Solving A System Of Equations: Step-by-Step Guide

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Hey guys! Today, we're diving into the world of algebra to tackle a common problem: solving a system of equations. We'll specifically be looking at the system:

  • 10x – 7y = 213
  • 8x + 9y = 201

Don't worry, it might look intimidating, but we'll break it down into manageable steps. We’re going to explore this topic in depth, making sure everyone understands how to approach and solve these types of problems. So grab your pencils, and let's get started!

Understanding Systems of Equations

Before we jump into solving, let’s quickly recap what a system of equations actually is. Basically, it's a set of two or more equations that share the same variables. Our goal is to find the values for those variables that make all the equations true simultaneously. In this case, we need to find the values for x and y that satisfy both 10x – 7y = 213 and 8x + 9y = 201.

There are several methods to solve systems of equations, but we're going to focus on two popular ones: the substitution method and the elimination method. Both have their strengths, and knowing both will make you a system-solving pro! Let's start by discussing the substitution method, which involves isolating one variable in one equation and substituting that expression into the other equation. This effectively reduces the system to a single equation with one variable, making it solvable. The elimination method, on the other hand, involves manipulating the equations so that when they are added or subtracted, one of the variables cancels out. This also results in a single equation with one variable, which can then be solved. Choosing the right method often depends on the specific structure of the equations, and with practice, you'll get a feel for which method is most efficient in different situations. Understanding the underlying principles and practicing both methods will give you a comprehensive toolkit for tackling systems of equations. So, whether you're facing a simple system or a more complex one, you'll be well-equipped to find the solutions.

Method 1: The Substitution Method

The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This eliminates one variable, leaving us with a single equation that we can solve. Let’s walk through this method step by step using our system:

Step 1: Solve one equation for one variable

Let's pick the first equation, 10x – 7y = 213, and solve it for x. We could solve for y instead, but solving for x seems a bit easier in this case.

  • 10x – 7y = 213
  • 10x = 213 + 7y
  • x = (213 + 7y) / 10

So, we've isolated x. This expression, x = (213 + 7y) / 10, tells us how x is related to y in the first equation. This is a crucial step in the substitution method because it allows us to express one variable in terms of the other, paving the way for substitution into the second equation. We could have chosen to solve for y instead, but by opting to solve for x, we’ve aimed for simplicity in this particular scenario. The flexibility to choose which variable to solve for is one of the advantages of the substitution method. It allows us to strategically select the variable that will result in the least complicated algebraic manipulations. Keep in mind, the goal is to make the subsequent steps as straightforward as possible. Therefore, practice recognizing which variable to isolate in which equation will make you more efficient in solving systems of equations. It’s all about choosing the path of least resistance and setting ourselves up for success in the later steps.

Step 2: Substitute into the other equation

Now we take our expression for x and substitute it into the other equation (the one we haven't used yet), which is 8x + 9y = 201.

  • 8((213 + 7y) / 10) + 9y = 201

See what we did there? We replaced the x in the second equation with the entire expression we found in step 1. This gives us a new equation that only has one variable, y. This substitution is the heart of the method, as it reduces the two-variable problem into a single-variable problem. By replacing x with its equivalent expression in terms of y, we’ve created an equation that we can solve directly for y. This is a significant step forward because once we find the value of y, we can easily substitute it back into one of the original equations (or the expression we found in step 1) to find the value of x. The power of substitution lies in its ability to simplify complex problems. By strategically eliminating one variable, we open the door to a straightforward algebraic solution. This method is particularly effective when one of the equations can be easily solved for one variable, making the substitution process less cumbersome. So, the next time you encounter a system of equations, remember the power of substitution and how it can transform a seemingly daunting problem into a manageable one.

