Solving Absolute Value Inequality |x-2| + 3 > 17
#Introduction
In this article, we will explore how to solve the absolute value inequality $|x-2| + 3 > 17$. Absolute value inequalities can seem tricky at first, but by understanding the properties of absolute values and following a systematic approach, we can find the solution set. We'll break down the problem step-by-step, providing a clear explanation of each step involved. This article aims to help anyone, regardless of their mathematical background, understand and solve similar problems. So, let's dive in and unravel the mystery of this inequality!
To effectively solve the inequality, it's crucial to first understand what absolute value represents. The absolute value of a number is its distance from zero on the number line. For example, $|3| = 3$ and $|-3| = 3$. This means that when dealing with absolute value inequalities, we need to consider two cases: the expression inside the absolute value can be either positive or negative. Understanding this concept is fundamental to solving the inequality $|x-2| + 3 > 17$. In essence, we are looking for all values of $x$ such that the distance between $x$ and 2, plus 3, is greater than 17. The two cases arise because the expression $x-2$ can be either positive or negative, and we need to account for both possibilities to find the complete solution set. The ability to handle these two cases separately and then combine the results is a key skill in solving absolute value problems. This approach ensures that we capture all the possible values of $x$ that satisfy the original inequality. Ignoring either case could lead to an incomplete or incorrect solution, which is why this dual-case approach is so important. Mastering this technique not only helps with this particular problem but also builds a solid foundation for tackling more complex mathematical challenges involving absolute values. Therefore, we will meticulously examine both scenarios to ensure a comprehensive and accurate solution.
Isolating the Absolute Value
The first crucial step in solving the inequality $|x-2| + 3 > 17$ is to isolate the absolute value term. This means we need to get the expression $|x-2|$ by itself on one side of the inequality. To do this, we subtract 3 from both sides of the inequality. This operation maintains the balance of the inequality and allows us to simplify the problem. When we subtract 3 from both sides, we get: $|x-2| + 3 - 3 > 17 - 3$, which simplifies to $|x-2| > 14$. This isolation step is paramount because it sets the stage for dealing with the two cases that arise from the absolute value. By isolating $|x-2|$, we now have a clearer picture of the condition that $x-2$ must satisfy: its absolute value must be greater than 14. This means that the distance of $x-2$ from zero must be greater than 14, which translates into two separate inequalities that we will address in the subsequent steps. If we hadn't isolated the absolute value term first, it would be significantly more difficult to correctly interpret and solve the inequality. The additional term of +3 would complicate the analysis of the two cases. Therefore, isolating the absolute value is not just a matter of algebraic manipulation; it's a critical strategic step that simplifies the problem and makes it solvable. This step highlights the importance of careful algebraic manipulation in solving mathematical problems, ensuring that each operation brings us closer to a clearer understanding and solution.
Case 1: When x-2 is Positive or Zero
Now that we have isolated the absolute value, $|x-2| > 14$, we need to consider the two cases. In the first case, we assume that the expression inside the absolute value, which is $x-2$, is positive or zero. This means $x-2 extgreater 14$. When $x-2$ is positive or zero, the absolute value of $x-2$ is simply $x-2$ itself. Therefore, the inequality $|x-2| > 14$ becomes $x-2 > 14$. To solve this inequality, we add 2 to both sides: $x-2 + 2 > 14 + 2$, which simplifies to $x > 16$. This result tells us that all values of $x$ greater than 16 satisfy the original inequality, provided that $x-2$ is positive. However, it is crucial to remember that we are only considering the case where $x-2$ is positive or zero. If $x-2$ is negative, the absolute value will change its sign, and we need to consider the second case. Therefore, this condition of $x > 16$ is only part of the solution. We must also ensure that we don't include values that might satisfy $x > 16$ but would lead to a contradiction when we consider the other case. The condition $x > 16$ represents a range of values on the number line, and we will combine this range with the solution from the second case to obtain the complete solution set. Therefore, while this case provides us with a significant part of the solution, it's not the entire picture. We need to keep in mind the initial assumption that $x-2$ is positive, and this condition helps us accurately interpret the results within the context of the overall problem.
