Solve For P: A Step-by-Step Math Guide

by Andrew McMorgan 39 views

Hey math whizzes! Ever feel like solving equations is like trying to crack a secret code? Well, you're not alone, guys. Today, we're diving deep into a problem that’ll test your algebra skills and show you how to nail it. We're tackling this beast: 2.5(βˆ’3pβˆ’8)+5p=4(2.25p+5.5)+15.52.5(-3 p-8)+5 p=4(2.25 p+5.5)+15.5. Don't let those decimals and parentheses scare you off. We're going to break it down, step by step, using some awesome properties of mathematics to find that elusive value of pp. Get ready to flex those brain muscles because by the end of this, you'll be a pro at simplifying and solving equations like this one. We’ll be using the distributive property and the concept of combining like terms, which are super fundamental tools in your algebra toolbox. Stick around, and let's make this equation crystal clear!

Unpacking the Equation: The First Steps to Solving for pp

Alright, let's get down to business with our equation: 2.5(βˆ’3pβˆ’8)+5p=4(2.25p+5.5)+15.52.5(-3 p-8)+5 p=4(2.25 p+5.5)+15.5. The first hurdle we often face with these kinds of problems is dealing with those pesky parentheses. This is where the distributive property comes in to save the day. Remember, the distributive property essentially says that a(b+c)=ab+aca(b+c) = ab + ac. We need to apply this to both sides of our equation. On the left side, we multiply 2.52.5 by each term inside the parentheses: 2.5Γ—βˆ’3p2.5 \times -3p and 2.5Γ—βˆ’82.5 \times -8. That gives us βˆ’7.5p-7.5p and βˆ’20-20. So, the left side becomes βˆ’7.5pβˆ’20+5p-7.5p - 20 + 5p. Now, for the right side, we do the same thing: multiply 44 by 2.25p2.25p and 44 by 5.55.5. This results in 9p9p and 2222. So, the right side transforms into 9p+22+15.59p + 22 + 15.5. After applying the distributive property, our equation now looks like this: βˆ’7.5pβˆ’20+5p=9p+22+15.5-7.5p - 20 + 5p = 9p + 22 + 15.5. See? We've already made significant progress just by tackling those parentheses. It's all about breaking down the problem into smaller, manageable steps. Don't rush; take your time with each multiplication to ensure accuracy. Mistakes here can lead to a totally wrong answer down the line, so double-check your work as you go. This initial step is crucial for setting up the rest of the solution.

Combining Like Terms: Simplifying the Equation

Now that we've conquered the distributive property, our equation is βˆ’7.5pβˆ’20+5p=9p+22+15.5-7.5p - 20 + 5p = 9p + 22 + 15.5. The next big move is to combine like terms on each side of the equals sign. Think of 'like terms' as terms that have the same variable raised to the same power, or constant terms (numbers without variables). On the left side, we have two terms with 'pp': βˆ’7.5p-7.5p and +5p+5p. When we combine them, βˆ’7.5p+5p-7.5p + 5p gives us βˆ’2.5p-2.5p. The constant term on the left is βˆ’20-20. So, the simplified left side is βˆ’2.5pβˆ’20-2.5p - 20. Now, let's move to the right side. Here, we have one term with 'pp', which is 9p9p. We also have two constant terms: 2222 and 15.515.5. Combining these constants, 22+15.522 + 15.5 equals 37.537.5. So, the simplified right side is 9p+37.59p + 37.5. After combining like terms, our equation is now a much cleaner: βˆ’2.5pβˆ’20=9p+37.5-2.5p - 20 = 9p + 37.5. This step is super important because it reduces the complexity of the equation, making it easier to isolate the variable 'pp'. It’s like tidying up your workspace before starting a big project; everything becomes clearer and more manageable. Always be careful with your signs (positive and negative) when combining terms; that’s a common spot where errors can creep in. We're getting closer to finding that value for pp!

