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 within the realm of real numbers (R). Inequalities might seem a bit tricky at first, but trust me, with a little practice, you'll be solving them like a pro. We'll be tackling five different inequalities, breaking down each step so you can follow along easily. So, grab your pencils and let's get started!

Understanding Inequalities

Before we jump into solving, let's quickly recap what inequalities are. Unlike equations that have a single solution, inequalities deal with ranges of values. Think of it like this: instead of finding the exact value of x, we're finding all the values of x that make the statement true. Common inequality symbols include:

  • > Greater than
  • < Less than
  • ≥ Greater than or equal to
  • ≤ Less than or equal to

With that in mind, let's dive into our first inequality!

a) 5x + 8 ≥ 5(x + 1) + 3

Let's start with the first inequality: 5x + 8 ≥ 5(x + 1) + 3. Our goal here is to isolate x on one side of the inequality. We'll do this by simplifying and rearranging the terms, just like we would with a regular equation. Remember, the key is to perform the same operation on both sides to maintain the balance.

First, distribute the 5 on the right side of the inequality:

5x + 8 ≥ 5x + 5 + 3

Next, combine the constants on the right side:

5x + 8 ≥ 5x + 8

Now, subtract 5x from both sides:

8 ≥ 8

Wait a minute! The x terms have canceled out. What does this mean? Well, we're left with the statement 8 ≥ 8, which is absolutely true. This tells us that the inequality is true for all real numbers. In other words, no matter what value you plug in for x, the inequality will hold. We call this an identity. So, the solution set for this inequality is all real numbers, often denoted as R.

Key Points for this inequality:

  • The distribution property is crucial for simplifying expressions.
  • When variables cancel out, pay attention to the remaining statement.
  • A true statement like 8 ≥ 8 indicates that the inequality is true for all real numbers.

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

Moving on to our second inequality: 2(x + 1) < 2(x - 1) + 4. Again, our mission is to isolate x by simplifying and rearranging. Let’s break it down step-by-step.

First, distribute the 2 on both sides of the inequality:

2x + 2 < 2x - 2 + 4

Next, combine the constants on the right side:

2x + 2 < 2x + 2

Now, subtract 2x from both sides:

2 < 2

Whoa, another cancellation! This time, we’re left with the statement 2 < 2. Is this true? Nope! 2 is definitely not less than itself. This means that there is no value of x that will make this inequality true. We call this a contradiction. So, the solution set for this inequality is the empty set, often denoted as ∅ or {}.

Key Points for this inequality:

  • Careful distribution is key to avoiding errors.
  • A false statement like 2 < 2 indicates that the inequality has no solution.
  • The empty set signifies that no value of x satisfies the inequality.

c) 2(x + 3) ≥ 4(x – 1) – 2(x - 5)

Let's tackle the third inequality: 2(x + 3) ≥ 4(x – 1) – 2(x - 5). This one looks a bit more complex, but don't worry, we'll handle it the same way: simplify and isolate x.

First, distribute the numbers on both sides of the inequality:

2x + 6 ≥ 4x - 4 - 2x + 10

Next, combine like terms on the right side:

2x + 6 ≥ 2x + 6

Now, subtract 2x from both sides:

6 ≥ 6

Just like in the first example, the x terms cancel out, leaving us with the statement 6 ≥ 6. This is true, which means this inequality is also an identity. The solution set is all real numbers, R.

Key Points for this inequality:

  • Remember to distribute negative signs carefully.
  • Combining like terms simplifies the inequality.
  • A true statement confirms that the inequality holds for all real numbers.

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

Now, let's move on to the fourth inequality: 2(x - 7) + 5 < 7(x - 2) - 5(x - 4). We're sticking to our strategy: simplify and isolate x. Let's break it down.

First, distribute the numbers on both sides:

2x - 14 + 5 < 7x - 14 - 5x + 20

Next, combine like terms on both sides:

2x - 9 < 2x + 6

Now, subtract 2x from both sides:

-9 < 6

Again, the x terms cancel out! We're left with the statement -9 < 6, which is true. This means that this inequality is true for all real numbers, R.

Key Points for this inequality:

  • Careful attention to signs during distribution and combination is essential.
  • When variables disappear, analyze the resulting numerical statement.
  • A true numerical statement indicates the inequality is valid for all real numbers.

e) 5(x - 3) ≤ 3x + 2(x - 7)

Finally, let's tackle the last inequality: 5(x - 3) ≤ 3x + 2(x - 7). We're in the home stretch! Let's follow our familiar pattern.

First, distribute the numbers on both sides:

5x - 15 ≤ 3x + 2x - 14

Next, combine like terms on the right side:

5x - 15 ≤ 5x - 14

Now, subtract 5x from both sides:

-15 ≤ -14

Once more, the x terms have vanished. We're left with the statement -15 ≤ -14, which is true. This inequality holds for all real numbers, R.

Key Points for this inequality:

  • Consistent application of distribution and simplification leads to the solution.
  • A true numerical statement signifies universal validity of the inequality.
  • The solution set is all real numbers.

Conclusion

And there you have it! We've solved five different inequalities in R. Remember, the key is to simplify, rearrange, and pay close attention to the signs. Don't be afraid to break down each problem step by step. With practice, you'll become a master of inequalities. Keep practicing, and you'll be amazed at what you can achieve! Keep up the great work, guys!