Case 2: When x-2 is Negative
In the second case, we consider what happens when the expression inside the absolute value, $x-2$, is negative. If $x-2$ is negative, then the absolute value of $x-2$ is $-(x-2)$, which is equivalent to $-x + 2$. So, the inequality $|x-2| > 14$ becomes $-x + 2 > 14$. To solve this inequality, we first subtract 2 from both sides: $-x + 2 - 2 > 14 - 2$, which simplifies to $-x > 12$. Now, to isolate $x$, we multiply both sides by -1. It is crucial to remember that when we multiply or divide an inequality by a negative number, we must reverse the inequality sign. Therefore, $-x > 12$ becomes $x extless -12$. This result shows that all values of $x$ less than -12 also satisfy the original inequality. This case is just as important as the first case, as it captures the other range of values that make the inequality true. Values of $x$ less than -12 are significantly far from 2 on the number line, ensuring that their absolute distance from 2, plus 3, is indeed greater than 17. Combining this case with the first case allows us to construct the complete solution set for the inequality. Without considering this case, we would only have a partial solution, missing a significant portion of the possible values of $x$. The importance of considering both cases highlights the nature of absolute value problems: they often require us to think about the possible scenarios and handle them separately. In this case, the negative scenario reveals a set of solutions that are quite different from those obtained in the positive scenario, emphasizing the need for a comprehensive approach.
Now that we have solved for both cases, we need to combine the solutions to find the complete solution set for the inequality $|x-2| + 3 > 17$. In Case 1, we found that $x > 16$. In Case 2, we found that $x extless -12$. To express this combined solution, we can use the “or” operator, which means that $x$ can be either greater than 16 or less than -12. This combination reflects the fact that there are two distinct ranges of values that satisfy the inequality. Values greater than 16 are sufficiently far to the right of 2 on the number line, and values less than -12 are sufficiently far to the left of 2, such that their absolute distance from 2, plus 3, exceeds 17. The “or” operator is crucial here because it acknowledges that $x$ cannot simultaneously be greater than 16 and less than -12. These are separate intervals on the number line. Graphically, we can represent the solution set on a number line by shading the regions to the right of 16 and to the left of -12, with open circles at 16 and -12 to indicate that these endpoints are not included in the solution. Understanding the use of “or” in this context is vital for accurately interpreting the solution. It indicates that there are two distinct sets of numbers that satisfy the condition, and considering both sets is necessary for a complete and accurate answer. If we were to use “and” instead of “or,” we would be implying that $x$ must satisfy both conditions simultaneously, which is impossible in this case. Therefore, the correct use of logical operators is essential in expressing mathematical solutions.
Therefore, the solution to the inequality $|x-2| + 3 > 17$ is $x extless -12$ or $x > 16$. This corresponds to option A in the given choices.
In this article, we have thoroughly explored the process of solving the absolute value inequality $|x-2| + 3 > 17$. We began by understanding the fundamental concept of absolute value and its implications for solving inequalities. We then systematically worked through the problem, isolating the absolute value term and considering the two cases that arise: when the expression inside the absolute value is positive or zero, and when it is negative. For each case, we solved the resulting inequality and obtained a partial solution. The key to solving absolute value inequalities lies in this methodical approach of considering all possible scenarios. By addressing each case separately and carefully applying the rules of algebra, we can unravel complex problems into manageable steps. We then combined the solutions from both cases using the “or” operator, recognizing that the solution set consists of two distinct intervals on the number line. This highlighted the importance of logical operators in accurately expressing mathematical solutions. Finally, we arrived at the complete solution: $x extless -12$ or $x > 16$. This comprehensive solution underscores the need for a thorough understanding of absolute values and inequalities. It also demonstrates the power of breaking down a problem into smaller, more manageable parts. By mastering these techniques, anyone can confidently tackle similar mathematical challenges. The process of solving this inequality provides valuable insights into algebraic problem-solving, emphasizing the significance of careful manipulation, logical reasoning, and a systematic approach. Ultimately, the ability to solve such problems strengthens mathematical skills and fosters a deeper understanding of the underlying concepts.
