Solving $5x + 34 = -2(1 - 7x)$ A Step-by-Step Guide

by THE IDEN 52 views

In this article, we will delve into the process of solving the linear equation 5x+34=βˆ’2(1βˆ’7x)5x + 34 = -2(1 - 7x). Linear equations are fundamental in mathematics and have wide applications in various fields, including physics, engineering, economics, and computer science. Mastering the techniques to solve them is crucial for anyone pursuing studies or careers in these areas. This comprehensive guide will not only provide a step-by-step solution to the given equation but also explain the underlying principles and strategies involved in solving linear equations in general. We will cover the essential algebraic manipulations, such as distribution, combining like terms, and isolating the variable, ensuring that you understand each step thoroughly. By the end of this article, you will be well-equipped to tackle similar linear equations with confidence and precision.

Understanding Linear Equations

Before we dive into the specific problem, it’s essential to understand what a linear equation is. A linear equation is an algebraic equation in which the highest power of the variable is one. It can be written in the general form:

ax+b=cax + b = c

where x is the variable, and a, b, and c are constants. The goal in solving a linear equation is to find the value of the variable x that makes the equation true. This involves a series of algebraic manipulations to isolate the variable on one side of the equation. These manipulations are based on the properties of equality, which state that if you perform the same operation on both sides of an equation, the equation remains balanced. For example, you can add, subtract, multiply, or divide both sides of the equation by the same number (except zero) without changing its solution. The strategy for solving linear equations typically involves simplifying both sides of the equation by removing parentheses and combining like terms, and then using inverse operations to isolate the variable. This systematic approach ensures that you can solve a wide variety of linear equations accurately and efficiently. Understanding the basic principles and strategies is crucial for tackling more complex problems in algebra and other areas of mathematics. Linear equations serve as the building blocks for more advanced concepts, making their mastery essential for anyone pursuing further studies in mathematics or related fields.

Step 1: Distribute the -2

Our first step in solving the equation 5x+34=βˆ’2(1βˆ’7x)5x + 34 = -2(1 - 7x) is to distribute the -2 on the right side of the equation. Distribution is a fundamental algebraic technique used to remove parentheses by multiplying a term outside the parentheses by each term inside the parentheses. In this case, we need to multiply -2 by both 1 and -7x. This process involves applying the distributive property, which states that for any numbers a, b, and c:

a(b + c) = ab + ac

Applying this property to our equation, we get:

-2 * 1 = -2

and

-2 * -7x = 14x

So, -2(1 - 7x) becomes -2 + 14x. Now, we substitute this back into our original equation, which gives us:

5x+34=βˆ’2+14x5x + 34 = -2 + 14x

This step is crucial because it simplifies the equation by removing the parentheses, making it easier to manipulate and solve. The distribution process ensures that we are correctly accounting for the terms inside the parentheses and their relationship to the term outside. By performing this step accurately, we set the stage for the subsequent steps in solving the equation. This foundational technique is essential for solving various algebraic equations and is a cornerstone of mathematical problem-solving. Mastering distribution allows for the simplification of complex expressions and is a critical skill for success in algebra and beyond. With the parentheses removed, we can now proceed to the next step, which involves combining like terms and isolating the variable.

Step 2: Rearrange the Equation

Now that we have distributed the -2, our equation is 5x+34=βˆ’2+14x5x + 34 = -2 + 14x. The next step is to rearrange the equation so that the terms with x are on one side and the constants are on the other. This is a crucial step in isolating the variable x and ultimately solving for its value. To do this, we need to perform inverse operations to move the terms across the equals sign. Remember, whatever operation we perform on one side of the equation, we must also perform on the other side to maintain the balance. Let's start by moving the 5x term from the left side to the right side. To do this, we subtract 5x from both sides of the equation:

5x+34βˆ’5x=βˆ’2+14xβˆ’5x5x + 34 - 5x = -2 + 14x - 5x

This simplifies to:

34=βˆ’2+9x34 = -2 + 9x

Next, we need to move the constant term -2 from the right side to the left side. To do this, we add 2 to both sides of the equation:

34+2=βˆ’2+9x+234 + 2 = -2 + 9x + 2

This simplifies to:

36=9x36 = 9x

By rearranging the equation in this way, we have successfully grouped the variable term (9x) on one side and the constant term (36) on the other. This makes the next step, which is to isolate the variable, much more straightforward. The process of rearranging equations using inverse operations is a fundamental skill in algebra and is essential for solving a wide variety of equations. It allows us to manipulate the equation into a form where the variable can be easily isolated and its value determined. With the equation rearranged, we are now ready to isolate x and find the solution.

