Solving Inequalities In R: A Step-by-Step Guide

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Hey guys! Today, we're diving into the world of inequalities and how to solve them using the power of algebra. We'll be tackling five different inequalities, walking through each step to make sure you grasp the concepts. So, grab your pencils and let's get started!

Understanding Inequalities and R

Before we jump into the problems, let's quickly recap what inequalities are. Unlike equations that have a single solution, inequalities represent a range of values. The symbols we'll be working with include: ≥ (greater than or equal to), < (less than), and ≤ (less than or equal to). When we say "solving in R", we're referring to finding the set of real numbers that satisfy the inequality. This means we're looking for all the possible values of x that make the inequality true.

To effectively solve inequalities, it's crucial to understand the underlying principles. We need to isolate the variable, much like we do with equations, but with a few extra rules to keep in mind. One key rule is that multiplying or dividing by a negative number flips the direction of the inequality sign. This is a crucial point that often trips people up, so always double-check when you perform these operations. Another important thing to remember is that the solution to an inequality is usually a range of values, rather than a single number. This range can be represented graphically on a number line or written in interval notation. Mastering these fundamentals is key to confidently tackling more complex problems. It's not just about getting the right answer; it's about understanding why the answer is correct and how the rules of algebra apply in this context. So, take your time, practice, and you'll become an inequality-solving pro in no time!

a) 5x + 8 ≥ 5(x + 1) + 3

Let's begin with our first inequality: 5x + 8 ≥ 5(x + 1) + 3. Our goal here is to isolate x on one side of the inequality. Here's how we'll do it:

  1. Distribute: First, we'll distribute the 5 on the right side of the inequality: 5x + 8 ≥ 5x + 5 + 3.
  2. Simplify: Combine the constants on the right side: 5x + 8 ≥ 5x + 8.
  3. Isolate x: Now, let's subtract 5x from both sides: 8 ≥ 8.

Wait a minute! The x terms canceled out. What does this mean? This leaves us with a statement that is always true. 8 is indeed greater than or equal to 8. This indicates that the solution to this inequality is all real numbers. No matter what value we plug in for x, the inequality will hold true. This is a special case where the variable disappears during the solving process, revealing either a universally true statement (like we have here) or a contradiction. Recognizing these cases is an important part of understanding how inequalities work. In this instance, the solution set is the entire set of real numbers, which can be written in interval notation as (-∞, ∞). Always remember to check for these kinds of outcomes as you solve, as they can give you valuable insights into the nature of the inequality. Think of it like a puzzle piece fitting perfectly no matter what – that's what's happening with this inequality and any real number you substitute for x.

b) 2(x + 1) < 2(x - 1) + 4

Next up, we have the inequality 2(x + 1) < 2(x - 1) + 4. Let's break it down step by step:

  1. Distribute: Distribute the 2 on both sides: 2x + 2 < 2x - 2 + 4.
  2. Simplify: Combine the constants on the right side: 2x + 2 < 2x + 2.
  3. Isolate x: Subtract 2x from both sides: 2 < 2.

Again, the x terms canceled out. But this time, we're left with the statement 2 < 2, which is false. 2 is not less than 2. This means there is no solution to this inequality. There's no value of x that we can plug in that will make the inequality true. This type of outcome is known as a contradiction. It's like trying to fit a square peg into a round hole – it just won't work. When you encounter a contradiction like this, it's a clear sign that the inequality has no solutions within the set of real numbers. In mathematical terms, the solution set is the empty set, often denoted by the symbol ∅. So, if you arrive at a false statement after simplifying an inequality, you know you've hit a dead end and there are no solutions to be found.

c) 2(x + 3) ≥ 4(x – 1) – 2(x - 5)

Now, let's tackle the inequality 2(x + 3) ≥ 4(x – 1) – 2(x - 5). This one looks a bit more involved, but don't worry, we'll take it one step at a time:

  1. Distribute: First, distribute the numbers outside the parentheses: 2x + 6 ≥ 4x - 4 - 2x + 10.
  2. Simplify: Combine like terms on the right side: 2x + 6 ≥ 2x + 6.
  3. Isolate x: Subtract 2x from both sides: 6 ≥ 6.

Just like in the first inequality, the x terms canceled out, and we're left with a true statement: 6 is greater than or equal to 6. This indicates that the solution to this inequality is all real numbers. Any value of x you choose will satisfy the inequality. The solution set, therefore, encompasses the entire number line, extending from negative infinity to positive infinity. This is another instance where the initial inequality, despite its appearance, simplifies to a universally true condition. It's a valuable reminder that sometimes, the complexity of an inequality can be deceptive, and the solution might be simpler than you initially anticipated. So, always persevere with the simplification process – you never know when you'll uncover a hidden truth!

d) 2(x - 7) + 5 < 7(x - 2) - 5(x - 4)

Let's move on to the inequality 2(x - 7) + 5 < 7(x - 2) - 5(x - 4). We'll follow the same steps as before:

  1. Distribute: Distribute the numbers outside the parentheses: 2x - 14 + 5 < 7x - 14 - 5x + 20.
  2. Simplify: Combine like terms on both sides: 2x - 9 < 2x + 6.
  3. Isolate x: Subtract 2x from both sides: -9 < 6.

Once again, the x terms canceled out. We're left with the statement -9 < 6, which is true. Since -9 is indeed less than 6, this inequality holds true for all real numbers. This is similar to the first and third examples, where the variable disappears and we're left with a universally true statement. This means that regardless of the value you substitute for x, the inequality will always be satisfied. The solution set, in this case, is the set of all real numbers, which can be expressed in interval notation as (-∞, ∞). It's important to recognize these situations where the variable cancels out, as they provide a clear indication of the nature of the solution – either all real numbers or no solution at all.

e) 5(x - 3) ≤ 3x + 2(x - 7)

Finally, let's solve the inequality 5(x - 3) ≤ 3x + 2(x - 7). This time, we'll hopefully get a more traditional solution with a specific range of values for x:

  1. Distribute: Distribute the numbers outside the parentheses: 5x - 15 ≤ 3x + 2x - 14.
  2. Simplify: Combine like terms on the right side: 5x - 15 ≤ 5x - 14.
  3. Isolate x: Subtract 5x from both sides: -15 ≤ -14.

Yet again, the x terms canceled out. We're left with the statement -15 ≤ -14, which is true. -15 is indeed less than or equal to -14. This means the solution to this inequality is all real numbers. No matter what value we plug in for x, the inequality will hold true. This highlights the importance of carefully simplifying both sides of the inequality before making any conclusions. In this instance, the solution encompasses the entire range of real numbers, denoted in interval notation as (-∞, ∞). This outcome serves as a reminder that inequalities can sometimes lead to unexpected results, where the variable disappears, and the truthfulness of the inequality depends solely on the constant terms. So, always be prepared for a variety of outcomes when solving inequalities.

Conclusion

So there you have it, guys! We've successfully tackled five different inequalities, each with its own unique twist. Remember, the key to solving inequalities is to isolate the variable while paying close attention to the rules, especially when multiplying or dividing by negative numbers. And don't forget to check for those special cases where the x terms cancel out, leaving you with either a true statement (all real numbers) or a false statement (no solution). Keep practicing, and you'll become a master of inequalities in no time!