Solving Absolute Value Inequalities: Finding Solutions In Set-Builder And Interval Notation For |2x-1| < 11

Hey guys! Let's dive into solving an absolute value inequality problem today. We're going to break down how to solve $|2x - 1| < 11$, and then express our solution in both set-builder and interval notations. Don't worry, it's not as intimidating as it sounds! We'll go through it step by step, making sure everyone understands the process.

Understanding Absolute Value Inequalities

Before we jump into the specifics, let’s quickly recap what absolute value means. The absolute value of a number is its distance from zero. So, $|x|$ is the distance of $x$ from 0, which means it could be $x$ itself (if $x$ is positive or zero) or $-x$ (if $x$ is negative). That's a fundamental concept to grasp.

When we deal with an absolute value inequality like $|2x - 1| < 11$, it means that the distance of the expression $2x - 1$ from zero is less than 11. This gives us two scenarios to consider, which is super important to remember. The expression $2x - 1$ could be less than 11, but it also could be greater than -11. This is because any number between -11 and 11 has an absolute value less than 11.

Think of it like this: if you're within 11 units of zero, you can be anywhere from -11 to 11. This principle is key to unraveling absolute value inequalities. We’re essentially creating a compound inequality from our original absolute value inequality. This conversion is the magic sauce that makes these problems solvable. By understanding the dual nature of absolute value, we’re setting ourselves up for success. We'll transform one absolute value problem into two linear inequalities, which are much easier to handle. This is a common theme in mathematics – breaking down complex problems into simpler components. So, keep this in mind as we proceed through the solution.

Step-by-Step Solution

So, how do we actually solve $|2x - 1| < 11$? Remember what we just discussed – we need to consider two cases to tackle this absolute value inequality. Absolute value inequalities might seem tricky, but they're totally manageable if you break them down into these two separate scenarios. Each scenario gives us a linear inequality to solve, which we're probably pretty comfortable with already.

Case 1: The expression inside the absolute value is positive or zero

In the first case, we assume that the expression inside the absolute value bars, which is $2x - 1$, is positive or zero. If $2x - 1$ is already non-negative, then the absolute value doesn't change anything, and we can just drop the absolute value bars. So, we have: $2x - 1 < 11$. This is the first inequality we need to solve.

To isolate $x$, we first add 1 to both sides of the inequality: $2x - 1 + 1 < 11 + 1$, which simplifies to $2x < 12$. Next, we divide both sides by 2 to get $x < 6$. Awesome, we've solved the first inequality. This tells us that one part of our solution includes all the numbers less than 6. But remember, this is just one piece of the puzzle. We still have another case to consider.

Case 2: The expression inside the absolute value is negative

Now, let's consider the second case: what if the expression $2x - 1$ is negative? If $2x - 1$ is negative, then the absolute value will change its sign. In other words, $|2x - 1|$ becomes $-(2x - 1)$. So, the inequality $|2x - 1| < 11$ turns into $-(2x - 1) < 11$. It's crucial to remember this sign change when dealing with the negative case.

To get rid of the negative sign outside the parentheses, we can multiply both sides of the inequality by -1. But, and this is super important, when we multiply or divide an inequality by a negative number, we have to flip the direction of the inequality sign. So, $-(2x - 1) < 11$ becomes $2x - 1 > -11$. Don't forget this flip; it's a common mistake! Now, we solve this inequality just like we did in the first case. Add 1 to both sides: $2x - 1 + 1 > -11 + 1$, which simplifies to $2x > -10$. Then, divide both sides by 2: $x > -5$. Excellent, we've found the second part of our solution: all numbers greater than -5.

Combining the Solutions

So, what does this all mean? We found that $x < 6$ and $x > -5$. To satisfy the original inequality $|2x - 1| < 11$, $x$ must satisfy both of these conditions. We can think of it as $x$ being trapped between -5 and 6. It has to be bigger than -5 but also smaller than 6. This is a range of values, and that’s why we’ll see how to express this in interval notation soon. For now, let's put it all together in a way that makes sense for the different notations.

Expressing the Solution

Now that we've nailed down the solution, let's talk about how to express it in set-builder and interval notations. These are just fancy ways of writing down the same answer, but they each have their own style and are useful in different situations. Think of them as different languages for the same mathematical concept. Knowing both is like being bilingual in math! Set-builder notation is great for describing the properties of the solution, while interval notation is super concise for showing the range of values.

Set-Builder Notation

Set-builder notation is a way to define a set by describing its properties. It looks a bit like a sentence written in math symbols. It generally follows the format {x | condition(s)}, which reads as "the set of all $x$ such that condition(s) is/are true". So, we're building a set by specifying the rules for membership. It's like a VIP club where you need to meet certain criteria to get in.

In our case, we want to describe all the numbers $x$ that are both greater than -5 and less than 6. So, the conditions are $x > -5$ and $x < 6$. We can write this in set-builder notation as: {x | -5 < x < 6}. This reads as “the set of all $x$ such that $x$ is greater than -5 and less than 6”. See how it captures the range of solutions perfectly? The vertical bar “|” is read as “such that,” and it's what separates the variable from the conditions it must meet. The inequalities -5 < x < 6 clearly define the boundaries of our solution set. This notation is particularly useful when dealing with more complex solution sets, as it allows us to precisely specify the conditions for membership. It might look a bit formal, but once you get the hang of it, it’s a powerful tool for expressing mathematical ideas.

Interval Notation

Interval notation is a more compact way to represent a set of numbers that form an interval on the number line. It uses parentheses and brackets to indicate whether the endpoints are included in the interval or not. Parentheses () mean that the endpoint is not included (it's an open interval), while brackets [] mean that the endpoint is included (it's a closed interval). Think of parentheses as saying “up to but not including,” and brackets as saying “up to and including.” This distinction is crucial for accurately representing the solution set.

For our problem, we have $x > -5$ and $x < 6$. This means that $x$ can be any number between -5 and 6, but it cannot be -5 or 6 themselves. So, we use parentheses to exclude the endpoints. The interval notation for this solution is $(-5, 6)$. This is a nice and clean way to express the solution set. The left parenthesis indicates that -5 is not included, and the right parenthesis indicates that 6 is not included. It visually represents the range of values that satisfy the inequality. Interval notation is especially handy when dealing with graphs and visualizing solutions on the number line. It gives a quick snapshot of the solution's boundaries without the need for lengthy descriptions. It's also commonly used in calculus and other advanced math topics, so getting comfortable with it now will definitely pay off later.

Final Answer

So, to wrap it all up, the solution to the inequality $|2x - 1| < 11$ is:

• In set-builder notation: {x | -5 < x < 6}
• In interval notation: $(-5, 6)$

There you have it! We've successfully solved the absolute value inequality and expressed the solution in both set-builder and interval notations. Remember, the key is to break down the absolute value inequality into two separate cases, solve each one, and then combine the solutions. Practice makes perfect, so try tackling similar problems to solidify your understanding. You got this!