Solving Linear Equations: A Step-by-Step Guide
Hey guys! Today, we're diving into the world of linear equations and tackling a specific problem. We've got a system of equations here, and we're going to walk through the steps to solve it. Don't worry, it's not as scary as it looks! We'll break it down and make it super easy to understand. So, let's jump right in and get those algebraic muscles working!
Understanding the System of Equations
Okay, so first things first, let's take a good look at our system of equations. We're given two equations, and our goal is to find the values of x and y that satisfy both of them simultaneously. Think of it like finding the point where two lines intersect on a graph. That point represents the solution to our system. The system we're working with looks like this:
In this system, the first equation is 2x - 3y = 1, and the second equation is 7x - 15y = -1. Each of these equations represents a straight line when graphed. The x and y are our variables, and we need to figure out what values they need to be to make both equations true. There are a few different methods we can use to solve systems of equations, but today we're focusing on one specific technique: elimination. Elimination involves manipulating the equations so that when we add or subtract them, one of the variables cancels out, making it easier to solve for the other.
Before we dive into the elimination method, let's quickly touch on why understanding these systems is so crucial. Linear equations pop up everywhere in real life, from calculating costs and distances to modeling complex systems in science and engineering. Mastering these skills opens doors to solving a wide range of problems, so stick with us, and let's conquer this algebraic challenge together! Remember, the key to understanding is breaking it down step by step, and that's exactly what we're going to do.
The Strategy: Elimination Method
The elimination method is our weapon of choice for this particular problem. It's a slick way to solve systems of equations by strategically manipulating them to eliminate one variable. The basic idea is to multiply one or both equations by a constant so that the coefficients of either x or y are opposites. This way, when we add the equations together, that variable disappears, leaving us with a single equation in one variable, which is much easier to solve.
In our case, we're focusing on eliminating the y variable. Looking at our system:
We notice that the coefficient of y in the first equation is -3, and in the second equation, it's -15. To eliminate y, we want these coefficients to be opposites (e.g., 15 and -15). The easiest way to achieve this is to multiply the first equation by 5. This will give us a y coefficient of -15 in the first equation, which is the same magnitude as the y coefficient in the second equation.
Once we've multiplied the first equation by 5, we'll have a modified system. The next step, as the problem states, is to subtract the modified first equation from the second equation. This is where the magic happens! The y terms will cancel each other out, leaving us with an equation in terms of x only. We can then easily solve for x. After finding the value of x, we can substitute it back into either of the original equations to solve for y. It’s like a domino effect – eliminate one variable, solve for the other, and then use that information to find the remaining variable. This method is super efficient and a fundamental technique in algebra.
Step-by-Step Solution: Multiplying and Subtracting
Alright, let's get our hands dirty and put the elimination method into action! The first step, as the problem instructs, is to multiply the first equation by 5. This is a crucial step in setting up the elimination of the y variable. Remember, whatever we do to one side of the equation, we have to do to the other to maintain the balance.
Our first equation is: 2x - 3y = 1. Multiplying both sides by 5, we get:
5(2x - 3y) = 5(1)
Distributing the 5 on the left side, we have:
10x - 15y = 5
So, our modified first equation is now 10x - 15y = 5. We'll keep this equation handy as we move on to the next step. Now, we have our original second equation and our modified first equation:
The next instruction is to subtract the modified first equation from the second equation. This is where the magic happens, and we eliminate the y variable. When subtracting equations, it’s super important to subtract corresponding terms. That means we subtract the x terms, the y terms, and the constant terms separately.
So, we have: (7x - 15y) - (10x - 15y) = -1 - 5. Now, let's simplify this expression. Distribute the negative sign in the second set of parentheses:
7x - 15y - 10x + 15y = -6
Notice how the -15y and +15y cancel each other out? That's exactly what we wanted! This leaves us with:
-3x = -6
We've successfully eliminated y and are one step closer to solving for x. The next step is to isolate x, which we'll tackle in the next section.
Isolating x: Solving for the First Variable
We've reached a pivotal point in our solution! After multiplying the first equation by 5 and subtracting it from the second, we've successfully eliminated the y variable. We're now left with a simple equation in terms of x: -3x = -6. Our next mission? To isolate x and find its value. This is a straightforward algebraic step, and we're going to nail it.
To get x by itself, we need to get rid of the -3 that's multiplying it. The opposite of multiplication is division, so we'll divide both sides of the equation by -3. Remember, whatever operation we perform on one side of the equation, we must perform on the other side to maintain balance. So, we have:
(-3x) / -3 = (-6) / -3
On the left side, the -3s cancel out, leaving us with just x. On the right side, -6 divided by -3 is 2. Therefore, we have:
x = 2
Boom! We've found the value of x. This is a major milestone in solving our system of equations. But we're not done yet – we still need to find the value of y. Now that we know x, we can substitute it back into one of our original equations and solve for y. This is the final piece of the puzzle, and we're going to fit it perfectly in the next section.
Finding y: Completing the Solution
Fantastic! We've successfully determined that x = 2. Now, it's time to find the value of y. To do this, we'll use a technique called substitution. We'll take the value of x that we just found and plug it back into one of the original equations. It doesn't matter which equation we choose; both will give us the same value for y. For simplicity, let's use the first original equation:
2x - 3y = 1
Now, substitute x = 2 into the equation:
2(2) - 3y = 1
Simplify the equation:
4 - 3y = 1
Now, we need to isolate y. First, let's subtract 4 from both sides of the equation:
4 - 3y - 4 = 1 - 4
This simplifies to:
-3y = -3
Next, divide both sides by -3 to solve for y:
(-3y) / -3 = (-3) / -3
This gives us:
y = 1
We've done it! We've found the value of y. Now we know that x = 2 and y = 1. This is the solution to our system of equations. But to be absolutely sure, it's always a good idea to check our solution.
Checking the Solution: Ensuring Accuracy
We've arrived at a solution: x = 2 and y = 1. But before we celebrate, let's make absolutely sure our answer is correct. The best way to do this is to check our solution by plugging the values of x and y back into both of the original equations. If both equations hold true, we know we've nailed it!
Let's start with the first original equation:
2x - 3y = 1
Substitute x = 2 and y = 1:
2(2) - 3(1) = 1
Simplify:
4 - 3 = 1
1 = 1
The first equation checks out! Now, let's move on to the second original equation:
7x - 15y = -1
Substitute x = 2 and y = 1:
7(2) - 15(1) = -1
Simplify:
14 - 15 = -1
-1 = -1
The second equation also checks out! Since our values for x and y satisfy both equations, we can confidently say that our solution is correct. We've successfully solved the system of linear equations. Great job, everyone!
Conclusion: Mastering Linear Equations
Woohoo! We did it, guys! We successfully solved a system of linear equations using the elimination method. We took a seemingly complex problem and broke it down into manageable steps. We learned how to multiply equations, eliminate variables, isolate unknowns, and, most importantly, check our solution to ensure accuracy. You've added a valuable tool to your algebraic arsenal today.
Linear equations are a fundamental concept in mathematics and have countless applications in the real world. By mastering these techniques, you're not just learning math; you're developing problem-solving skills that will serve you well in various aspects of life. Remember, practice makes perfect! The more you work with these types of problems, the more comfortable and confident you'll become.
So, keep practicing, keep exploring, and never stop asking questions. Algebra can be challenging, but it's also incredibly rewarding. You've taken a big step today, and I'm super proud of your progress. Keep up the awesome work, and who knows, maybe you'll be teaching someone else how to solve linear equations in the future! Keep those brains buzzing and happy solving!