Solve For P: Linear Equation 24p + 12 - 18p = 10 + 2p - 6

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Solving for 'p' in a Linear Equation: A Step-by-Step Guide

Hey math enthusiasts! Ever stared at an equation and wondered, "What the heck is 'p' trying to tell me?" Well, today, guys, we're diving deep into a linear equation that might look a little jumbled at first glance: 24p+12βˆ’18p=10+2pβˆ’624 p+12-18 p=10+2 p-6. Our mission, should we choose to accept it, is to find the value of pp that makes this equation true. Think of it like a puzzle where we need to isolate that mysterious 'p' and discover its numerical worth. Linear equations are the building blocks of algebra, and understanding how to solve them is super crucial, whether you're tackling homework, prepping for exams, or just flexing those brain muscles. We're going to break this down piece by piece, making sure every step is crystal clear, so by the end of this, you'll be a pro at simplifying and solving these kinds of problems. So, grab your pencils, maybe a snack, and let's get this math party started!

Understanding the Equation and Our Goal

Alright, let's get down to business with our equation: 24p+12βˆ’18p=10+2pβˆ’624 p+12-18 p=10+2 p-6. What we're trying to do here is find the value of pp. In simpler terms, we want to get 'p' all by itself on one side of the equals sign. This is the fundamental concept behind solving algebraic equations. The equals sign is like a balanced scale; whatever you do to one side, you must do to the other to keep it balanced. Our equation has terms with 'p' and constant terms (just numbers) on both the left side and the right side. Before we can isolate 'p', we need to simplify both sides of the equation as much as possible. This means combining like terms. For instance, on the left side, we have '24p' and '-18p'. These are 'like terms' because they both involve the variable 'p'. We can combine them. Similarly, '12' is a constant term. On the right side, we have '10' and '-6' as constant terms, and '2p' as a term with the variable 'p'. Our main goal is to rearrange the equation so that all the 'p' terms are on one side and all the constant terms are on the other, and then finally, figure out what 'p' equals. It’s like tidying up a messy room – you put all the clothes in the closet and all the books on the shelf before you can really see what you have. So, let's start by tidying up each side of our equation.

Step 1: Simplify the Left Side of the Equation

Let's focus on the left side first, which is 24p+12βˆ’18p24 p+12-18 p. Our primary task here is to combine the 'like terms'. Remember, 'like terms' are terms that have the same variable raised to the same power. In this case, '24p' and '-18p' are like terms because they both contain the variable 'p' raised to the power of 1 (which we usually don't write). The number '12' is a constant term; it doesn't have a 'p' attached to it. So, we'll combine the 'p' terms: 24pβˆ’18p24p - 18p. If you have 24 'p's and you take away 18 'p's, you're left with (24βˆ’18)p(24 - 18)p, which simplifies to 6p6p. Now, we still have the constant term '+12' hanging around. So, the simplified left side of our equation becomes 6p+126p + 12. It's amazing how much cleaner it looks already, right? This simplification is a crucial step because it makes the equation much easier to handle in the subsequent steps. Always tackle simplification first, guys! It’s like preparing your ingredients before you start cooking – it makes the whole process smoother and less prone to errors. Take a moment to appreciate this simpler form: 6p+126p + 12. This represents the exact same value as the original 24p+12βˆ’18p24 p+12-18 p, just presented in a more organized way. It’s all about making complex things manageable, and in math, simplification is our best friend for achieving that.

Step 2: Simplify the Right Side of the Equation

Now, let's turn our attention to the right side of the original equation: 10+2pβˆ’610+2 p-6. Just like we did on the left, our goal is to combine any like terms here. We have a 'p' term, which is 2p2p, and we have two constant terms: +10+10 and βˆ’6-6. The 'p' term, 2p2p, doesn't have any other 'p' terms to combine with on this side, so it stays as is for now. However, we can combine the constant terms: 10βˆ’610 - 6. Doing this simple subtraction gives us 44. So, the simplified right side of our equation becomes 2p+42p + 4. See? Another side tidied up! This is why combining like terms is so powerful. We've taken a slightly messy expression and turned it into a much cleaner one: 2p+42p + 4. This new form makes it much easier to see what we need to do next to get 'p' by itself. Always remember to look for those like terms on each side before you start moving things across the equals sign. It saves a ton of confusion and potential mistakes down the line. So, after simplifying both sides, our original equation 24p+12βˆ’18p=10+2pβˆ’624 p+12-18 p=10+2 p-6 has now transformed into the much more manageable 6p+12=2p+46p + 12 = 2p + 4. Pretty neat, huh?

