 15 is not included in the solution because it makes the denominator of the original expression equal to zero, resulting in an undefined value. The symbol -∞ represents negative infinity, indicating that the interval extends indefinitely in the negative direction. Therefore, the solution to the inequality 25/(x-15) ≤ 0 is the interval (-∞, 15). This means that any value of x less than 15 will satisfy the original inequality. It is crucial to exclude x = 15 from the solution because it leads to division by zero, which is undefined in mathematics. In summary, expressing the solution involves carefully considering the intervals where the inequality is satisfied, as well as any points that must be excluded due to mathematical constraints. Interval notation provides a precise way to represent the solution set, ensuring clarity and accuracy. The solution (-∞, 15) concisely captures the range of values that make the inequality 25/(x-15) ≤ 0 true. This final step completes the process of solving the rational inequality, providing a comprehensive and accurate answer. The ability to express solutions in interval notation is a fundamental skill in algebra and calculus. It allows for a clear and unambiguous representation of the range of values that satisfy a given condition. This skill is essential for various mathematical applications and problem-solving scenarios.

Therefore, the solution to the inequality (x+10)/(x-15) ≤ 1 is (-∞, 15).