Isolating the Variable: Bringing pp Together

We've simplified our equation to βˆ’2.5pβˆ’20=9p+37.5-2.5p - 20 = 9p + 37.5. Our goal now is to get all the terms containing 'pp' on one side of the equation and all the constant terms on the other. This process is often called isolating the variable. Let's decide to move all the 'pp' terms to the right side to keep the coefficient of 'pp' positive, which can sometimes make calculations a bit easier. To get rid of the βˆ’2.5p-2.5p on the left side, we need to add 2.5p2.5p to both sides of the equation. Remember, whatever you do to one side, you must do to the other to maintain the balance. So, βˆ’2.5pβˆ’20+2.5p=9p+37.5+2.5p-2.5p - 20 + 2.5p = 9p + 37.5 + 2.5p. This simplifies the left side to just βˆ’20-20. The right side becomes 9p+2.5p+37.59p + 2.5p + 37.5, which combines to 11.5p+37.511.5p + 37.5. So now our equation is βˆ’20=11.5p+37.5-20 = 11.5p + 37.5. Next, we need to move the constant term 37.537.5 from the right side to the left side. To do this, we subtract 37.537.5 from both sides: βˆ’20βˆ’37.5=11.5p+37.5βˆ’37.5-20 - 37.5 = 11.5p + 37.5 - 37.5. This gives us βˆ’57.5-57.5 on the left side and 11.5p11.5p on the right side. Our equation is now βˆ’57.5=11.5p-57.5 = 11.5p. This is a huge step! We've successfully gathered all the 'pp' terms on one side and all the constants on the other. It’s really satisfying to see the equation transform like this. Keep an eye on those signs and ensure you're performing the inverse operation correctly to move terms across the equals sign.

The Final Solution: Finding the Value of pp

We've reached the home stretch, guys! Our equation currently stands at βˆ’57.5=11.5p-57.5 = 11.5p. The final step to solve for pp is to isolate it completely. Currently, pp is being multiplied by 11.511.5. To undo multiplication, we use its inverse operation: division. So, we need to divide both sides of the equation by 11.511.5. This gives us βˆ’57.511.5=11.5p11.5\frac{-57.5}{11.5} = \frac{11.5p}{11.5}. Performing the division on the left side, βˆ’57.5Γ·11.5-57.5 \div 11.5, results in βˆ’5-5. On the right side, 11.5p11.5\frac{11.5p}{11.5} simplifies to just pp. Therefore, we have found our answer: p=βˆ’5p = -5. Bingo! You've successfully solved the equation. It’s always a good idea to check your answer by plugging this value of pp back into the original equation to make sure both sides are equal. Let's do that quickly. Original equation: 2.5(βˆ’3pβˆ’8)+5p=4(2.25p+5.5)+15.52.5(-3 p-8)+5 p=4(2.25 p+5.5)+15.5. Substituting p=βˆ’5p=-5: Left side: 2.5(βˆ’3(βˆ’5)βˆ’8)+5(βˆ’5)=2.5(15βˆ’8)βˆ’25=2.5(7)βˆ’25=17.5βˆ’25=βˆ’7.52.5(-3(-5)-8)+5(-5) = 2.5(15-8)-25 = 2.5(7)-25 = 17.5-25 = -7.5. Right side: 4(2.25(βˆ’5)+5.5)+15.5=4(βˆ’11.25+5.5)+15.5=4(βˆ’5.75)+15.5=βˆ’23+15.5=βˆ’7.54(2.25(-5)+5.5)+15.5 = 4(-11.25+5.5)+15.5 = 4(-5.75)+15.5 = -23+15.5 = -7.5. Since both sides equal βˆ’7.5-7.5, our solution p=βˆ’5p = -5 is correct! Mastering these steps – the distributive property, combining like terms, isolating the variable, and checking your work – will make you a formidable force in algebra. Keep practicing, and you'll solve even tougher equations in no time!