Solving 3x + 2y = 6 And 2x + 3y = 5: A Step-by-Step Guide
Hey guys! Let's dive into solving this system of linear equations. We've got two equations here: 3x + 2y = 6 and 2x + 3y = 5. Don't worry, it might look a little intimidating at first, but we'll break it down step by step so it's super easy to understand. Systems of equations are a fundamental concept in algebra, showing up in various real-world scenarios, from balancing chemical equations to determining costs in business. Mastering these techniques will not only help you ace your math tests but also equip you with problem-solving skills applicable in many different fields. The approach we're going to use, the substitution method, is a powerful tool, especially when you can easily isolate one variable in terms of the other. So, buckle up, and let's get started!
Understanding Systems of Equations
So, what exactly is a system of equations? Simply put, it's a set of two or more equations that share variables. In our case, we have two equations, and they both have the variables 'x' and 'y'. The goal here is to find the values of 'x' and 'y' that satisfy both equations simultaneously. Think of it like finding a secret code that unlocks two doors at the same time! There are several methods to tackle these systems, but today, we're going to focus on the substitution method. This method is particularly handy when one of the equations can be easily rearranged to isolate one variable. Solving systems of equations is a cornerstone of algebra and appears frequently in various mathematical contexts. It's not just about manipulating numbers and symbols; it's about understanding relationships between different equations and finding solutions that work across the board. In real-world scenarios, systems of equations can model everything from supply and demand in economics to the flow of traffic in a city. So, let's roll up our sleeves and see how the substitution method works in practice!
Method 1: The Substitution Method
The substitution method is all about isolating one variable in one equation and then substituting that expression into the other equation. This effectively turns our two-variable problem into a one-variable problem, which is much easier to solve. Let's walk through it with our equations.
Step 1: Isolate a Variable
First, we need to pick one equation and one variable to isolate. Looking at our equations:
- 3x + 2y = 6
- 2x + 3y = 5
It seems easier to isolate 'y' in the first equation. Let's do that. We'll rearrange equation (1) to solve for 'y':
3x + 2y = 6
Subtract 3x from both sides:
2y = 6 - 3x
Now, divide both sides by 2:
y = (6 - 3x) / 2
So, we've got 'y' isolated! Great job so far! This is a crucial step because it sets us up perfectly for the next part of the substitution method. Isolating a variable isn't just a mechanical step; it's about strategically simplifying the problem. By expressing one variable in terms of the other, we're paving the way to eliminate one variable from the equation and make it solvable. You can choose to isolate either 'x' or 'y' from either equation, but the key is to pick the one that seems the easiest to manipulate. This can save you time and reduce the chances of making errors along the way. Remember, the goal is to make the problem as straightforward as possible, and isolating a variable effectively streamlines the solution process.
Step 2: Substitute
Now that we have y = (6 - 3x) / 2, we're going to substitute this expression for 'y' into the other equation (equation 2):
2x + 3y = 5
Replace 'y' with our expression:
2x + 3((6 - 3x) / 2) = 5
See what we did there? We've swapped 'y' for its equivalent expression in terms of 'x'. This is the heart of the substitution method. By replacing 'y' with (6 - 3x) / 2, we've transformed the second equation from an equation with two variables (x and y) into an equation with just one variable (x). This is a huge step forward because we know how to solve equations with a single variable. We're one step closer to finding the value of 'x', and once we have 'x', we can easily find 'y' as well. This substitution step is like a bridge that connects the two equations, allowing us to transfer information from one equation to the other. It's a clever way of simplifying the system and making it solvable. Keep going, you're doing awesome! The next step is to simplify this equation and solve for 'x'.
Step 3: Solve for x
Let's simplify and solve the equation we got in the last step:
2x + 3((6 - 3x) / 2) = 5
First, let's get rid of that fraction. Multiply the entire equation by 2:
2 * [2x + 3((6 - 3x) / 2)] = 2 * 5
This gives us:
4x + 3(6 - 3x) = 10
Now, distribute the 3:
4x + 18 - 9x = 10
Combine like terms:
-5x + 18 = 10
Subtract 18 from both sides:
-5x = -8
Finally, divide by -5:
x = 8/5
Woohoo! We found the value of 'x'! It's a fraction, but don't let that scare you. Fractions are just numbers too! Solving for 'x' in this step is where the algebraic skills really come into play. We've used a combination of distribution, combining like terms, and isolating the variable to arrive at our solution. Each of these steps is a fundamental algebraic technique that you'll use again and again in math. The key is to take it one step at a time, making sure to perform the same operation on both sides of the equation to maintain balance. Now that we have 'x', we're in the home stretch. The next step is to plug this value back into one of our equations to find 'y'. You're doing fantastic; let's keep the momentum going!
