Solving Absolute Value Equations: 3|x-2|+6=18

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of mathematics, specifically tackling absolute value equations. If you've ever stared at an equation like $f(x)=3|x-2|+6$ and wondered, "For what values of $x$ is $f(x)=18$?" you're in the right place. We're going to break down this problem step-by-step, making sure you guys understand every bit of it. So grab your notebooks, put on your thinking caps, and let's get started on this awesome math adventure! Understanding absolute value is key here, so let's quickly recap what it means. The absolute value of a number is its distance from zero, and it's always positive. For example, $|-5|$ is 5, and $|5|$ is also 5. This 'always positive' nature is what makes absolute value equations sometimes have two possible solutions. In our case, we have $f(x) = 3|x-2|+6$, and we want to find the $x$ values when $f(x)$ equals 18. So, we're essentially setting up the equation: $3|x-2|+6 = 18$. Our goal is to isolate that absolute value part, $|x-2|$, so we can then split it into its two possible cases. This process is fundamental for solving any equation involving absolute values, and by the end of this article, you'll be a pro at it! We'll go through each manipulation carefully, ensuring that no stone is left unturned in our quest for the correct values of $x$. So, let's roll up our sleeves and get to it!

Isolating the Absolute Value Term

Alright, team, the first crucial step in solving our absolute value equation, $3|x-2|+6 = 18$, is to isolate the absolute value expression, which is $|x-2|$. Think of it like unwrapping a present – you have to get rid of the outer layers first to get to the good stuff inside. Our current equation has the absolute value term being multiplied by 3 and then having 6 added to it. To isolate $|x-2|$, we need to reverse these operations. First, let's tackle that '+6'. The opposite of adding 6 is subtracting 6. So, we'll subtract 6 from both sides of the equation to maintain balance. This gives us: $3|x-2| + 6 - 6 = 18 - 6$, which simplifies to $3|x-2| = 12$. Nice one! We're one step closer. Now, we have the absolute value term being multiplied by 3. The opposite of multiplying by 3 is dividing by 3. So, we'll divide both sides of the equation by 3: $\frac{3|x-2|}{3} = \frac{12}{3}$. This leaves us with the beautifully isolated absolute value expression: $|x-2| = 4$. See? We've successfully peeled away the outer layers, and now we have the core of our problem right in front of us. This isolated form, $|x-2| = 4$, tells us that the distance between $x$ and 2 on the number line is exactly 4 units. This is where the 'two solutions' magic of absolute value comes into play. Because distance is always positive, there are two numbers that are 4 units away from 2: one on the right and one on the left. We're going to explore those two possibilities in the next section. Keep up the great work, everyone!

Splitting into Two Cases

Now that we've got our equation down to $|x-2| = 4$, it's time to unleash the power of absolute value and split this into two separate, simpler equations. Remember, the absolute value of something being equal to 4 means that 'something' is either 4 or -4. This is the core concept we're applying here. Our 'something' in this case is the expression $(x-2)$. So, we can set up our two equations:

Case 1: The expression inside the absolute value is positive. In this scenario, we simply remove the absolute value bars, assuming the expression $x-2$ is equal to 4. $x-2 = 4$

Case 2: The expression inside the absolute value is negative. Here, we consider the possibility that the expression $x-2$ is equal to -4. When you take the absolute value of -4, you get 4, which is why this case is valid. $x-2 = -4$

Why is this so important, you ask? Because the original absolute value equation $|x-2| = 4$ is satisfied by any value of $x$ that makes $x-2$ equal to either 4 or -4. By splitting it into these two distinct linear equations, we transform a problem that might seem tricky at first into two straightforward ones that we can solve easily. This technique is fundamental for mastering absolute value equations, and it's a skill that will serve you well in all your future math endeavors. We're now just one step away from finding our values for $x$. Let's solve these two cases in the next section and see what answers we get!

Solving for x in Each Case

We're on the home stretch, guys! We've successfully isolated the absolute value and split our problem into two manageable equations: $x-2 = 4$ and $x-2 = -4$. Now, all we need to do is solve each of these linear equations for $x$. This is the easy part!

Solving Case 1: We have the equation $x-2 = 4$. To get $x$ by itself, we need to undo the '-2'. The opposite of subtracting 2 is adding 2. So, we add 2 to both sides of the equation: $x - 2 + 2 = 4 + 2$ $x = 6$

So, one possible value for $x$ is 6. Pretty straightforward, right?

Solving Case 2: Next, we tackle the second equation: $x-2 = -4$. Again, to isolate $x$, we need to get rid of that '-2'. We do this by adding 2 to both sides: $x - 2 + 2 = -4 + 2$ $x = -2$

And there you have it! The second possible value for $x$ is -2. So, our two solutions are $x=6$ and $x=-2$. Fantastic job! You've navigated through the complexities of absolute value, isolated the term, split the equation into two cases, and solved each one to find the values of $x$. This process demonstrates a solid understanding of how absolute value functions work and how to solve equations involving them. We've arrived at our final answer by systematically applying mathematical principles. Let's quickly check our answers in the original equation to make sure they are correct.

Verification of Solutions

To really nail this down and be 100% sure we've got the right answers, let's plug our found values of $x$ back into the original function $f(x)=3|x-2|+6$ and see if we indeed get $f(x)=18$. This step is super important in math – it's like double-checking your work before submitting it!

Let's check $x = 6$: $f(6) = 3|6-2| + 6$ $f(6) = 3|4| + 6$

Since $|4| = 4$ (the absolute value of 4 is just 4): $f(6) = 3(4) + 6$ $f(6) = 12 + 6$ $f(6) = 18$

Boom! $x=6$ works perfectly. It satisfies the condition $f(x)=18$. Now, let's test our other solution.

Let's check $x = -2$: $f(-2) = 3|-2-2| + 6$ $f(-2) = 3|-4| + 6$

Remember, the absolute value of -4 is 4: $f(-2) = 3(4) + 6$ $f(-2) = 12 + 6$ $f(-2) = 18$

Awesome! $x=-2$ also works, giving us $f(x)=18$. This verification process confirms that our two solutions, $x=-2$ and $x=6$, are indeed the correct values for which $f(x)=18$. This confirms our thorough understanding and application of absolute value principles. It's always a good practice to verify your solutions, especially in exams, to avoid silly mistakes and ensure you get full credit. You guys have crushed this problem!

Conclusion and Final Answer

So, there you have it, math whizzes! We started with the function $f(x)=3|x-2|+6$ and were asked to find the values of $x$ for which $f(x)=18$. Through a systematic process of isolating the absolute value, splitting the equation into two cases, and solving each case, we found our solutions. We meticulously worked through each step, ensuring clarity and accuracy, and even verified our answers by plugging them back into the original equation. This journey through solving absolute value equations highlights the importance of understanding the definition of absolute value and applying algebraic techniques correctly. The two solutions we found, $x = -2$ and $x = 6$, are the only values that satisfy the given condition. Looking back at the multiple-choice options provided:

A. $x=-2, x=-8$ B. $x=-2, x=-6$ C. $x=-2, x=6$ D. $x=-2, x=8$

Our verified solutions perfectly match option C. You guys have successfully solved this absolute value problem, demonstrating a strong grasp of the concepts. Keep practicing, and you'll become even more confident in tackling these types of math challenges. Remember, every problem solved is a step towards mastering mathematics. Keep that curiosity alive and keep exploring the amazing world of numbers! Until next time, stay sharp and keep those mathematical minds engaged here at Plastik Magazine!