Step 3: Solve for y

Now, let's solve this equation for y:

  • 8(213 + 7y) / 10 + 9y = 201
  • (1704 + 56y) / 10 + 9y = 201
  • 1704 + 56y + 90y = 2010 (Multiply both sides by 10)
  • 146y = 306
  • y = 306 / 146
  • y = 153 / 73

So, we've found our y value! This step involves algebraic manipulation to isolate y. We started by distributing the 8 in the numerator and then multiplied the entire equation by 10 to eliminate the fraction. This is a common technique to clear denominators and simplify equations. Next, we combined like terms, specifically the y terms, and then isolated the y variable by subtracting the constant term from both sides. Finally, we divided by the coefficient of y to solve for y. Each of these steps is a fundamental algebraic operation, and mastering them is crucial for solving equations effectively. The value we obtained for y, 153/73, might look a bit unusual, but it's a perfectly valid solution. It highlights the importance of being comfortable working with fractions and not being intimidated by non-integer solutions. In many real-world applications, solutions aren't always neat whole numbers, so it’s essential to be prepared to handle fractional or decimal answers. Now that we have the value of y, the next step is to use it to find the value of x, which will complete our solution to the system of equations.

Step 4: Substitute y back to find x

We now know the value of y. To find x, we can substitute this value back into either of the original equations or the expression we found in step 1. Let's use the expression we found in step 1, as it's already solved for x:

  • x = (213 + 7y) / 10
  • x = (213 + 7(153 / 73)) / 10
  • x = (213 + 1071 / 73) / 10
  • x = (15549 / 73 + 1071 / 73) / 10
  • x = (16620 / 73) / 10
  • x = 1662 / 73

So, we've found our x value as well! Substituting the value of y back into the equation to find x is a critical step in solving systems of equations. It completes the process and gives us the full solution. We chose to use the expression we found in step 1 because it was already solved for x, making the calculation more straightforward. However, we could have also substituted the value of y into either of the original equations, and we would have arrived at the same value for x. This flexibility is a key feature of the substitution method. The arithmetic in this step might seem a bit involved, especially with the fractions, but it's essential to work through it carefully. Each step, from multiplying and adding fractions to dividing, is a fundamental operation that builds towards the final answer. The resulting value for x, 1662/73, is another fraction, reinforcing the idea that solutions to systems of equations aren't always integers. The important thing is that we followed the correct procedure and arrived at the accurate value for x. Now that we have both x and y, we have successfully solved the system of equations using the substitution method.

Step 5: Verify the solution

It's always a good idea to check your solution by plugging the x and y values back into both original equations to make sure they hold true. Let's do that:

  • 10x – 7y = 213
  • 10(1662 / 73) – 7(153 / 73) = 213
  • 16620 / 73 – 1071 / 73 = 213
  • 15549 / 73 = 213
  • 213 = 213 (This checks out!)
  • 8x + 9y = 201
  • 8(1662 / 73) + 9(153 / 73) = 201
  • 13296 / 73 + 1377 / 73 = 201
  • 14673 / 73 = 201
  • 201 = 201 (This checks out too!)

Our solution is correct! This verification step is a crucial safeguard against errors. It involves plugging the values we found for x and y back into the original equations to ensure they satisfy both equations. This process helps to catch any mistakes we might have made in our calculations. If the values don't satisfy both equations, it indicates that there's an error somewhere in our solution, and we need to go back and check our work. In our case, when we plugged in the values x = 1662/73 and y = 153/73 into both original equations, they both held true. This gives us confidence that our solution is correct. Verification is not just a formality; it's an integral part of the problem-solving process. It reinforces the accuracy of our work and ensures that we're presenting the correct solution. So, always remember to verify your solutions, especially in exams or when dealing with real-world applications where accuracy is paramount.

Method 2: The Elimination Method

The elimination method, also known as the addition method, involves manipulating the equations so that when you add or subtract them, one of the variables cancels out. This leaves you with a single equation in one variable, which is much easier to solve.

Step 1: Multiply equations to match coefficients

The goal here is to make the coefficients of either x or y the same (but with opposite signs) in both equations. Let's eliminate x. To do this, we'll multiply the first equation by 8 and the second equation by -10:

  • (10x – 7y = 213) * 8 -> 80x - 56y = 1704
  • (8x + 9y = 201) * -10 -> -80x - 90y = -2010

Notice that now the x coefficients are 80 and -80. This step is crucial in the elimination method because it sets up the equations for the next step, where we will eliminate one of the variables by adding the equations together. We chose to eliminate x in this case, but we could have also chosen to eliminate y. The decision often depends on which variable will result in the simplest calculations. To match the coefficients, we multiplied each equation by a constant. The key is to choose constants that will make the coefficients of one of the variables equal in magnitude but opposite in sign. This ensures that when we add the equations, that variable will cancel out. In our example, multiplying the first equation by 8 and the second equation by -10 achieved this goal for the x variable. This step requires careful attention to detail to avoid errors in multiplication. Once the coefficients are matched, we are ready to move on to the next step, where the magic of elimination happens.