Step 4: Solve for y
Now that we know x = 8/5, we can plug this value back into any of our equations to solve for 'y'. Let's use the equation we got when we isolated 'y' earlier:
y = (6 - 3x) / 2
Substitute x = 8/5:
y = (6 - 3(8/5)) / 2
Simplify:
y = (6 - 24/5) / 2
To subtract the fraction, we need a common denominator. So, we'll rewrite 6 as 30/5:
y = (30/5 - 24/5) / 2
y = (6/5) / 2
To divide by 2, we can multiply by 1/2:
y = (6/5) * (1/2)
y = 3/5
Awesome! We've found the value of 'y'! So, y = 3/5. Finding 'y' is like the second half of the puzzle. We already had 'x', and by plugging it back into one of our original equations, we were able to solve for the remaining variable. This back-substitution is a common technique in solving systems of equations, and it's a powerful way to complete the solution once you've found one of the variables. We chose to use the equation y = (6 - 3x) / 2 because it was already solved for 'y', making the substitution process a bit simpler. However, we could have used either of the original equations and arrived at the same answer. The key is to pick the equation that seems the easiest to work with. Now that we have both 'x' and 'y', we're ready for the final step: checking our solution.
Step 5: Check Your Solution
It's always a good idea to check our solution to make sure we didn't make any mistakes. We'll plug our values for 'x' and 'y' into both original equations:
Equation 1: 3x + 2y = 6
3(8/5) + 2(3/5) = 6
24/5 + 6/5 = 6
30/5 = 6
6 = 6 (This checks out!)
Equation 2: 2x + 3y = 5
2(8/5) + 3(3/5) = 5
16/5 + 9/5 = 5
25/5 = 5
5 = 5 (This checks out too!)
Fantastic! Both equations hold true. This means our solution is correct. Checking our solution is a crucial step in the problem-solving process. It's like the final seal of approval, ensuring that our values for 'x' and 'y' satisfy both equations in the system. By plugging our solution back into the original equations, we can catch any errors we might have made along the way. If the equations don't balance, we know we need to go back and review our work. However, since both of our equations checked out, we can confidently say that we've found the correct solution. This step is not just about getting the right answer; it's about building confidence in our solution and ensuring accuracy. So, always remember to check your work – it's a habit that will serve you well in mathematics and beyond!
Solution
So, the solution to the system of equations is:
x = 8/5 y = 3/5
We can also write this as an ordered pair: (8/5, 3/5). You nailed it! We've successfully navigated through the system of equations using the substitution method. We isolated a variable, substituted it into the other equation, solved for one variable, and then used that value to find the other variable. Finally, we checked our solution to make sure everything was correct. The ordered pair (8/5, 3/5) represents the point where the two lines represented by our equations intersect on a graph. In other words, it's the unique point that satisfies both equations simultaneously. This ordered pair is the key to unlocking the system, and we found it! Remember, solving systems of equations is a fundamental skill in algebra, and it's applicable in a wide range of real-world scenarios. By mastering these techniques, you're not just solving math problems; you're developing problem-solving skills that will benefit you in many areas of life.
Method 2: Elimination Method
Another common method for solving systems of equations is the elimination method. In this method, we manipulate the equations so that when we add or subtract them, one of the variables is eliminated. Let's apply this to our system:
- 3x + 2y = 6
- 2x + 3y = 5
Step 1: Manipulate the Equations
We want to make either the 'x' or 'y' coefficients opposites. Let's eliminate 'x'. To do this, we'll multiply equation (1) by 2 and equation (2) by -3:
Equation (1) * 2: 6x + 4y = 12 Equation (2) * -3: -6x - 9y = -15
See how the 'x' coefficients are now 6 and -6? This is exactly what we wanted! Manipulating the equations in this step is a strategic move that sets us up for eliminating one of the variables. By multiplying each equation by a carefully chosen constant, we can ensure that the coefficients of one variable are opposites. This allows us to add the equations together in the next step, effectively canceling out that variable. This method is particularly useful when the coefficients of the variables are not easily isolated. The key is to identify the variable you want to eliminate and then find the appropriate multipliers to make its coefficients opposites. Remember, whatever you do to one side of the equation, you must do to the other side to maintain balance. This step is like setting the stage for the grand finale, where we'll combine the equations and watch one of the variables disappear!
Step 2: Eliminate x
Now, we'll add the two modified equations together:
(6x + 4y) + (-6x - 9y) = 12 + (-15)
The 'x' terms cancel out:
-5y = -3
Look at that! 'x' is gone! Adding the equations together is where the magic of the elimination method happens. By carefully manipulating the equations in the previous step, we created a situation where the 'x' terms had opposite coefficients. When we add the equations, these terms cancel each other out, leaving us with an equation in just one variable ('y'). This is a significant step forward because we know how to solve equations with a single variable. It's like we've cleared a major hurdle and are now on the path to finding the solution. The elimination method is a powerful technique because it systematically reduces the complexity of the system, making it solvable. This step is the culmination of our efforts, and it demonstrates the elegance and efficiency of this method. Now that we have an equation in 'y', the next step is to solve for 'y'.
