Solving Inequalities: A & B Sets Explained

Hey guys, let's dive into some math fun! We're going to break down two sets, A and B, and explore how to solve inequalities and find their intersection. This is super helpful for understanding intervals and set theory. So, buckle up and let's get started!

Understanding Set A and Converting to Interval

Our first challenge is to unravel set A. It's defined as all real numbers x where the absolute value of (2x + 1) divided by 3 is less than or equal to 1. That looks a bit intimidating at first, but trust me, it's not as scary as it seems. We'll break it down step by step. The problem states that A = {x ∈ ℝ | |2x + 1|/3 ≤ 1}. The main keyword is Set A, and we want to write this set in interval form.

First things first, let's get rid of that absolute value sign. Remember that the absolute value of an expression means its distance from zero. So, if the absolute value of (2x + 1)/3 is less than or equal to 1, that means (2x + 1)/3 is between -1 and 1, inclusive. We can write this as:

-1 ≤ (2x + 1)/3 ≤ 1

To solve this, we need to isolate x. Let's start by multiplying everything by 3 to get rid of the fraction:

-3 ≤ 2x + 1 ≤ 3

Next, subtract 1 from all parts of the inequality:

-4 ≤ 2x ≤ 2

Finally, divide everything by 2:

-2 ≤ x ≤ 1

So, we've found that x must be greater than or equal to -2 and less than or equal to 1. This is the solution for A. Thus, in interval notation, set A is written as A = [-2, 1]. That means all the numbers between -2 and 1, including -2 and 1 themselves. Easy peasy, right? By carefully following these steps, we've successfully converted the initial inequality into a simple interval, making it much easier to work with and visualize. The key here is to remember the properties of absolute values and how they translate into compound inequalities. Remember to be extra careful and to check your work to avoid careless errors.

Unraveling Set B and Expressing as Interval

Now let's turn our attention to set B. Set B is defined as all real numbers x where x² - 4 ≤ 0. This involves a quadratic inequality, which is a little different from what we did with set A, but still manageable. The main keyword here is Set B and our goal is to find this set in interval form. Let's find out the interval.

First things first, let's rearrange the inequality: x² - 4 ≤ 0. Then, add 4 to both sides: x² ≤ 4. Now we need to find the values of x that satisfy this condition. We can solve this inequality in a couple of ways. The most straightforward approach is to recognize that x² ≤ 4 means that the square of x is less than or equal to 4. Take the square root of both sides, we get |x| ≤ 2. Remember that when you take the square root of a variable squared, you need to consider both positive and negative values. This gives us -2 ≤ x ≤ 2. Therefore, x must be between -2 and 2, inclusive. Thus, set B in interval notation is B = [-2, 2].

Another method is to factor the original inequality x² - 4 ≤ 0. We can factor this difference of squares into (x - 2)(x + 2) ≤ 0. To solve this, we find the zeros of the quadratic, which are x = 2 and x = -2. These are the points where the expression equals zero. Then, we can test intervals to determine where the expression is less than or equal to zero. If x < -2, for example, we can test x = -3. We find that (x - 2)(x + 2) = (-5)(-1) > 0. This means that the expression is positive for x < -2. If -2 < x < 2, we can test x = 0. We find that (x - 2)(x + 2) = (-2)(2) < 0. This means the expression is negative in this interval. Finally, if x > 2, we can test x = 3. We find that (x - 2)(x + 2) = (1)(5) > 0. This means the expression is positive for x > 2. Since we're looking for where the expression is less than or equal to zero, we're looking at the interval between -2 and 2, including -2 and 2. Which again gives us the answer B = [-2, 2]. Remember, understanding quadratic inequalities is all about finding the roots, testing intervals, and knowing when to include or exclude the endpoints.

Determining the Intersection of Sets A and B

Alright, now for the grand finale: finding the intersection of A and B. The intersection of two sets is the set of all elements that are common to both sets. In simpler terms, we're looking for the values of x that satisfy both inequalities. Now we have A = [-2, 1] and B = [-2, 2].

We can visualize this on a number line. Draw a number line and mark the intervals for A and B. A goes from -2 to 1, and B goes from -2 to 2. The values that belong to both sets are those that fall within both intervals. Looking at the number line, we can clearly see that the overlapping region goes from -2 to 1. Thus, the intersection of A and B, which we denote as A ∩ B, is [-2, 1]. Notice how the intersection is the set containing the values that satisfy both original inequalities. In this case, it is the interval that contains all of the numbers that satisfy the requirements of both A and B. This is a classic example of how set theory works and how to solve mathematical problems using basic concepts.

This tells us that the intersection of A and B includes all the numbers that are greater than or equal to -2 and less than or equal to 1. Therefore, the solution for the intersection A ∩ B is exactly the same as set A! We can notice this because set A is entirely contained within set B. Any time one set is contained within another, the intersection will simply be the smaller of the two sets. Isn't math wonderful?

Summary and Key Takeaways

So, to recap, here's what we've done:

1. We looked at set A and transformed the inequality |2x + 1|/3 ≤ 1 into interval notation, which is [-2, 1].
2. We examined set B and found its interval representation, which is [-2, 2].
3. Finally, we found the intersection of A and B, which is [-2, 1].

Key takeaways: Always remember how to work with absolute values and quadratic inequalities. Also, the importance of interval notation and how it simplifies the representation of solutions. Practice makes perfect, so try some similar problems on your own to cement your understanding. Now you are ready to take on more complex problems and explore the world of mathematics. Keep it up!