Solving Absolute Value Equations: A Step-by-Step Guide

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Solving Absolute Value Equations: A Step-by-Step Guide

Hey math enthusiasts! Ever stumbled upon an absolute value equation and felt a little lost? Don't sweat it! We're gonna break down how to solve equations like this one: 34+∣5βˆ’x∣=134\frac{3}{4}+|5-x|=\frac{13}{4}. It's all about understanding a few key concepts and following a systematic approach. By the end of this guide, you'll be tackling these problems with confidence. Let's dive in and unlock the secrets of absolute value equations! This guide is designed to be super easy to follow, even if you're just starting out with algebra. We'll go through each step carefully, explaining the why behind the what. Ready to get started? Awesome! Let's get our math on.

First off, let's understand what absolute value actually means. The absolute value of a number is its distance from zero on the number line. It's always a non-negative value. Think of it like this: the absolute value of 5, written as |5|, is 5 because 5 is 5 units away from zero. Similarly, the absolute value of -5, written as |-5|, is also 5 because -5 is also 5 units away from zero. Pretty straightforward, right? This concept is the cornerstone of solving absolute value equations. Now that we've got the basics down, let's get into the nitty-gritty of solving these equations. The main goal here is to isolate the absolute value expression. We want to get the part that looks like |something| all by itself on one side of the equation. This is the crucial first step that sets us up for success. Remember, our goal here is to isolate that absolute value expression. This means we want to get the term like ∣5βˆ’x∣|5-x| all alone on one side of the equation, without any other numbers or terms hanging around. Let's get started!

Isolating the Absolute Value Expression

Alright, guys, let's get down to business and start solving our equation: 34+∣5βˆ’x∣=134\frac{3}{4}+|5-x|=\frac{13}{4}. The first thing we need to do, as mentioned earlier, is to isolate the absolute value part. How do we do that? Well, we need to get rid of that pesky 34\frac{3}{4} hanging out on the left side. What's the opposite of adding 34\frac{3}{4}? That's right, subtracting 34\frac{3}{4}. So, we'll subtract 34\frac{3}{4} from both sides of the equation. This is super important because it keeps the equation balanced. Remember, whatever we do to one side, we must do to the other. Now, let's do the math:

34+∣5βˆ’xβˆ£βˆ’34=134βˆ’34\frac{3}{4}+|5-x|-\frac{3}{4}=\frac{13}{4}-\frac{3}{4}

The 34\frac{3}{4} terms on the left side cancel each other out, leaving us with:

∣5βˆ’x∣=134βˆ’34|5-x|=\frac{13}{4}-\frac{3}{4}

Now, let's simplify the right side of the equation:

∣5βˆ’x∣=104|5-x|=\frac{10}{4}

We can simplify 104\frac{10}{4} to 52\frac{5}{2}, so our equation now looks like this:

∣5βˆ’x∣=52|5-x|=\frac{5}{2}

See? We've successfully isolated the absolute value expression! It’s all alone on the left side, and we have a regular number on the right side. Great job, everyone! We're making awesome progress. Next up, we'll explore the two possible scenarios that arise from this absolute value setup. This is where things get a bit more interesting, but don't worry, we'll walk through it step by step. Keep up the amazing work.

Setting Up Two Equations

Okay, team, here's where the magic of absolute values really shines. Remember how we said the absolute value of a number is its distance from zero? This means that whatever's inside the absolute value bars, in our case (5βˆ’x)(5-x), could be either a positive number or a negative number. Both would have the same absolute value. Let's look at our equation again: ∣5βˆ’x∣=52|5-x|=\frac{5}{2}. This means that (5βˆ’x)(5-x) could equal 52\frac{5}{2} or (5βˆ’x)(5-x) could equal βˆ’52-\frac{5}{2}. We need to consider both possibilities because both will satisfy the original absolute value equation. This is why we're going to set up two separate equations to solve. Now, let's create our two equations based on the two possible values inside the absolute value:

Equation 1: 5βˆ’x=525-x=\frac{5}{2}

This equation assumes that the expression inside the absolute value is positive (or zero).

Equation 2: 5βˆ’x=βˆ’525-x=-\frac{5}{2}

This equation assumes that the expression inside the absolute value is negative. See how we've essentially removed the absolute value bars and created two new equations? Each equation represents a possible scenario for the value of (5βˆ’x)(5-x). Now, let's solve each of these equations independently. Remember to treat them as separate problems. This is a very important part of solving absolute value equations, so make sure you understand why we need to do this. We're getting closer to our final answer. Just hang tight, and we will get there.

Solving for x in Each Equation

Alright, let's roll up our sleeves and tackle these two equations one by one. Remember, our goal is to isolate x in each case. Let's start with Equation 1: 5βˆ’x=525-x=\frac{5}{2}.