Step 3: Solve for y
Divide both sides by -5:
y = 3/5
Nice! We have the value for 'y'. This step is the straightforward algebraic process of isolating the variable 'y' in the equation -5y = -3. By dividing both sides of the equation by -5, we effectively undo the multiplication and reveal the value of 'y'. It's a simple but essential step in the elimination method, as it gives us one piece of the puzzle – the value of one of the variables. This step highlights the importance of basic algebraic operations in solving more complex problems. Once we have 'y', we can use it to find the value of 'x'. It's like we've found one key that unlocks the system, and now we can use it to find the other key. Let's move on to the next step and see how we can use 'y' to solve for 'x'.
Step 4: Solve for x
Substitute y = 3/5 into either original equation. Let's use equation (1):
3x + 2(3/5) = 6
Simplify:
3x + 6/5 = 6
Subtract 6/5 from both sides:
3x = 6 - 6/5
3x = 30/5 - 6/5
3x = 24/5
Divide by 3:
x = 8/5
Excellent! We found the value of 'x'. Substituting 'y' back into one of the original equations is a crucial step in the elimination method. Once we've solved for one variable, we can plug its value into any of the equations in the system to find the value of the other variable. This back-substitution allows us to leverage the information we've already obtained and complete the solution. The choice of which equation to use is often a matter of convenience; pick the one that seems the easiest to work with. In this case, we chose equation (1), but we could have used equation (2) and arrived at the same answer. Solving for 'x' involves a series of algebraic manipulations, including simplifying fractions and isolating the variable. Each step builds upon the previous one, leading us closer to the solution. Now that we have both 'x' and 'y', we're ready for the final step: checking our solution.
Step 5: Check Your Solution
Just like before, let's check our solution in both original equations:
Equation 1: 3x + 2y = 6
3(8/5) + 2(3/5) = 6
24/5 + 6/5 = 6
30/5 = 6
6 = 6 (It checks out!)
Equation 2: 2x + 3y = 5
2(8/5) + 3(3/5) = 5
16/5 + 9/5 = 5
25/5 = 5
5 = 5 (It checks out!)
Perfect! Our solution is correct. Checking our solution is just as important in the elimination method as it is in the substitution method. It's our way of ensuring that we haven't made any errors along the way and that our values for 'x' and 'y' truly satisfy both equations in the system. By plugging our solution back into the original equations, we can verify that the equations balance, confirming the accuracy of our work. If the equations don't check out, we know we need to revisit our steps and identify any mistakes. However, since both of our equations checked out perfectly, we can confidently conclude that we've found the correct solution using the elimination method. This step is not just a formality; it's a crucial part of the problem-solving process that reinforces our understanding and builds confidence in our results. So, always remember to check your solutions – it's a habit that will pay off in the long run.
Solution
Again, the solution is:
x = 8/5 y = 3/5
Or (8/5, 3/5). We did it again! We've successfully solved the same system of equations using both the substitution and elimination methods. This demonstrates the flexibility of algebraic techniques and how different approaches can lead to the same solution. The elimination method, with its strategic manipulation of equations, provides an alternative pathway to solving systems of equations. It's like having another tool in our toolkit, ready to be used when the situation calls for it. The fact that we arrived at the same solution using both methods reinforces the accuracy of our work and deepens our understanding of the underlying concepts. Solving systems of equations is a fundamental skill in algebra, and mastering different methods allows us to tackle a wider range of problems with confidence. So, congratulations on mastering both the substitution and elimination methods – you're becoming a true algebra whiz!
Conclusion
So, there you have it! We've solved the system of equations 3x + 2y = 6 and 2x + 3y = 5 using both the substitution and elimination methods. Awesome job, guys! We've journeyed through the world of systems of equations, exploring two powerful techniques for finding solutions. We've seen how the substitution method allows us to isolate a variable and replace it in another equation, while the elimination method enables us to strategically manipulate equations to cancel out variables. Both methods are valuable tools in the algebraic toolbox, and mastering them will equip you with the skills to tackle a wide range of problems. Solving systems of equations is not just about finding numbers; it's about understanding relationships between different equations and finding solutions that satisfy multiple conditions. These skills are applicable in various fields, from science and engineering to economics and finance. So, the next time you encounter a system of equations, remember the steps we've covered, and approach it with confidence. Keep practicing, keep exploring, and keep unlocking the power of algebra!