Step 2: Add the equations

Now we add the two modified equations together:

  • 80x - 56y = 1704
  • -80x - 90y = -2010
  • -146y = -306

The x terms cancel out, leaving us with an equation in just y. This is the core idea behind the elimination method: to strategically eliminate one variable so we can solve for the other. By adding the equations, we've effectively reduced the system to a single equation with one unknown. This significantly simplifies the problem. The cancellation of the x terms is a direct result of our careful manipulation in the previous step, where we made the coefficients of x equal in magnitude but opposite in sign. The resulting equation, -146y = -306, is now straightforward to solve for y. This step highlights the efficiency of the elimination method, especially when dealing with systems where the coefficients can be easily manipulated. It’s a powerful technique for simplifying complex problems and making them more manageable. The key is to recognize when the elimination method is the most efficient approach and to execute the steps accurately. With practice, you'll become adept at identifying opportunities to use this method and streamlining your problem-solving process.

Step 3: Solve for y

Let's solve for y:

  • -146y = -306
  • y = -306 / -146
  • y = 153 / 73

We got the same y value as before, which is a good sign! Solving for y in this step is a straightforward algebraic process. We simply divided both sides of the equation by the coefficient of y, which is -146. The negative signs cancel out, giving us a positive value for y. The result, y = 153/73, is the same value we obtained when using the substitution method. This consistency is a strong indication that our calculations are correct. It's reassuring to see that different methods lead to the same solution, reinforcing the accuracy of our work. This step underscores the importance of basic algebraic skills, such as dividing and simplifying fractions. These skills are fundamental to solving equations and are used extensively in various mathematical contexts. The fact that we arrived at the same y value using two different methods highlights the reliability of these methods and the consistency of mathematical principles. Now that we have the value of y, the next step is to substitute it back into one of the original equations to find the value of x.

Step 4: Substitute y back to find x

Now we substitute the value of y back into either of the original equations. Let's use the second equation, 8x + 9y = 201:

  • 8x + 9(153 / 73) = 201
  • 8x + 1377 / 73 = 201
  • 8x = 201 - 1377 / 73
  • 8x = 14673 / 73 - 1377 / 73
  • 8x = 13296 / 73
  • x = 13296 / (73 * 8)
  • x = 1662 / 73

Again, we found the same x value as before! This step is the reverse of the elimination process. We take the value we found for y and plug it back into one of the original equations to solve for x. We chose the second equation in this case, but we could have also used the first equation. The key is to substitute the value of y accurately and then solve the resulting equation for x. The calculations in this step involve fractions, so it's essential to be careful with the arithmetic. We first multiplied 9 by 153/73, then subtracted the result from 201. Next, we divided both sides of the equation by 8 to isolate x. The final result, x = 1662/73, matches the value we obtained using the substitution method. This consistency is a strong indicator that our solution is correct. It reinforces the reliability of both methods and the accuracy of our calculations. Now that we have both x and y, we have successfully solved the system of equations using the elimination method. As a final check, it's always a good idea to verify our solution by plugging the values back into the original equations.

Step 5: Verify the solution

Just like before, let's verify our solution:

  • 10x – 7y = 213
  • 10(1662 / 73) – 7(153 / 73) = 213
  • 16620 / 73 – 1071 / 73 = 213
  • 15549 / 73 = 213
  • 213 = 213 (It checks out!)
  • 8x + 9y = 201
  • 8(1662 / 73) + 9(153 / 73) = 201
  • 13296 / 73 + 1377 / 73 = 201
  • 14673 / 73 = 201
  • 201 = 201 (It checks out too!)

Yep, our solution is correct! This verification step is a critical part of the problem-solving process. It ensures that the values we found for x and y actually satisfy both original equations. By plugging the values back in, we can confirm that our calculations are accurate and that we have indeed found the correct solution. In our case, when we substituted x = 1662/73 and y = 153/73 into both equations, they both held true. This gives us confidence that our solution is correct. Verification is especially important in situations where a mistake could have significant consequences, such as in engineering or finance. It's also a good habit to develop for academic work, as it can help you catch errors before submitting your work. The process of verification reinforces the understanding of the problem and the solution. It's not just about getting the right answer; it's about understanding why the answer is correct. So, always remember to verify your solutions, and you'll be well on your way to becoming a proficient problem solver.