Step 3: Move 'p' Terms to One Side

Okay, team, we've simplified both sides and our equation now looks like this: 6p+12=2p+46p + 12 = 2p + 4. Our next big mission is to gather all the terms containing 'p' on one side of the equation and all the constant terms on the other. It doesn't really matter which side you choose for the 'p' terms, but I usually like to keep them on the side where they'll end up positive, if possible, to avoid extra steps later. Here, we have 6p6p on the left and 2p2p on the right. Since 6p6p is larger than 2p2p, let's aim to move the 'p' terms to the left side. To do this, we need to get rid of the 2p2p on the right side. How do we do that? By subtracting 2p2p from the right side. But remember our golden rule: whatever you do to one side, you MUST do to the other side to keep the equation balanced. So, we'll subtract 2p2p from both sides of the equation:

(6p+12)βˆ’2p=(2p+4)βˆ’2p(6p + 12) - 2p = (2p + 4) - 2p

Let's simplify both sides. On the left, we combine the 'p' terms: 6pβˆ’2p=4p6p - 2p = 4p. So the left side becomes 4p+124p + 12. On the right side, the 2p2p terms cancel each other out (2pβˆ’2p=02p - 2p = 0), leaving us with just 44.

So, our equation has now evolved into: 4p+12=44p + 12 = 4. Look at that! All the 'p' terms are now on the left side. This is a huge step towards finding the value of 'p'. We're getting closer, folks!

Step 4: Move Constant Terms to the Other Side

We're on the home stretch, guys! Our equation is currently 4p+12=44p + 12 = 4. We've successfully grouped the 'p' terms on the left. Now, we need to isolate the 4p4p term by moving the constant term, +12+12, from the left side to the right side. To eliminate the +12+12 from the left, we perform the opposite operation, which is subtracting 12. And, of course, to maintain the balance of our equation, we must subtract 12 from the right side as well.

So, let's do it:

(4p+12)βˆ’12=4βˆ’12(4p + 12) - 12 = 4 - 12

On the left side, the +12+12 and βˆ’12-12 cancel each other out (12βˆ’12=012 - 12 = 0), leaving us with just 4p4p. On the right side, we perform the subtraction: 4βˆ’12=βˆ’84 - 12 = -8.

Our equation now simplifies to: 4p=βˆ’84p = -8. We are so close to finding the value of 'p' now. This step is all about isolating the term that contains our variable. It’s like clearing the path so our variable can finally stand alone.

Step 5: Solve for 'p'

We've reached the final step, the grand finale! Our equation is now 4p=βˆ’84p = -8. This means 4 times the value of 'p' equals -8. To find out what a single 'p' is worth, we need to undo the multiplication. The opposite of multiplying by 4 is dividing by 4. So, we'll divide both sides of the equation by 4 to isolate 'p':

rac{4p}{4} = rac{-8}{4}

On the left side, rac{4p}{4} simplifies to just pp. On the right side, rac{-8}{4} equals βˆ’2-2.

Therefore, the value of pp is βˆ’2-2. We did it! We cracked the code and found the solution to our linear equation. It’s incredibly satisfying when you get to this point, isn't it? This means that if you were to substitute βˆ’2-2 back into the original equation for every 'p', both sides would be equal. Pretty awesome!

Verifying the Solution

Now, to be absolutely sure we nailed it, let's verify our solution by plugging p=βˆ’2p = -2 back into the original equation: 24p+12βˆ’18p=10+2pβˆ’624 p+12-18 p=10+2 p-6. This step is super important for confirming your answer and building confidence in your algebraic skills. Let's substitute βˆ’2-2 wherever we see pp:

Left Side: 24(βˆ’2)+12βˆ’18(βˆ’2)24(-2) + 12 - 18(-2) =βˆ’48+12+36= -48 + 12 + 36 =βˆ’36+36= -36 + 36 =0= 0

Right Side: 10+2(βˆ’2)βˆ’610 + 2(-2) - 6 =10βˆ’4βˆ’6= 10 - 4 - 6 =6βˆ’6= 6 - 6 =0= 0

Since the left side (0) equals the right side (0), our solution p=βˆ’2p = -2 is correct! High five! This verification process confirms that our step-by-step simplification and solving were accurate. It’s like double-checking your work before handing in a test – it ensures you haven’t missed anything. So, remember to always try and verify your answers when you can. It’s a fantastic habit to get into for mastering algebra.

Conclusion

And there you have it, folks! We’ve successfully navigated the twists and turns of the linear equation 24p+12βˆ’18p=10+2pβˆ’624 p+12-18 p=10+2 p-6 and discovered that the value of pp is βˆ’2-2. We broke it down by first simplifying both sides of the equation, then strategically moving the variable terms to one side and the constant terms to the other, and finally, isolating 'p' through division. Remember, the key to solving linear equations is to simplify, balance, and isolate. By combining like terms and performing inverse operations on both sides, you can tackle even the most intimidating-looking equations. Keep practicing these steps, and you'll find yourself becoming more and more comfortable with algebra. It’s all about practice, persistence, and a little bit of mathematical grit! Until next time, happy solving!