Inequality Proof And Solution: A ∈ (1/5, ∞) And (2b-1)/3

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Inequality Proof and Solution: a ∈ (1/5, ∞) and (2b-1)/3

Hey guys! Let's dive into some cool inequality problems today. We've got two main challenges: first, proving an inequality given a specific interval for a variable, and second, solving an inequality involving another variable. Buckle up, because we're about to break it all down in a way that's super easy to understand. We will focus on providing a detailed and comprehensive explanation, ensuring a human-friendly approach. Let’s get started!

Part 1: Proving 15 - a - 3 > 0 when a ∈ (1/5, ∞)

Okay, so the first part of our mission is to show that 15 - a - 3 > 0, given that a belongs to the interval (1/5, ∞). What does this interval even mean? Well, it simply means that a is any number greater than 1/5, all the way up to infinity. Think of it like a number line stretching out forever to the right, starting just past 1/5. Let's make sure we nail the core concepts from the get-go.

Understanding the Problem

Before we start crunching numbers, let's really understand what we're trying to prove. We need to show that if we take any number a from that infinite range (1/5 to infinity), plug it into the expression 15 - a - 3, the result will always be greater than zero. Essentially, we're aiming to demonstrate a consistent mathematical truth across a range of values. This requires a bit of algebraic manipulation and logical deduction.

Step-by-Step Solution

  1. Simplify the Expression: First things first, let's simplify the expression 15 - a - 3. This is basic algebra, guys. We can combine the constants 15 and -3 to get:

    15 - a - 3 = 12 - a

    So, now our inequality to prove is 12 - a > 0. This already looks a little friendlier, doesn't it?

  2. Isolate a: Next, we want to isolate a to see how it relates to the rest of the inequality. We can add a to both sides of the inequality:

    12 - a + a > 0 + a

    This simplifies to:

    12 > a

    Or, we can rewrite it as:

    a < 12

    Now we know that we need to show that a is less than 12.

  3. Consider the Given Interval: Remember, we know that a is in the interval (1/5, ∞). This means 1/5 < a. We need to combine this information with what we just found (a < 12) to complete our proof.

  4. Combine the Inequalities: We have two key pieces of information about a:

    • a > 1/5
    • a < 12

    We need to show that these two facts together prove that 12 - a > 0. Think about it this way: a is bigger than 1/5, but it's also smaller than 12. This is crucial because it sets the boundaries for a.

  5. The Final Step: We already know a < 12. This directly implies that if we subtract a from 12, the result will be positive. In other words:

    12 - a > 0

    Which is exactly what we needed to prove! See, guys? We did it!

Conclusion for Part 1

So, we’ve successfully shown that given a ∈ (1/5, ∞), the inequality 15 - a - 3 > 0 holds true. We simplified the expression, isolated a, considered the given interval, and combined the inequalities to reach our conclusion. High five for that! This type of problem highlights how important it is to understand intervals and manipulate inequalities. Remember, it's all about breaking it down step by step and thinking logically.

Part 2: Proving -3b + 16 > 1 when (2b - 1)/3 ∈ (-∞, 3)

Alright, now for the second part of our mathematical adventure! This time, we're dealing with a different variable, b, and a slightly more complex initial condition. We're given that (2b - 1)/3 belongs to the interval (-∞, 3), and we need to show that -3b + 16 > 1. Don't worry, we've got this! Let's tackle it with the same clear, step-by-step approach we used before. Ready to roll?

Understanding the New Challenge

Firstly, let's clarify the givens. When we say (2b - 1)/3 ∈ (-∞, 3), we're stating that the expression (2b - 1)/3 can take any value less than 3. It could be 2, 0, -5, or any other number stretching infinitely to the left on the number line, but it cannot be 3 or greater. Our mission is to use this information to prove that -3b + 16 is greater than 1. Think of it as a domino effect: the interval for (2b - 1)/3 should lead us to the inequality for -3b + 16.

Step-by-Step Solution

  1. Express the Interval as an Inequality: Our first move is to translate the interval (2b - 1)/3 ∈ (-∞, 3) into a mathematical inequality. This is straightforward: it simply means that:

    (2b - 1)/3 < 3

    This is our starting point. Everything else will flow from here.

  2. Isolate b: We need to manipulate this inequality to isolate b. Let's start by getting rid of the fraction. We can multiply both sides of the inequality by 3:

    3 * [(2b - 1)/3] < 3 * 3

    This simplifies to:

    2b - 1 < 9

    Next, let's add 1 to both sides:

    2b - 1 + 1 < 9 + 1

    Which gives us:

    2b < 10

    Finally, we divide both sides by 2:

    2b / 2 < 10 / 2

    Resulting in:

    b < 5

    Awesome! We now know that b must be less than 5.