Final Solution

So, the solution to the system of equations is:

  • x = 1662 / 73
  • y = 153 / 73

We successfully solved the system using both the substitution and elimination methods. Both methods led us to the same solution, which reinforces the accuracy of our work. This demonstrates the versatility of algebraic techniques and how different approaches can converge on the same result. The values x = 1662/73 and y = 153/73 represent the point of intersection of the two lines represented by the equations. This is a fundamental concept in algebra and has applications in various fields, including graphing and linear programming. Understanding how to solve systems of equations is a valuable skill that can be applied to many real-world problems. Whether you're balancing a budget, planning a project, or analyzing data, the ability to solve for unknowns is essential. The solution we found might seem like a pair of unusual fractions, but it's a perfectly valid solution. It highlights the importance of being comfortable working with non-integer values and not expecting every solution to be a neat whole number. The key takeaway is that we systematically applied algebraic methods to find the solution, and we verified our results to ensure accuracy. This approach can be applied to a wide range of mathematical problems, making it a valuable skill to master.

Choosing the Right Method

So, which method is better, substitution or elimination? Well, it depends on the specific system of equations. Some systems are easier to solve using one method over the other. The substitution method is particularly useful when one of the equations can be easily solved for one variable. This makes the substitution process more straightforward and less prone to errors. For example, if one equation is already in the form y = ... or x = ..., then substitution is often the most efficient choice. On the other hand, the elimination method shines when the coefficients of one of the variables are either the same or easily made the same by multiplication. In such cases, the elimination method can quickly eliminate one variable and simplify the system. The choice between the two methods often comes down to personal preference and the specific structure of the equations. With practice, you'll develop a sense for which method is likely to be more efficient in a given situation. It's also worth noting that some systems might require a combination of both methods. For example, you might use elimination to simplify the system and then use substitution to find the final solution. The key is to be flexible and adaptable in your approach. Mastering both methods gives you a powerful toolkit for solving systems of equations and allows you to tackle a wide range of problems with confidence.

Practice Makes Perfect

The best way to become comfortable with solving systems of equations is to practice, practice, practice! Try solving different systems using both methods to get a feel for which one works best in different situations. You can find plenty of practice problems in textbooks, online resources, or even by creating your own. The more you practice, the more confident and proficient you'll become. Solving systems of equations is a fundamental skill in algebra and has applications in various fields, from science and engineering to economics and finance. It's a skill that builds upon other algebraic concepts, such as solving linear equations and manipulating expressions. By mastering systems of equations, you'll strengthen your overall mathematical foundation and be better prepared for more advanced topics. Practice not only improves your speed and accuracy but also deepens your understanding of the underlying concepts. It helps you develop problem-solving strategies and recognize patterns that can simplify the process. So, don't be afraid to tackle challenging problems and make mistakes along the way. Each mistake is an opportunity to learn and improve. With consistent practice, you'll become a system-solving expert and be able to confidently tackle any system of equations that comes your way. So keep practicing, and you'll see your skills grow.

Conclusion

And there you have it! We've walked through solving a system of equations using both the substitution and elimination methods. Remember, the key is to break the problem down into smaller, manageable steps and to double-check your work along the way. Solving systems of equations is a fundamental skill in mathematics, and mastering it will open doors to more advanced concepts and applications. Whether you're working on a math assignment, preparing for an exam, or tackling a real-world problem, the ability to solve for unknowns is a valuable asset. We've covered two powerful methods, substitution and elimination, and discussed how to choose the most efficient method for a given system. We've also emphasized the importance of verification to ensure the accuracy of your solutions. But perhaps the most important takeaway is the value of practice. The more you practice, the more comfortable and confident you'll become in solving systems of equations. So, keep honing your skills, and don't be afraid to challenge yourself with increasingly complex problems. With dedication and perseverance, you'll become a proficient problem solver and be well-equipped to tackle any mathematical challenge that comes your way. Keep up the great work, and happy solving!