Solving Linear Equations: A Step-by-Step Guide

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Let's dive into solving the linear equation 5x8βˆ’3x2=βˆ’2816\frac{5 x}{8}-\frac{3 x}{2}=\frac{-28}{16}. Linear equations might seem daunting at first, but with a systematic approach, they become quite manageable. This guide will walk you through each step, ensuring you understand the underlying principles and can confidently tackle similar problems in the future. Our goal is to isolate the variable x on one side of the equation to find its value. We'll achieve this by performing operations on both sides of the equation to maintain balance and simplify the expression. Understanding these fundamental concepts is crucial not only for solving this particular equation but also for building a solid foundation in algebra. So, let's begin this mathematical journey together, breaking down each step and illuminating the path to the solution. Remember, practice makes perfect, so don't hesitate to try out similar problems on your own to reinforce your understanding. By the end of this guide, you'll have a clear grasp of how to solve linear equations effectively.

Step 1: Simplify the Fractions

To begin, let’s simplify the fractions in the equation 5x8βˆ’3x2=βˆ’2816\frac{5 x}{8}-\frac{3 x}{2}=\frac{-28}{16}. Simplifying fractions involves reducing them to their lowest terms. This makes the equation easier to work with and reduces the chances of making errors in subsequent steps. In our equation, we can simplify the fraction on the right-hand side. The fraction βˆ’2816\frac{-28}{16} can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 4. Doing so, we get βˆ’28Γ·416Γ·4=βˆ’74\frac{-28 Γ· 4}{16 Γ· 4} = \frac{-7}{4}. Therefore, our equation now becomes 5x8βˆ’3x2=βˆ’74\frac{5 x}{8}-\frac{3 x}{2}=\frac{-7}{4}. Simplifying fractions is a fundamental skill in algebra, and it's always a good practice to simplify any fraction before proceeding with other operations. This not only makes the equation easier to manage but also helps in avoiding potential calculation errors. Remember to always look for opportunities to simplify fractions throughout the problem-solving process. This step lays the groundwork for the subsequent steps and ensures that the equation is in its simplest form, making it easier to solve. By simplifying the fractions early on, we set ourselves up for a smoother and more efficient problem-solving experience. This initial simplification is a crucial step in mastering the art of solving linear equations.

Step 2: Find a Common Denominator

Now, let's find a common denominator for the fractions on the left side of the equation: 5x8βˆ’3x2=βˆ’74\frac{5 x}{8}-\frac{3 x}{2}=\frac{-7}{4}. To combine these fractions, they need to have the same denominator. The least common multiple (LCM) of 8 and 2 is 8. So, we'll convert the fraction 3x2\frac{3 x}{2} to an equivalent fraction with a denominator of 8. To do this, we multiply both the numerator and the denominator of 3x2\frac{3 x}{2} by 4: 3xβˆ—42βˆ—4=12x8\frac{3 x * 4}{2 * 4} = \frac{12 x}{8}. Now our equation looks like this: 5x8βˆ’12x8=βˆ’74\frac{5 x}{8}-\frac{12 x}{8}=\frac{-7}{4}. Finding a common denominator is a critical step in adding or subtracting fractions. It ensures that we are working with comparable units, allowing us to combine the numerators correctly. The LCM is the smallest number that is a multiple of both denominators, which helps keep the numbers manageable. In this case, 8 is the LCM of 8 and 2, making it the ideal common denominator. By converting 3x2\frac{3 x}{2} to 12x8\frac{12 x}{8}, we can now easily combine the fractions on the left side of the equation. This process of finding a common denominator is a fundamental skill in algebra and is essential for solving equations involving fractions. By mastering this step, you'll be well-equipped to tackle more complex algebraic problems. Remember, the key is to find the LCM and then convert each fraction to an equivalent fraction with that denominator.

Step 3: Combine Like Terms

With a common denominator in place, we can now combine the like terms on the left side of the equation: 5x8βˆ’12x8=βˆ’74\frac{5 x}{8}-\frac{12 x}{8}=\frac{-7}{4}. Combining the terms involves subtracting the numerators while keeping the common denominator. So, 5xβˆ’12x=βˆ’7x5x - 12x = -7x. Therefore, the left side of the equation becomes βˆ’7x8\frac{-7 x}{8}. Now our equation is: βˆ’7x8=βˆ’74\frac{-7 x}{8}=\frac{-7}{4}. Combining like terms is a fundamental step in simplifying algebraic expressions and equations. It allows us to consolidate similar terms into a single term, making the equation easier to solve. In this case, we combined the terms involving x on the left side of the equation, resulting in a single term, βˆ’7x8\frac{-7 x}{8}. This simplification is crucial for isolating the variable x and finding its value. By combining like terms, we reduce the complexity of the equation and make it more manageable. This skill is essential for solving a wide range of algebraic problems and is a key component of algebraic manipulation. Remember to always look for opportunities to combine like terms to simplify expressions and equations. This step not only makes the equation easier to solve but also helps in avoiding potential errors. By mastering this technique, you'll be well-equipped to tackle more complex algebraic challenges. The ability to combine like terms is a cornerstone of algebraic proficiency.