  3. Work with the Target Inequality: Now let's turn our attention to the inequality we need to prove: -3b + 16 > 1. Our goal is to use the information we have about b (that b < 5) to show that this inequality is true.

  4. Manipulate the Target Inequality: Let's subtract 16 from both sides of the inequality:

    -3b + 16 - 16 > 1 - 16

    This simplifies to:

    -3b > -15

  5. Divide by -3 (and Flip the Inequality): This is a crucial step. Remember, whenever we multiply or divide an inequality by a negative number, we need to flip the direction of the inequality sign. So, we divide both sides by -3:

    (-3b) / -3 < (-15) / -3

    This gives us:

    b < 5

    Wait a minute... That's exactly what we found earlier! This is fantastic! It means we're on the right track.

  6. Connect the Pieces: We've shown that if (2b - 1)/3 ∈ (-∞, 3), then b < 5. And we've also shown that if -3b + 16 > 1, then b < 5. Now we need to put it all together to complete our proof.

  7. The Final Argument: We know b < 5. Let's go back to the inequality -3b + 16 > 1. We want to show that this holds true given b < 5. Let's add 3b to both sides and subtract 1 from both sides:

    16 > 3b + 1

    Subtract 1 from both sides:

    15 > 3b

    Divide both sides by 3:

    5 > b

    This is the same as b < 5, which we already know is true. Therefore, -3b + 16 > 1 must also be true. Boom! We've nailed it!

Conclusion for Part 2

We've successfully proven that if (2b - 1)/3 ∈ (-∞, 3), then -3b + 16 > 1. We started by expressing the interval as an inequality, isolated b, manipulated the target inequality, and connected the pieces to form our final argument. Give yourself a pat on the back! This problem showcases the power of algebraic manipulation and logical reasoning in proving inequalities.

Part 3: Solving Inequalities

Now, let's shift gears and talk about solving inequalities. Solving inequalities is super similar to solving equations, but with one critical difference: remember to flip the inequality sign when you multiply or divide by a negative number! This little rule can be a real game-changer. Let's look at some examples to get the hang of it.

Example 1: A Simple Inequality

Let's solve the inequality:

2x + 3 < 7

  1. Isolate the Variable Term: Subtract 3 from both sides:

    2x < 4

  2. Solve for x: Divide both sides by 2:

    x < 2

    So, the solution is x < 2. This means any number less than 2 will satisfy the inequality. We can represent this solution graphically on a number line with an open circle at 2 and shading to the left.

Example 2: Inequality with a Negative Coefficient

Now, let's tackle an inequality with a negative coefficient:

-3x + 5 ≥ 14

  1. Isolate the Variable Term: Subtract 5 from both sides:

    -3x ≥ 9

  2. Solve for x: Divide both sides by -3 (and remember to flip the inequality sign!):

    x ≤ -3

    The solution here is x ≤ -3. Any number less than or equal to -3 will make the inequality true. On a number line, we'd use a closed circle at -3 and shade to the left.

Example 3: Compound Inequality

Let's look at a compound inequality, which is essentially two inequalities joined together:

-5 < 2x - 1 ≤ 7

  1. Isolate the Variable Term: We need to isolate x in the middle. Add 1 to all parts of the inequality:

    -4 < 2x ≤ 8

  2. Solve for x: Divide all parts by 2:

    -2 < x ≤ 4

    So, the solution is -2 < x ≤ 4. This means x is greater than -2 but less than or equal to 4. On a number line, we'd use an open circle at -2, a closed circle at 4, and shade the region in between.

Key Takeaways for Solving Inequalities

  • Treat it Like an Equation: For the most part, you solve inequalities just like you solve equations. Use the same operations (addition, subtraction, multiplication, division) to isolate the variable.
  • The Flip Rule: The big exception is when you multiply or divide by a negative number. Always flip the inequality sign in this case!
  • Graphing the Solution: Visualizing the solution on a number line can be super helpful, especially for understanding compound inequalities.

Final Thoughts

So, guys, we've covered a lot of ground today! We've proven inequalities using given intervals, and we've brushed up on our inequality-solving skills. Remember, the key to mastering inequalities is practice, practice, practice! The more problems you solve, the more comfortable you'll become with the concepts and the techniques. And don’t forget that friendly reminder to flip that inequality sign when dividing or multiplying by a negative number. Keep up the great work, and happy problem-solving!