Solve |1/3x + 3| < 15: A Step-by-Step Guide

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Hey everyone! Today, we're diving into the exciting world of inequalities, specifically tackling the absolute value inequality ${\left|\frac{1}{3}x + 3\right| < 15}$. Inequalities might seem a bit intimidating at first, but trust me, with a systematic approach, they're totally manageable. We're going to break it down step by step, so you'll not only understand how to solve it but also why each step works. So, grab your pencils, and let’s get started!

Understanding Absolute Value Inequalities

Before we jump into the nitty-gritty of solving this specific inequality, let's quickly recap what absolute value means. The absolute value of a number is its distance from zero on the number line. It's always non-negative. For instance, ${|5| = 5}$ and ${|-5| = 5}$. This concept is crucial when dealing with inequalities because it introduces two possibilities: the expression inside the absolute value can be either positive or negative, but its distance from zero must still satisfy the inequality.

When you encounter an absolute value inequality like ${|ax + b| < c}$, where a, b, and c are constants, you're essentially saying that the distance between ${ax + b}$ and zero is less than c. This translates into two separate inequalities that you need to solve. For ${|ax + b| < c}$, these inequalities are:

1. ${ax + b < c}$
2. ${-(ax + b) < c}$, which is equivalent to ${ax + b > -c}$

Why do we split it like this? Think about it: if a number's absolute value is less than, say, 5, that number must be between -5 and 5. This is precisely what these two inequalities capture. The first one, ${ax + b < c}$, deals with the case where ${ax + b}$ is positive or zero, while the second one, ${ax + b > -c}$, takes care of the case where ${ax + b}$ is negative. By solving both, we cover all possible scenarios.

Remember, absolute value inequalities of the form ${|ax + b| > c}$ are handled slightly differently, leading to a union of two intervals rather than an intersection. But we'll focus on the "less than" case for this problem. Understanding this fundamental principle is key to successfully navigating absolute value inequalities. It’s not just about memorizing steps; it’s about grasping the underlying concept of distance and how it relates to inequalities. Once you get that, you’ll be solving these problems like a pro!

Breaking Down the Inequality: ${\left|\frac{1}{3}x + 3\right| < 15}$

Okay, let’s get our hands dirty with the actual problem! We have the inequality ${\left|\frac{1}{3}x + 3\right| < 15}$. The first thing we need to do, as we discussed, is to split this absolute value inequality into two separate inequalities. This is the golden rule for dealing with absolute values: eliminate the absolute value bars by considering both positive and negative scenarios of the expression inside.

Based on our understanding of absolute value, this inequality is equivalent to the following two inequalities:

1. ${\frac{1}{3}x + 3 < 15}$
2. ${\frac{1}{3}x + 3 > -15}$

Notice how the first inequality simply removes the absolute value bars, keeping the expression and the inequality sign the same. The second inequality, on the other hand, flips the inequality sign and negates the constant term on the right side. This is because we're considering the case where the expression inside the absolute value is negative, and its negative must still be greater than -15 for its absolute value to be less than 15.

Now, we have two linear inequalities that we can solve individually. This is a significant simplification! We've transformed a potentially confusing absolute value problem into two straightforward algebraic exercises. The next step involves isolating x in each inequality, which means getting rid of the constants and the coefficient attached to x. We'll do this by performing the same operations on both sides of each inequality, ensuring we maintain the balance and don't accidentally flip the inequality sign (which only happens when multiplying or dividing by a negative number, something we won't be doing in this case).

Remember, the goal here is to find the range of x values that satisfy both inequalities simultaneously. This means we're looking for the intersection of the solutions to each individual inequality. This concept of intersection is crucial for understanding the final solution. We're not just finding any values that work for one or the other; we need values that work for both. This is the essence of solving compound inequalities derived from absolute values.

So, with our two inequalities in hand, we're ready to roll up our sleeves and start isolating x. The journey might involve a bit of arithmetic, but the destination – the solution set – is well worth the effort! Let's dive into solving each inequality separately.

Solving the First Inequality: ${\frac{1}{3}x + 3 < 15}$

Alright, let's tackle the first inequality: ${\frac{1}{3}x + 3 < 15}$. Our mission here is to isolate x on one side of the inequality. The first step in this isolation process is to get rid of the constant term, which is +3 in this case. We can do this by subtracting 3 from both sides of the inequality. Remember, whatever we do to one side, we must do to the other to maintain the balance.

