Solving The Equation 4(px + 1) = 64 Find X In Terms Of P

Introduction

Hey guys! Today, we're diving into a fun math problem that involves solving for x in an equation. This is a common type of problem you'll encounter in algebra, and mastering it will definitely boost your math skills. We're going to break down the equation 4(px + 1) = 64 step-by-step, so you can confidently find the value of x in terms of p and also determine the specific value of x when p is -5. Let's get started and make math a little less intimidating and a lot more enjoyable!

Step-by-Step Solution: Finding x in Terms of p

In this section, we'll walk through the process of isolating x in the equation 4(px + 1) = 64. Our main goal here is to express x as a function of p. This means we want to rearrange the equation so that x is on one side and an expression involving p is on the other. Ready? Let's dive in!

First, let's start with the original equation:

4(px + 1) = 64

The first thing we need to do is get rid of those parentheses. We can do this by distributing the 4 across the terms inside the parentheses. This means we multiply both px and 1 by 4:

4 * px + 4 * 1 = 64

This simplifies to:

4px + 4 = 64

Now, we want to isolate the term with x in it (which is 4px). To do this, we need to get rid of the +4 on the left side. We can do this by subtracting 4 from both sides of the equation. Remember, whatever we do to one side, we have to do to the other to keep the equation balanced:

4px + 4 - 4 = 64 - 4

This simplifies to:

4px = 60

Great! We're one step closer. Now we have 4px on the left side. To finally isolate x, we need to get rid of the 4p that's multiplying it. We can do this by dividing both sides of the equation by 4p. Again, keep the balance in mind:

(4px) / (4p) = 60 / (4p)

On the left side, the 4p in the numerator and denominator cancel out, leaving us with just x. On the right side, we can simplify the fraction by dividing both the numerator and denominator by 4:

x = 60 / (4p)
x = 15 / p

And there you have it! We've successfully solved for x in terms of p. The value of x is 15/ p. This is a crucial step, guys, because it gives us a general formula for x that works for any value of p (except p = 0, because we can't divide by zero!).

Key Takeaway: Remember the steps: distribute, isolate the x term, and then divide to solve for x. Practice makes perfect, so try working through similar problems to solidify your understanding. In this comprehensive guide, we've meticulously detailed the process of expressing x in terms of p, emphasizing the core algebraic manipulations required to isolate the variable. By distributing the constant, isolating the term containing x, and subsequently dividing by the coefficient of x, we've arrived at the expression x = 15/p. This expression not only provides a solution for x but also underscores the fundamental principles of equation solving in algebra. Understanding these principles is crucial for tackling more complex equations and problem-solving scenarios. Moreover, this exercise highlights the significance of careful algebraic manipulation and attention to detail in arriving at the correct solution. It's essential to grasp the step-by-step process to effectively apply similar techniques in future mathematical endeavors. Through practice and diligent application, mastering these skills becomes second nature, paving the way for success in advanced algebraic concepts and beyond. The ability to express variables in terms of others is a cornerstone of mathematical literacy, empowering individuals to analyze relationships between quantities and solve real-world problems with confidence and precision. Furthermore, this skill extends beyond the realm of mathematics, finding applications in various fields such as physics, engineering, and economics, where understanding the interplay of variables is paramount for modeling and predicting outcomes.

Substituting p = -5: Finding the Value of x

Now that we've found the value of x in terms of p (which is x = 15/p), the next step is to find the specific value of x when p is -5. This is a straightforward substitution problem. We'll simply replace p with -5 in our equation and then simplify. Let's see how it's done!

We start with the expression we found for x:

x = 15 / p

Now, we substitute p with -5:

x = 15 / (-5)

This is a simple division problem. 15 divided by -5 is -3:

x = -3

So, when p is -5, the value of x is -3. That wasn't so hard, was it? This shows how useful it is to first find a general expression for x; once you have that, finding specific values is just a matter of plugging in the given value for p.

Key Takeaway: Substitution is a powerful tool in algebra. Once you have an equation or expression, you can find specific values by simply replacing variables with their given values. Remember to pay attention to signs (positive and negative) when substituting and simplifying! This section demonstrates the practical application of the expression x = 15/p by substituting p = -5 to find the corresponding value of x. This step is crucial in reinforcing the understanding of how variables relate within an equation and how specific solutions can be obtained. The process of substitution involves replacing a variable with its given value and then simplifying the resulting expression. In this case, substituting p = -5 into x = 15/p yields x = 15/(-5), which simplifies to x = -3. This simple arithmetic operation demonstrates the power of algebraic expressions in solving real-world problems. By determining the value of x when p is -5, we gain a concrete understanding of the relationship between these two variables within the given equation. This practical application solidifies the theoretical concepts and enhances problem-solving skills. Moreover, the ability to substitute values into algebraic expressions is a fundamental skill in mathematics and is widely used in various fields, including science, engineering, and economics. Mastering this skill enables individuals to analyze and interpret data, make predictions, and solve complex problems with confidence and accuracy. Furthermore, it highlights the importance of careful attention to detail and precise calculation in arriving at the correct solution. In this context, understanding the rules of division with negative numbers is essential for accurate simplification and interpretation of results. Therefore, this exercise serves as a valuable learning opportunity, reinforcing both algebraic manipulation skills and arithmetic proficiency.

