Solving X + 2y = 4 And X + Y = 3 By Elimination Method
Hey guys! Today, we're diving into a super important concept in algebra: solving systems of equations. Specifically, we're going to tackle the following system using the elimination method:
- x + 2y = 4
- x + y = 3
The elimination method, also known as the addition method, is a powerful technique to solve a set of linear equations by removing one of the variables. This method is particularly useful when the coefficients of one variable in the equations are the same or easy to make the same. The goal is to manipulate the equations so that when you add or subtract them, one variable cancels out, leaving you with a single equation in one variable. This makes it way easier to solve! Once you find the value of one variable, you can easily substitute it back into one of the original equations to find the value of the other variable. Sounds like a plan? Let's jump right in and break down how to use the elimination method step by step to solve the system above. Trust me, by the end of this, you'll feel like a total pro at solving systems of equations!
Understanding the Elimination Method
So, what's the big idea behind the elimination method? The core principle is that if you have two equations, you can add or subtract them in a way that eliminates one of the variables. This is possible when the coefficients of one variable in both equations are either the same or opposites. If they are the same (e.g., both are 2), subtracting the equations will eliminate that variable. If they are opposites (e.g., 2 and -2), adding the equations will do the trick. Now, sometimes, the coefficients aren't initially the same or opposites. That's where the magic happens! You can multiply one or both equations by a constant to make the coefficients match or become opposites. This is perfectly legal because you're essentially multiplying both sides of the equation by the same number, maintaining the equality. Once you've eliminated a variable, you're left with a simple equation in one variable, which you can easily solve. After finding the value of one variable, you just plug it back into any of the original equations to solve for the other variable. It's like a puzzle, and each step gets you closer to the final solution. The beauty of the elimination method is its systematic approach. It provides a clear pathway to the solution, making it less prone to errors than other methods like substitution or graphing. Plus, it's super efficient for systems with larger coefficients or fractions, where other methods can get a bit messy. So, let's keep this in mind as we move on to apply this method to our specific problem.
Step-by-Step Solution
Okay, let's get down to business and solve our system of equations using the elimination method. Remember, we have:
- x + 2y = 4
- x + y = 3
The first thing we want to do is look at the coefficients of our variables. Notice that the coefficients of x in both equations are the same (both are 1). This is perfect because it means we can eliminate x quite easily. How? By subtracting one equation from the other!
Let's subtract equation (2) from equation (1):
(x + 2y) - (x + y) = 4 - 3
When we distribute the negative sign and combine like terms, we get:
x + 2y - x - y = 1
The x terms cancel each other out (x - x = 0), which is exactly what we wanted! Now we're left with:
y = 1
Awesome! We've found the value of y. Now that we know y = 1, we can substitute this value into either equation (1) or (2) to solve for x. Let's use equation (2) because it looks a bit simpler:
x + y = 3
Substitute y = 1:
x + 1 = 3
Subtract 1 from both sides:
x = 2
And there we have it! We've found that x = 2. So, the solution to our system of equations is x = 2 and y = 1. This means the point (2, 1) is the intersection of the two lines represented by our equations. We've successfully used the elimination method to solve the system, and hopefully, you're starting to see how powerful this method can be. Remember, the key is to make the coefficients of one variable the same or opposites so that you can eliminate that variable by adding or subtracting the equations. Now, let's take a moment to verify our solution to make sure everything checks out.
Verifying the Solution
Alright, we've got our solution: x = 2 and y = 1. But before we declare victory, it's always a good idea to verify our solution. This step is super important because it helps us catch any potential errors we might have made along the way. To verify, we simply plug our values of x and y back into the original equations and see if they hold true.
Let's start with equation (1):
x + 2y = 4
Substitute x = 2 and y = 1:
2 + 2(1) = 4
Simplify:
2 + 2 = 4
4 = 4
Great! The equation holds true. Now, let's check equation (2):
x + y = 3
Substitute x = 2 and y = 1:
2 + 1 = 3
Simplify:
3 = 3
Perfect! This equation also holds true. Since our solution satisfies both original equations, we can confidently say that x = 2 and y = 1 is the correct solution to the system. Verifying your solution is a fantastic habit to get into, especially when you're dealing with more complex systems of equations. It gives you peace of mind knowing that you've got the right answer. Plus, it's a great way to reinforce your understanding of the problem and the method you used to solve it. So, always take that extra minute to verify – you'll thank yourself later! Now that we've nailed the solution and verified it, let's think about when the elimination method really shines.
When to Use the Elimination Method
So, we've successfully used the elimination method to solve our system of equations, but you might be wondering, "When is this method the best choice?" Well, the elimination method is particularly effective when the coefficients of one of the variables in the system are the same or easily made the same. In our case, the coefficients of x were already the same, which made elimination a breeze. But what if the coefficients aren't the same? No problem! You can multiply one or both equations by a constant to make the coefficients match or become opposites. This is a crucial step in many elimination problems. For example, if you have equations like 2x + 3y = 7 and x - y = 1, you could multiply the second equation by 2 to make the x coefficients match (2x). Then, you can subtract the equations to eliminate x. The elimination method is also super handy when dealing with systems that have larger coefficients or fractions. In these situations, substitution can get a bit messy with fractions and extra steps, but elimination often provides a cleaner and more straightforward approach. Think about it this way: if you can easily manipulate the equations to eliminate a variable without creating a ton of fractions or complex expressions, elimination is likely a good bet. However, it's not always the perfect method for every situation. Sometimes, if one equation is already solved for one variable (e.g., y = 3x + 2), substitution might be quicker. Ultimately, the best method depends on the specific system of equations you're dealing with. But having the elimination method in your toolkit gives you a powerful option for tackling a wide range of problems. So, keep practicing, and you'll get a feel for when to use it effectively. Let's wrap things up with a quick recap of what we've learned.
Conclusion
Alright guys, we've covered a lot in this article! We tackled a system of equations using the elimination method, and hopefully, you've got a solid understanding of how this technique works. We started with the equations:
- x + 2y = 4
- x + y = 3
We walked through the step-by-step process of subtracting the equations to eliminate x, solving for y, and then substituting y back into one of the original equations to find x. Our solution was x = 2 and y = 1, which we then verified to make sure we were spot-on. Remember, the key to the elimination method is to manipulate the equations so that the coefficients of one variable are the same or opposites. This allows you to eliminate that variable by adding or subtracting the equations, leaving you with a single equation in one variable. We also discussed when the elimination method is particularly useful – especially when coefficients are the same or easily made the same, and when dealing with larger coefficients or fractions. But don't forget, it's just one tool in your algebraic toolbox. Sometimes substitution or graphing might be more efficient, depending on the problem. The more you practice, the better you'll become at recognizing which method is best suited for each situation. So, keep up the great work, keep solving those equations, and you'll be an algebra whiz in no time! If you've got any questions or want to dive deeper into systems of equations, don't hesitate to explore more resources or ask for help. Happy solving!