Step 3: Isolate the Variable

After rearranging the equation, we have 36=9x36 = 9x. To isolate the variable x, we need to get x by itself on one side of the equation. Currently, x is being multiplied by 9. To undo this multiplication, we perform the inverse operation, which is division. We divide both sides of the equation by 9:

rac{36}{9} = rac{9x}{9}

This simplifies to:

4=x4 = x

So, we have found that x = 4. This means that the value of x that makes the original equation true is 4. Isolating the variable is a key step in solving any algebraic equation. It involves using inverse operations to undo any operations that are being performed on the variable, such as addition, subtraction, multiplication, or division. By performing these operations on both sides of the equation, we maintain the balance and ensure that we are finding the correct solution. In this case, dividing both sides by 9 was the final step needed to isolate x and determine its value. This process is fundamental to solving equations and is a skill that is used extensively in algebra and other areas of mathematics. With x now isolated, we have successfully solved the equation. However, it is always a good practice to check our solution to ensure that it is correct. The next step will demonstrate how to check our solution by substituting it back into the original equation.

Step 4: Check the Solution

To ensure our solution is correct, it’s crucial to check the solution by substituting the value we found for x back into the original equation. This step helps us verify that our calculations are accurate and that the value we obtained for x indeed satisfies the equation. Our original equation was:

5x+34=βˆ’2(1βˆ’7x)5x + 34 = -2(1 - 7x)

We found that x = 4. Now, we substitute 4 for x in the equation:

5(4)+34=βˆ’2(1βˆ’7(4))5(4) + 34 = -2(1 - 7(4))

First, let's simplify the left side of the equation:

5(4)=205(4) = 20

So, the left side becomes:

20+34=5420 + 34 = 54

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

7(4)=287(4) = 28

So, the expression inside the parentheses becomes:

1βˆ’28=βˆ’271 - 28 = -27

Now, multiply -2 by -27:

-2(-27) = 54

So, the right side of the equation is also 54. Comparing both sides:

54=5454 = 54

Since both sides of the equation are equal, our solution x = 4 is correct. Checking the solution is an essential step in the problem-solving process. It not only confirms the accuracy of our calculations but also helps us catch any errors that might have occurred during the process. By substituting the solution back into the original equation, we can verify that it satisfies the equation and that we have indeed found the correct value for the variable. This practice builds confidence in our problem-solving abilities and ensures that we are providing accurate solutions. In this case, our check confirmed that x = 4 is the correct solution. This final step completes the solution process and provides assurance that we have solved the equation accurately.

Conclusion

In conclusion, we have successfully solved the linear equation 5x+34=βˆ’2(1βˆ’7x)5x + 34 = -2(1 - 7x) by following a step-by-step approach. We began by understanding the basic principles of linear equations and the importance of maintaining balance while performing algebraic manipulations. The first step involved distributing the -2 on the right side of the equation to remove the parentheses, which simplified the equation and made it easier to work with. Next, we rearranged the equation by moving the terms with x to one side and the constants to the other, using inverse operations to maintain the equation's balance. This step was crucial in isolating the variable x. Then, we isolated the variable by dividing both sides of the equation by the coefficient of x, which gave us the solution x = 4. Finally, we checked our solution by substituting x = 4 back into the original equation, verifying that both sides of the equation were equal and confirming the accuracy of our solution. This process demonstrates the importance of each step in solving linear equations and the need for careful and systematic algebraic manipulations. By mastering these techniques, you can confidently solve a wide range of linear equations. Linear equations are a fundamental concept in mathematics, and their understanding is essential for success in various fields. This comprehensive guide has provided you with the knowledge and skills to tackle such problems effectively. Remember to practice regularly to reinforce your understanding and improve your problem-solving abilities. With consistent effort, you will become proficient in solving linear equations and be well-prepared for more advanced mathematical concepts.