To solve for x, we first want to get the x term by itself. Let's subtract 5 from both sides of the equation:

5βˆ’xβˆ’5=52βˆ’55-x-5=\frac{5}{2}-5

This simplifies to:

βˆ’x=52βˆ’102-x=\frac{5}{2}-\frac{10}{2}

Remember, we converted 5 into a fraction with a denominator of 2 to make the subtraction easier. Simplifying further, we get:

βˆ’x=βˆ’52-x=-\frac{5}{2}

Now, to solve for x, we need to get rid of that negative sign. We can do this by multiplying both sides of the equation by -1:

(βˆ’1)βˆ—βˆ’x=(βˆ’1)βˆ—βˆ’52(-1) * -x=(-1) * -\frac{5}{2}

Which gives us:

x=52x=\frac{5}{2}

Great! We've found the solution for the first equation. Now, let's move on to Equation 2: 5βˆ’x=βˆ’525-x=-\frac{5}{2}.

We'll follow the same steps. First, subtract 5 from both sides:

5βˆ’xβˆ’5=βˆ’52βˆ’55-x-5=-\frac{5}{2}-5

This simplifies to:

βˆ’x=βˆ’52βˆ’102-x=-\frac{5}{2}-\frac{10}{2}

Which gives us:

βˆ’x=βˆ’152-x=-\frac{15}{2}

Again, multiply both sides by -1:

(βˆ’1)βˆ—βˆ’x=(βˆ’1)βˆ—βˆ’152(-1) * -x=(-1) * -\frac{15}{2}

Which gives us:

x=152x=\frac{15}{2}

Awesome! We've solved both equations, and we have two possible solutions for x. Now, what do we do with these solutions? We need to make sure we didn't make any errors, and make sure we have the correct answers. We're almost there! Let's check our solutions and then wrap things up.

Checking the Solutions

Before we declare victory, it's always a good idea to check our solutions. This helps us catch any errors we might have made along the way. We do this by plugging each solution back into the original equation and seeing if it makes the equation true. Let's start by checking our first solution, x=52x = \frac{5}{2}, in the original equation: 34+∣5βˆ’x∣=134\frac{3}{4}+|5-x|=\frac{13}{4}.

Plug in x=52x = \frac{5}{2}:

34+∣5βˆ’52∣=134\frac{3}{4}+|5-\frac{5}{2}|=\frac{13}{4}

Simplify the expression inside the absolute value:

34+∣102βˆ’52∣=134\frac{3}{4}+|\frac{10}{2}-\frac{5}{2}|=\frac{13}{4}

34+∣52∣=134\frac{3}{4}+|\frac{5}{2}|=\frac{13}{4}

Take the absolute value:

34+52=134\frac{3}{4}+\frac{5}{2}=\frac{13}{4}

Convert 52\frac{5}{2} to a fraction with a denominator of 4:

34+104=134\frac{3}{4}+\frac{10}{4}=\frac{13}{4}

Add the fractions:

134=134\frac{13}{4}=\frac{13}{4}

It checks out! Now, let's check our second solution, x=152x = \frac{15}{2}, in the original equation:

34+∣5βˆ’x∣=134\frac{3}{4}+|5-x|=\frac{13}{4}

Plug in x=152x = \frac{15}{2}:

34+∣5βˆ’152∣=134\frac{3}{4}+|5-\frac{15}{2}|=\frac{13}{4}

Simplify the expression inside the absolute value:

34+∣102βˆ’152∣=134\frac{3}{4}+|\frac{10}{2}-\frac{15}{2}|=\frac{13}{4}

34+βˆ£βˆ’52∣=134\frac{3}{4}+|-\frac{5}{2}|=\frac{13}{4}

Take the absolute value:

34+52=134\frac{3}{4}+\frac{5}{2}=\frac{13}{4}

Convert 52\frac{5}{2} to a fraction with a denominator of 4:

34+104=134\frac{3}{4}+\frac{10}{4}=\frac{13}{4}

Add the fractions:

134=134\frac{13}{4}=\frac{13}{4}

It checks out too! Both of our solutions are correct. That's the beauty of checking our work. We've ensured that we've found the correct values for x. The final result is in sight, so let's write it down.

Conclusion

Fantastic work, everyone! We've successfully navigated through the process of solving the absolute value equation 34+∣5βˆ’x∣=134\frac{3}{4}+|5-x|=\frac{13}{4}. We started by isolating the absolute value expression, then we set up two separate equations, and finally, we solved for x in each equation. After that, we verified our solutions by plugging them back into the original equation. We found that the solutions are x=52x = \frac{5}{2} and x=152x = \frac{15}{2}. Remember, the key to mastering these types of problems is practice! Keep working through different examples, and you'll become more and more confident with absolute value equations. Keep practicing and keep learning! You've totally got this! Feel free to revisit this guide whenever you need a refresher. Now go out there and conquer those math problems! Remember the steps: isolate, split, solve, and check. You're now equipped with the knowledge and skills to tackle similar problems with confidence. Well done!