Unlocking Solutions: Solving For X And Y In Equations
Hey there, math enthusiasts! Today, we're diving into the exciting world of solving systems of equations, specifically focusing on how to find the values of x and y. It might seem a little daunting at first, but trust me, with a few simple steps, you'll be cracking these problems like a pro. We'll be tackling four different systems, walking through each one step-by-step so you can totally grasp the process. Ready to jump in? Let's go!
Understanding Systems of Equations
Before we start, let's quickly recap what a system of equations actually is. Basically, it's a set of two or more equations, each containing the same variables (in our case, x and y). The goal is to find values for x and y that satisfy all the equations in the system. Think of it like a puzzle where you need to find the pieces that fit perfectly in multiple places. These equations often represent lines on a graph, and the solution (the values of x and y) is the point where those lines intersect. So, when we solve a system, we're essentially finding that intersection point!
There are several methods to solve systems of equations, but we'll focus on the elimination method because it's super effective for these particular examples. The elimination method involves manipulating the equations (multiplying them by constants, adding or subtracting them) to eliminate one of the variables. This leaves you with a single equation with only one variable, which is easy to solve. Once you have the value of that variable, you can plug it back into one of the original equations to find the value of the other variable. It's all about strategic manipulation and a bit of detective work! The key is to look for opportunities to make the coefficients of either x or y opposites so that when you add the equations, that variable disappears, voila!
To become fluent in solving systems of equations, you'll want to practice. That means the more problems you work through, the more comfortable and confident you'll become. Each problem presents a slightly different arrangement of terms. Some equations might need a little tweaking before they're ready to be solved, and that's okay! We'll start with the first example, where we have a system of two equations, and we want to find the values of x and y that satisfy both.
Solving System A: x + y = 10 and 2x - y = 8
Let's get started with our first system of equations. We've got:
- Equation 1: x + y = 10
- Equation 2: 2x - y = 8
Notice something cool? In this system, the 'y' terms already have opposite signs (+y and -y). This makes our lives much easier! We can simply add the two equations together. When we do that, the 'y' terms will cancel each other out because +y - y = 0. So, let's do it:
(x + y) + (2x - y) = 10 + 8
Simplifying this, we get:
3x = 18
Now, to solve for x, we divide both sides by 3:
x = 6
Awesome! We've found the value of x. Now, let's plug this value back into either Equation 1 or Equation 2 to find y. I'm going to use Equation 1 (x + y = 10) because it looks simpler:
6 + y = 10
Subtract 6 from both sides:
y = 4
And there you have it! The solution to system A is x = 6 and y = 4. This means that if you were to graph the lines represented by these equations, they would intersect at the point (6, 4).
To recap: The key here was that the 'y' terms were already opposites. By adding the equations, we eliminated 'y' and could solve for 'x' directly. Then, we used the value of 'x' to find 'y'.
Solving System B: 3x + y = 1 and y - x = 5
Alright, let's move on to System B. Here's what we've got:
- Equation 1: 3x + y = 1
- Equation 2: y - x = 5
In this system, we don't have terms with opposite signs. But no worries! We can rearrange Equation 2 to make it easier to work with. Let's rewrite it as -x + y = 5. Now, we could multiply either equation by a constant to make the 'y' terms have opposite coefficients. Another approach is to multiply Equation 2 by -1. But, let's try a different strategy. Let's solve Equation 2 for 'y':
y = x + 5
Now, substitute this value of 'y' into Equation 1:
3x + (x + 5) = 1
Simplify and solve for x:
4x + 5 = 1
4x = -4
x = -1
Great! We have x = -1. Now, substitute this value back into y = x + 5:
y = -1 + 5
y = 4
Therefore, the solution to System B is x = -1 and y = 4. Again, this solution represents the point of intersection of the two lines represented by the equations.
Key Takeaway: In this system, we solved for 'y' in one equation and substituted it into the other. This eliminated 'y' and allowed us to solve for 'x'. Then, we used the value of 'x' to find 'y'.
Solving System C: x + y = 2 and 3x + 4y = 3
Let's tackle System C:
- Equation 1: x + y = 2
- Equation 2: 3x + 4y = 3
In this system, we don't have any terms that are immediately ready for elimination. However, we can use the elimination method again. This time, let's multiply Equation 1 by -3. This will make the 'x' terms have opposite coefficients. Here's what that looks like:
-3(x + y) = -3(2)
Which simplifies to:
-3x - 3y = -6
Now, let's rewrite the system with the modified Equation 1:
- -3x - 3y = -6
- 3x + 4y = 3
Now, add the two equations together. The 'x' terms will cancel out:
(-3x - 3y) + (3x + 4y) = -6 + 3
Simplifying, we get:
y = -3
We've found y! Now, substitute this value into Equation 1 (x + y = 2):
x + (-3) = 2
x - 3 = 2
Add 3 to both sides:
x = 5
So, the solution to System C is x = 5 and y = -3. We multiplied one equation by a constant, then added the equations to eliminate a variable. This allowed us to find the solution.
Important Note: Always double-check your work. After solving for x and y, plug your answers back into the original equations to make sure they are correct. If the values don't work in both equations, there's likely an error somewhere in your calculations. It's a great way to catch mistakes!
Solving System D: x + 2y = 8 and y - 3x = 1
Finally, let's solve System D:
- Equation 1: x + 2y = 8
- Equation 2: y - 3x = 1
First, let's rewrite Equation 2 as -3x + y = 1. We could solve for 'y' in one equation and substitute, or we could multiply by a constant. Let's multiply Equation 2 by -2. Doing so will allow us to eliminate the 'y' term:
-2(-3x + y) = -2(1)
Which simplifies to:
6x - 2y = -2
Now, let's rewrite the system with the modified Equation 2:
- x + 2y = 8
- 6x - 2y = -2
Now, add the equations together. The 'y' terms will cancel out:
(x + 2y) + (6x - 2y) = 8 + (-2)
Simplifying:
7x = 6
Now solve for x:
x = 6/7
Substitute the value of 'x' back into Equation 1:
6/7 + 2y = 8
2y = 8 - 6/7
2y = 56/7 - 6/7
2y = 50/7
y = 25/7
So, the solution to System D is x = 6/7 and y = 25/7.
In Summary: We've tackled four different systems of equations, each requiring a slightly different approach. The elimination method is a powerful tool, and with a little practice, you'll be able to solve these problems with confidence. Remember to always double-check your answers and don't be afraid to experiment with different strategies. Keep practicing, and you'll become a systems-of-equations whiz in no time!