Solving 1 < |x-2| < 5: Your Math Solution

Hey guys, let's dive into a totally rad math problem today that's all about inequalities and absolute values. We're going to figure out the solution set for the inequality $1 < |x-2| < 5$. This might look a little intimidating at first, but trust me, once we break it down, it's totally manageable. We're aiming to fill in those blanks: (-3, oxed{ } ) igcup (oxed{ }, oxed{ }). So, buckle up, grab your favorite beverage, and let's get this math party started! We'll go step-by-step, making sure everyone's on the same page, and by the end, you'll be a pro at tackling these kinds of problems. Remember, the key to mastering these is understanding the properties of absolute values and how they play with inequalities. We're going to explore how the absolute value affects the possible values of x, and how the compound inequality $1 < |x-2| < 5$ actually represents two separate conditions that x must satisfy simultaneously. It’s like solving two puzzles and then finding the pieces that fit in both.

Understanding Absolute Value Inequalities

Alright, let's get down to business with the core concept: absolute value. You guys know that the absolute value of a number, like $|a|$, is its distance from zero on the number line. So, $|a|$ is always positive or zero. When we see an inequality involving an absolute value, like $|x-2| < 5$, it means the distance between x and 2 is less than 5. To solve this, we can think of it as $-5 < x-2 < 5$. This is a crucial step, guys, because it transforms the absolute value problem into a standard compound inequality. Adding 2 to all parts gives us $-3 < x < 7$. Easy peasy, right? Now, what about when the absolute value is on the other side of the inequality, like $|x-2| > 1$? This means the distance between x and 2 is greater than 1. This translates into two separate inequalities: $x-2 > 1$ or $x-2 < -1$. Solving these gives us $x > 3$ or $x < 1$. So, for $|x-2| > 1$, our solution is x is in (-\infty, 1) igcup (3, \infty). See how the absolute value splits the number line into different regions? This is the fundamental idea we'll use to crack our main problem. It's like having two separate paths to consider, and we need to find the values of x that satisfy both conditions within our given range. The visual representation on a number line can be super helpful here, showing you exactly where these ranges lie and how they overlap or don't overlap. We're essentially dissecting the absolute value expression into its positive and negative cases, and then applying the inequality rules to each case.

Breaking Down the Compound Inequality

Now, let's tackle our specific inequality: $1 < |x-2| < 5$. This bad boy is a compound inequality, meaning it combines two inequalities into one. It's basically saying two things must be true at the same time: First, $|x-2| > 1$, AND second, $|x-2| < 5$. We've already figured out how to solve these individual pieces, so let's put it all together. For the first part, $|x-2| > 1$, we found that this means $x-2 > 1$ or $x-2 < -1$. Solving these gives us $x > 3$ or $x < 1$. In interval notation, this is (-\infty, 1) igcup (3, \infty). Now, for the second part, $|x-2| < 5$, we found that this means $-5 < x-2 < 5$. Adding 2 to all parts, we get $-3 < x < 7$. In interval notation, this is $(-3, 7)$.

So, we need the values of x that satisfy BOTH conditions. This means we need to find the intersection of the solution sets. We need x to be in (-\infty, 1) igcup (3, \infty) AND x to be in $(-3, 7)$. Let's visualize this on a number line, guys. We have the first set covering everything less than 1 and everything greater than 3. The second set covers everything between -3 and 7. Where do these overlap?

Think about it:

• For the part where $x < 1$ (from the first condition), it needs to also be within $(-3, 7)$. So, the overlap here is $(-3, 1)$.
• For the part where $x > 3$ (from the first condition), it also needs to be within $(-3, 7)$. So, the overlap here is $(3, 7)$.

Therefore, the solution set for $1 < |x-2| < 5$ is the combination of these two overlapping intervals. We're joining these two valid regions together because x can fall into either of them. It's crucial to remember that the 'AND' condition in a compound inequality means we're looking for the overlap, the common ground, between the individual solution sets. This intersection step is where many people can get tripped up, so taking a moment to sketch out a number line or carefully list the intervals is super important. We're not just combining everything; we're finding what x values satisfy all parts of the original inequality simultaneously. The structure of the absolute value being 'sandwiched' between two numbers is key here – it defines a specific range of distances from the center point (which is 2 in this case).

Finding the Final Solution Set

Okay, we've done the heavy lifting, guys! We've broken down the inequality $1 < |x-2| < 5$ into its core components and solved each part. We found that $|x-2| > 1$ leads to $x < 1$ or $x > 3$, which in interval notation is (-\infty, 1) igcup (3, \infty). We also found that $|x-2| < 5$ leads to $-3 < x < 7$, or the interval $(-3, 7)$. Now, we need the values of x that satisfy both conditions. This means we're looking for the intersection of these two solution sets. Let's visualize this again. We need the numbers that are less than 1 and between -3 and 7. That intersection is $(-3, 1)$. Then, we need the numbers that are greater than 3 and between -3 and 7. That intersection is $(3, 7)$.

Since x can satisfy either of these combined conditions, we use the union symbol to put them together. So, the complete solution set for $1 < |x-2| < 5$ is (-3, 1) igcup (3, 7).

This means that any number x between -3 and 1 (exclusive), OR any number x between 3 and 7 (exclusive), will make the original inequality true. Let's quickly test a value. How about $x = 0$? Then $|0-2| = |-2| = 2$. Is $1 < 2 < 5$? Yes, it is! Okay, what about $x = 5$? Then $|5-2| = |3| = 3$. Is $1 < 3 < 5$? Yep, that works too! Now, let's try a value outside the set, like $x = -4$. Then $|-4-2| = |-6| = 6$. Is $1 < 6 < 5$? Nope, 6 is not less than 5. How about $x = 2$? Then $|2-2| = |0| = 0$. Is $1 < 0 < 5$? Nope, 0 is not greater than 1. This confirms our solution set is correct.

So, to fill in the blanks in your problem: (-3, oxed{1} ) igcup (oxed{3}, oxed{7}). You guys totally crushed it! Understanding how to split absolute value inequalities and then finding the intersection of the resulting conditions is the key. Keep practicing, and these will become second nature. Remember, the absolute value is all about distance, and when it's bounded between two numbers, it describes a specific range of possible distances from a central point. This geometric interpretation can make the algebra feel much more intuitive. Don't hesitate to draw out the number lines; it's a super powerful tool for visualizing these inequalities and ensuring you've got the correct intersections and unions. It’s about seeing where your potential solutions actually 'live' on the number line and how they fit the constraints of the original problem.