Subtracting 3 from both sides gives us:

${\frac{1}{3}x + 3 - 3 < 15 - 3}$

Simplifying this, we get:

${\frac{1}{3}x < 12}$

Great! We've successfully eliminated the constant term. Now, x is closer to being all by itself. The next hurdle is the coefficient of x, which is ${\frac{1}{3}}$. To get rid of this fraction, we need to multiply both sides of the inequality by the reciprocal of ${\frac{1}{3}}$, which is 3. This is a common technique when dealing with fractional coefficients.

Multiplying both sides by 3, we have:

${3 \cdot \frac{1}{3}x < 12 \cdot 3}$

This simplifies to:

${x < 36}$

Fantastic! We've isolated x in the first inequality. This tells us that all values of x less than 36 satisfy this part of our original absolute value inequality. But remember, this is only half the battle. We still need to solve the second inequality and then find the intersection of the two solution sets.

So, what does this ${x < 36}$ actually mean? It means that if you were to pick any number less than 36 and plug it into the expression ${\frac{1}{3}x + 3}$, the absolute value of the result would be less than 15. That's a pretty cool insight! But before we get too excited, let's not forget our other inequality. The complete solution to the absolute value inequality requires that x satisfies both inequalities. So, let's move on to the next one and see what it has in store for us.

Solving the Second Inequality: ${\frac{1}{3}x + 3 > -15}$

Now, let's dive into the second inequality: ${\frac{1}{3}x + 3 > -15}$. Just like before, our goal is to isolate x. We'll follow a similar process as we did with the first inequality, but this time, we're dealing with a “greater than” sign and a negative number on the right side. But don't worry, the principles remain the same!

The first step is, once again, to get rid of the constant term, which is +3. We do this by subtracting 3 from both sides of the inequality:

${\frac{1}{3}x + 3 - 3 > -15 - 3}$

Simplifying, we get:

${\frac{1}{3}x > -18}$

Excellent! The constant term is gone, and we're one step closer to isolating x. Now, we need to deal with the coefficient of x, which is ${\frac{1}{3}}$. As before, we'll multiply both sides of the inequality by the reciprocal of ${\frac{1}{3}}$, which is 3.

Multiplying both sides by 3, we have:

${3 \cdot \frac{1}{3}x > -18 \cdot 3}$

This simplifies to:

${x > -54}$

There we have it! We've successfully isolated x in the second inequality. This tells us that all values of x greater than -54 satisfy this part of our original absolute value inequality. So, we now know that x must be greater than -54 for the absolute value of ${\frac{1}{3}x + 3}$ to be less than 15.

Think about what we've accomplished: we've found two boundaries for x. From the first inequality, we learned that x must be less than 36, and from the second, we learned that x must be greater than -54. This means that x is trapped between these two values. But the solution isn't complete until we combine these two pieces of information.

So, we have ${x < 36}$ and ${x > -54}$. What does this mean in the context of our original problem? It means that any value of x that falls within this range will make the absolute value inequality true. But how do we express this combined solution in a clear and concise way? That’s what we'll tackle in the next section.

Combining the Solutions: Finding the Intersection

Okay, we've done the heavy lifting! We've solved both inequalities separately and found that ${x < 36}$ and ${x > -54}$. Now comes the crucial step: combining these solutions to find the complete solution set for our original absolute value inequality. This is where the concept of intersection comes into play.

Remember, we're looking for the values of x that satisfy both inequalities simultaneously. This is the key idea behind finding the intersection. If a value of x satisfies one inequality but not the other, it's not part of the final solution set. We need values that work for both.

So, how do we visualize this intersection? One of the best ways is to use a number line. Imagine a number line stretching from negative infinity to positive infinity. Let's mark the points -54 and 36 on this number line. The inequality ${x > -54}$ tells us that all values to the right of -54 are solutions. We can represent this on the number line with an open circle at -54 (since -54 itself is not included) and an arrow extending to the right.

Similarly, the inequality ${x < 36}$ tells us that all values to the left of 36 are solutions. We can represent this with an open circle at 36 and an arrow extending to the left. Now, the intersection is the region of the number line where these two arrows overlap. It's the segment between -54 and 36.

This means our solution set consists of all numbers that are both greater than -54 and less than 36. We can express this in several ways. One common way is to use inequality notation:

${-54 < x < 36}$

This notation is very concise and clearly shows the range of values that x can take. Another way to represent this solution is using interval notation. In interval notation, we use parentheses and brackets to indicate whether the endpoints are included or not. Since our endpoints, -54 and 36, are not included (due to the strict inequalities), we use parentheses:

${(-54, 36)}$

This interval notation means the same thing as ${-54 < x < 36}$: x can be any number between -54 and 36, but it cannot be -54 or 36 themselves.