Practice Problems

Alright, guys, now that we've walked through the solution step-by-step, it's time to put your skills to the test! Here are a couple of practice problems that are similar to the one we just solved. Working through these will help you solidify your understanding and build confidence. Remember, the key to mastering math is practice, practice, practice!

1. Solve for y in terms of q:

   5(qy - 2) = 35
   

   Then, find the value of y when q = 3.
2. Consider the equation:

   2(ar + 3) = 24
   

   Find the value of r in terms of a, and then find the value of r when a = -2.

Tips for Solving:

• Remember to distribute first to get rid of the parentheses.
• Isolate the term with the variable you're solving for.
• Divide to get the variable by itself.
• When substituting, be careful with negative signs.

Grab a pen and paper, and give these problems a try. Don't be afraid to make mistakes – that's how we learn! If you get stuck, go back and review the steps we used in the example problem. You got this! This section provides an opportunity for readers to reinforce their understanding of the concepts discussed earlier through practice problems. By offering two similar equations to solve, readers can apply the step-by-step methods demonstrated in the previous sections. This active engagement enhances learning and retention, as individuals internalize the problem-solving strategies through hands-on experience. The practice problems are designed to challenge readers to independently apply the algebraic techniques of distribution, isolation, and division to solve for variables in terms of others. This not only reinforces the procedural aspects of equation solving but also encourages critical thinking and problem-solving skills. Furthermore, the inclusion of substitution as a subsequent step in each problem ensures that readers can translate their algebraic solutions into specific numerical answers. The tips provided offer guidance and reminders of key steps and common pitfalls, such as careful handling of negative signs. This support encourages self-reliance while minimizing potential errors. Moreover, these practice problems promote a growth mindset, emphasizing that mistakes are a natural part of the learning process. By encouraging readers to view errors as opportunities for improvement, this section fosters resilience and persistence in problem-solving. The emphasis on practice aligns with the pedagogical principle that mastery of mathematical concepts requires consistent effort and application. Therefore, this section serves as a valuable tool for self-assessment and skill development, enabling readers to build confidence and competence in algebraic problem-solving. Additionally, the structure of the problems closely mirrors real-world scenarios where mathematical equations are used to model and solve practical challenges, thus enhancing the relevance and applicability of the learning experience.

Conclusion

Great job, guys! We've tackled a tricky equation and learned how to solve for x in terms of another variable, p. We also learned how to substitute a value for p to find a specific value for x. These are fundamental skills in algebra, and you've taken a big step towards mastering them. Remember, math is like any other skill – the more you practice, the better you'll get. Keep practicing, keep asking questions, and most importantly, keep having fun with it! You're doing awesome! In conclusion, this comprehensive guide has provided a detailed walkthrough of solving for x in the equation 4(px + 1) = 64, both in terms of p and when p is equal to -5. By systematically applying algebraic techniques such as distribution, isolation, and substitution, we've demonstrated the power and elegance of mathematical problem-solving. This exercise not only reinforces core algebraic skills but also highlights the importance of conceptual understanding and procedural fluency. Throughout the guide, emphasis has been placed on the step-by-step process, encouraging readers to follow along and actively engage with the material. This approach fosters a deeper understanding of the underlying concepts and builds confidence in tackling similar problems. Moreover, the inclusion of practice problems provides readers with an opportunity to apply their knowledge and refine their skills. By working through these exercises, individuals can solidify their understanding and identify areas for further improvement. The tips and guidance offered throughout the guide aim to support learners at every stage of the problem-solving process. Furthermore, the positive and encouraging tone used throughout the guide is intended to foster a growth mindset and promote a positive attitude towards mathematics. By emphasizing the importance of practice, perseverance, and seeking help when needed, this guide encourages learners to embrace challenges and view mistakes as opportunities for growth. Ultimately, the goal of this guide is to empower readers with the knowledge and skills necessary to confidently tackle algebraic equations and problem-solving scenarios. By mastering these fundamental concepts, individuals can unlock new opportunities for learning and success in mathematics and beyond. The emphasis on both conceptual understanding and practical application ensures that learners are well-equipped to apply their mathematical skills in real-world contexts. Therefore, this guide serves as a valuable resource for students, educators, and anyone seeking to enhance their algebraic proficiency and problem-solving abilities.