Solving The Equation 5u + 3 = -2 A Step By Step Guide

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Introduction

In this article, we will delve into the process of solving the equation 5u+3=βˆ’25u + 3 = -2, where uu represents a real number. This is a fundamental algebraic problem that involves isolating the variable uu to find its value. We will break down each step in detail, ensuring clarity and understanding for readers of all backgrounds. Our goal is to provide a comprehensive guide that not only solves this specific equation but also equips you with the skills to tackle similar problems with confidence. Solving algebraic equations is a cornerstone of mathematics, and mastering these techniques opens doors to more advanced concepts. We'll explore the underlying principles, common pitfalls, and strategies for verifying your solutions. So, let's embark on this mathematical journey together and unravel the mystery of this equation. This article aims to be more than just a solution; it's a learning experience that enhances your problem-solving abilities and mathematical intuition. We will also discuss the importance of simplifying your answer as much as possible, a crucial step in ensuring your solution is presented in its most concise and understandable form. Whether you're a student learning algebra for the first time or someone looking to refresh your skills, this guide will provide you with the knowledge and tools you need to succeed. By the end of this article, you will not only know the solution to 5u+3=βˆ’25u + 3 = -2 but also understand the why behind each step, fostering a deeper appreciation for the elegance and power of algebra. Remember, mathematics is not just about finding answers; it's about the journey of discovery and the development of logical reasoning.

Step-by-Step Solution

To solve the equation 5u+3=βˆ’25u + 3 = -2, our primary objective is to isolate the variable uu on one side of the equation. This involves performing a series of algebraic manipulations that maintain the equality while simplifying the expression. Let's break down the solution into a step-by-step process:

Step 1: Isolate the term with the variable

The first step in solving for uu is to isolate the term that contains the variable, which in this case is 5u5u. To do this, we need to eliminate the constant term, which is +3+3, from the left side of the equation. We can achieve this by subtracting 33 from both sides of the equation. Remember, the golden rule of algebra is that whatever operation you perform on one side of the equation, you must perform the same operation on the other side to maintain balance. This ensures that the equality remains valid throughout the solution process. Subtracting 33 from both sides gives us:

5u+3βˆ’3=βˆ’2βˆ’35u + 3 - 3 = -2 - 3

This simplifies to:

5u=βˆ’55u = -5

Now, we have successfully isolated the term with the variable, 5u5u. This is a crucial step because it brings us closer to our goal of finding the value of uu. By eliminating the constant term, we have simplified the equation and made it easier to work with. This step highlights the importance of inverse operations in solving equations. Subtraction is the inverse operation of addition, and by subtracting 33, we effectively "undid" the addition of 33 on the left side of the equation. This principle of using inverse operations is fundamental to solving a wide range of algebraic equations. As we move on to the next step, we will continue to apply this principle to further isolate the variable uu. Remember, each step we take is a deliberate attempt to simplify the equation and bring us closer to the solution.

Step 2: Solve for the variable

Now that we have isolated the term 5u5u and our equation is simplified to 5u=βˆ’55u = -5, the next step is to solve for the variable uu itself. To do this, we need to get uu by itself on one side of the equation. Currently, uu is being multiplied by 55. To isolate uu, we need to perform the inverse operation of multiplication, which is division. We will divide both sides of the equation by 55. This will effectively "undo" the multiplication and leave us with uu on the left side. Dividing both sides by 55 gives us:

rac{5u}{5} = rac{-5}{5}

This simplifies to:

u=βˆ’1u = -1

Therefore, we have found the solution to the equation. The value of uu that satisfies the equation 5u+3=βˆ’25u + 3 = -2 is βˆ’1-1. This step demonstrates the power of using inverse operations to isolate variables and solve equations. By dividing both sides by the coefficient of uu, we were able to effectively "undo" the multiplication and reveal the value of uu. This principle is a cornerstone of algebraic manipulation and is applicable to a wide variety of equations. It's important to remember that whatever operation you perform on one side of the equation, you must perform the same operation on the other side to maintain balance and ensure the equality remains valid. Now that we have found the solution, it's always a good practice to verify it to ensure accuracy. We will do this in the next step.

Verification of the Solution

After finding a solution to an equation, it's crucial to verify its correctness. This step ensures that the value we obtained for the variable actually satisfies the original equation. To verify our solution u=βˆ’1u = -1 for the equation 5u+3=βˆ’25u + 3 = -2, we will substitute βˆ’1-1 for uu in the original equation and check if the left-hand side (LHS) equals the right-hand side (RHS). This process involves replacing every instance of the variable uu in the equation with the value we found, which is βˆ’1-1. Then, we will simplify both sides of the equation and compare the results. If the LHS and RHS are equal, then our solution is correct. If they are not equal, it indicates that we may have made an error in our solution process and need to re-examine our steps. Verification is not just about checking our work; it's about building confidence in our mathematical abilities and ensuring the accuracy of our results. It's a fundamental aspect of problem-solving that promotes carefulness and attention to detail. So, let's proceed with the verification process and confirm whether our solution is indeed correct. This step is a valuable opportunity to reinforce our understanding of the equation and the solution we have found.