So, there you have it! We've successfully combined the solutions to our two inequalities and expressed the result in both inequality and interval notation. This is the complete solution to the original absolute value inequality ${\left|\frac{1}{3}x + 3\right| < 15}$. We've come a long way, from understanding absolute values to breaking down the problem into smaller parts and then piecing everything back together. But the journey isn't quite over yet. Let's take a moment to verify our solution and make sure everything checks out.

Verifying the Solution: Does It Hold Up?

We've arrived at our solution: ${-54 < x < 36}$, or in interval notation, ${(-54, 36)}$. But before we celebrate, it's always a good idea to verify our solution. This step is crucial because it helps us catch any potential errors we might have made along the way. Verification gives us confidence that our answer is correct and reinforces our understanding of the problem.

So, how do we verify a solution to an inequality? The basic idea is to pick a few test values within our solution set and plug them back into the original inequality. If the inequality holds true for these values, it gives us strong evidence that our solution is correct. It's also a good idea to test values outside our solution set to see if they make the inequality false, as they should.

Let's start by picking a value inside our solution set, say, ${x = 0}$. This is a nice, easy number to work with, and it clearly falls between -54 and 36. Plugging ${x = 0}$ into the original inequality, we get:

${\left|\frac{1}{3}(0) + 3\right| < 15}$

Simplifying, we have:

${|3| < 15}$

Which is clearly true, since 3 is indeed less than 15. This gives us some confidence that our solution is on the right track.

Now, let's pick another value inside our solution set, perhaps a bit closer to the boundaries. How about ${x = 30}$? Plugging this in, we get:

${\left|\frac{1}{3}(30) + 3\right| < 15}$

Simplifying:

${|10 + 3| < 15}$

${|13| < 15}$

Again, this is true, as 13 is less than 15. Our solution seems to be holding up well.

Now, let's test a value outside our solution set. Let's try ${x = 40}$, which is greater than 36. Plugging this in:

${\left|\frac{1}{3}(40) + 3\right| < 15}$

Simplifying:

${\left|\frac{40}{3} + 3\right| < 15}$

${\left|\frac{49}{3}\right| < 15}$

${16.333... < 15}$

This is false! As expected, a value outside our solution set does not satisfy the original inequality. This further strengthens our confidence in our solution.

Finally, let's test a value less than -54, say ${x = -60}$:

${\left|\frac{1}{3}(-60) + 3\right| < 15}$

Simplifying:

${|-20 + 3| < 15}$

${|-17| < 15}$

${17 < 15}$

This is also false, as it should be. We've thoroughly tested our solution with values inside and outside the solution set, and it checks out perfectly. We can confidently say that our solution, ${-54 < x < 36}$ or ${(-54, 36)}$, is correct!

Conclusion: Mastering Absolute Value Inequalities

Wow, we've covered a lot of ground! We started with the absolute value inequality ${\left|\frac{1}{3}x + 3\right| < 15}$ and, through a step-by-step process, arrived at the solution ${-54 < x < 36}$. We not only solved the problem but also delved into the why behind each step, ensuring a solid understanding of the underlying concepts.

We began by understanding the fundamental meaning of absolute value and how it relates to inequalities. We learned that an absolute value inequality of the form ${|ax + b| < c}$ can be split into two separate inequalities: ${ax + b < c}$ and ${ax + b > -c}$. This is the cornerstone of solving these types of problems.

Next, we meticulously solved each inequality individually, carefully isolating x by performing the same operations on both sides. We saw how subtracting constants and multiplying by reciprocals helped us get closer to our goal. We then emphasized the crucial concept of intersection, understanding that the solution to the original absolute value inequality is the set of values that satisfy both individual inequalities.

We visualized the intersection using a number line and learned how to express our solution in both inequality notation (${-54 < x < 36}$) and interval notation (${(-54, 36)}$). This gave us flexibility in how we communicate our results.

Finally, we stressed the importance of verification. We picked test values inside and outside our solution set and plugged them back into the original inequality. This not only confirmed our solution but also deepened our understanding of the problem.

Solving absolute value inequalities might seem daunting at first, but with a clear understanding of the concepts and a systematic approach, they become much more manageable. The key is to break down the problem into smaller, more digestible parts, solve each part carefully, and then combine the results thoughtfully. And, of course, always verify your solution!

So, the next time you encounter an absolute value inequality, remember the steps we've discussed. You've got this! Keep practicing, and you'll become a master of inequalities in no time. Happy solving!