Step 4: Isolate the Variable

Next, we want to isolate the variable x in the equation βˆ’7x8=βˆ’74\frac{-7 x}{8}=\frac{-7}{4}. To do this, we need to get x by itself on one side of the equation. Since x is being multiplied by βˆ’78\frac{-7}{8}, we can undo this operation by multiplying both sides of the equation by the reciprocal of βˆ’78\frac{-7}{8}, which is βˆ’87\frac{-8}{7}. Multiplying both sides by βˆ’87\frac{-8}{7}, we get: βˆ’7x8βˆ—βˆ’87=βˆ’74βˆ—βˆ’87\frac{-7 x}{8} * \frac{-8}{7} = \frac{-7}{4} * \frac{-8}{7}. On the left side, the fractions cancel out, leaving us with just x. On the right side, we have βˆ’74βˆ—βˆ’87=5628\frac{-7}{4} * \frac{-8}{7} = \frac{56}{28}. Simplifying the fraction 5628\frac{56}{28}, we get 2. Therefore, x=2x = 2. Isolating the variable is the ultimate goal in solving any equation. It involves performing operations on both sides of the equation to get the variable by itself on one side. In this case, we isolated x by multiplying both sides of the equation by the reciprocal of the coefficient of x. This process effectively cancels out the coefficient, leaving x by itself. This step is crucial for finding the value of the variable and solving the equation. By mastering the technique of isolating the variable, you'll be able to solve a wide range of algebraic equations. Remember to always perform the same operation on both sides of the equation to maintain balance and ensure that the equation remains true. This skill is essential for algebraic manipulation and is a key component of mathematical problem-solving.

Step 5: Verify the Solution

Finally, to ensure that our solution is correct, we should verify it by substituting the value of x back into the original equation: 5x8βˆ’3x2=βˆ’2816\frac{5 x}{8}-\frac{3 x}{2}=\frac{-28}{16}. We found that x=2x = 2, so we substitute 2 for x in the original equation: 5(2)8βˆ’3(2)2=βˆ’2816\frac{5 (2)}{8}-\frac{3 (2)}{2}=\frac{-28}{16}. Simplifying, we get: 108βˆ’62=βˆ’2816\frac{10}{8}-\frac{6}{2}=\frac{-28}{16}. Now, let's simplify the fractions: 54βˆ’3=βˆ’74\frac{5}{4} - 3 = \frac{-7}{4}. Converting 3 to a fraction with a denominator of 4, we get: 54βˆ’124=βˆ’74\frac{5}{4} - \frac{12}{4} = \frac{-7}{4}. Combining the fractions on the left side, we have: βˆ’74=βˆ’74\frac{-7}{4} = \frac{-7}{4}. Since both sides of the equation are equal, our solution is correct. Verifying the solution is a crucial step in the problem-solving process. It ensures that the value we found for the variable satisfies the original equation. By substituting the solution back into the equation, we can check whether both sides of the equation are equal. If they are, then our solution is correct. This step helps in avoiding errors and ensures that we have found the correct value for the variable. Remember to always verify your solution to ensure accuracy. This practice not only helps in avoiding errors but also reinforces your understanding of the problem-solving process. By mastering the technique of verifying the solution, you'll be able to confidently solve algebraic equations and ensure that your answers are correct.

Conclusion

In conclusion, solving the equation 5x8βˆ’3x2=βˆ’2816\frac{5 x}{8}-\frac{3 x}{2}=\frac{-28}{16} involves simplifying fractions, finding a common denominator, combining like terms, isolating the variable, and verifying the solution. By following these steps carefully, we found that x=2x = 2. Remember to always double-check your work and practice solving similar equations to reinforce your understanding. Linear equations are a fundamental part of algebra, and mastering them will set you up for success in more advanced math courses. Keep practicing, and you'll become more confident in your ability to solve these types of problems.

For additional resources and practice problems, you can visit Khan Academy's Algebra I section on linear equations: Khan Academy Algebra I