Substituting the value of u

To verify the solution, substitute u=βˆ’1u = -1 into the original equation:

5(βˆ’1)+3=βˆ’25(-1) + 3 = -2

This substitution replaces the variable uu with its numerical value, allowing us to evaluate the expression and determine if it satisfies the equation. This step is a direct application of the concept of substitution, a fundamental technique in algebra and other branches of mathematics. By substituting the value, we transform the equation from an abstract statement about a variable into a concrete numerical calculation. This allows us to directly compare the two sides of the equation and see if they are equal. The substitution step is a critical bridge between the symbolic representation of the equation and the numerical reality of the solution. It's a powerful tool for verifying solutions and understanding the relationship between variables and their values. Now that we have substituted the value of uu, the next step is to simplify the expression and determine if the equation holds true. This will involve performing the arithmetic operations in the correct order and comparing the results on both sides of the equation.

Simplifying the equation

Now, let's simplify the left-hand side (LHS) of the equation:

5(βˆ’1)+3=βˆ’5+35(-1) + 3 = -5 + 3

βˆ’5+3=βˆ’2-5 + 3 = -2

So, the left-hand side simplifies to βˆ’2-2, which is equal to the right-hand side (RHS) of the equation, which is also βˆ’2-2. This simplification process involves performing the arithmetic operations in the correct order, following the rules of precedence. In this case, we first performed the multiplication, 5(βˆ’1)5(-1), which resulted in βˆ’5-5. Then, we added 33 to βˆ’5-5, which gave us βˆ’2-2. This step demonstrates the importance of following the order of operations to ensure accurate calculations. The simplification process transforms the expression into its simplest form, making it easier to compare with the right-hand side of the equation. The fact that the left-hand side simplified to βˆ’2-2, which is exactly the same as the right-hand side, provides strong evidence that our solution is correct. This verification step is a crucial confirmation that our solution satisfies the original equation and that we have solved the problem accurately. Now that we have verified the solution, we can confidently state that u=βˆ’1u = -1 is the correct answer.

Conclusion of verification

Since the left-hand side (LHS) equals the right-hand side (RHS), our solution u=βˆ’1u = -1 is correct.

Final Answer

Therefore, the solution to the equation 5u+3=βˆ’25u + 3 = -2 is:

u=βˆ’1u = -1

This is the final answer, and we have successfully solved the equation. We have not only found the value of uu that satisfies the equation but also verified its correctness through a rigorous substitution and simplification process. This journey through solving the equation has reinforced the importance of algebraic manipulation, inverse operations, and the order of operations. We have also highlighted the crucial role of verification in ensuring the accuracy of our solutions. Solving algebraic equations is a fundamental skill in mathematics, and mastering these techniques opens doors to more advanced concepts. The process we have followed in this article provides a framework for tackling similar problems with confidence and clarity. Remember, mathematics is not just about finding the answer; it's about the process of problem-solving and the development of logical reasoning. By understanding the steps involved and the principles behind them, you can approach mathematical challenges with greater understanding and success. We hope this comprehensive guide has been helpful in your learning journey.

Discussion of Mathematical Category

The category of this problem falls under the broad umbrella of algebra, which is a fundamental branch of mathematics dealing with symbols and the rules for manipulating those symbols. More specifically, this problem is an example of solving a linear equation in one variable. Linear equations are characterized by having a highest power of the variable equal to 1. In this case, the equation 5u+3=βˆ’25u + 3 = -2 involves the variable uu raised to the power of 1, making it a linear equation. Solving linear equations is a cornerstone of algebra and a crucial skill for tackling more advanced mathematical concepts. These equations appear in various contexts, from simple word problems to complex scientific models. Understanding how to solve them is essential for anyone pursuing studies in mathematics, science, engineering, or related fields. The techniques used to solve this equation, such as isolating the variable and using inverse operations, are applicable to a wide range of algebraic problems. Linear equations form the basis for more complex equations and systems of equations, making their mastery a key building block in mathematical proficiency. Furthermore, the concepts of equality, balance, and performing operations on both sides of an equation are fundamental principles that extend beyond linear equations and into other areas of mathematics. The ability to solve linear equations efficiently and accurately is a valuable skill that empowers individuals to tackle real-world problems and pursue further mathematical studies. In essence, this problem serves as a gateway to understanding more complex algebraic concepts and their applications.