Hey guys! Ever get stumped by a system of linear equations? Don't worry, it happens to the best of us. Let's break down how to solve these problems, step by step, so you can confidently find the correct ordered pair that satisfies the equations. We'll focus on a specific example, but the techniques we'll cover are applicable to a wide range of similar problems. So, let's dive in and conquer those equations!
Understanding Systems of Linear Equations
Before we jump into solving our specific problem, let's take a moment to understand what a system of linear equations actually is. At its core, a system of linear equations is simply a set of two or more linear equations that share the same variables. A linear equation, as you probably know, is an equation that can be written in the form ax + by = c, where a, b, and c are constants, and x and y are variables. When you graph a linear equation, you get a straight line – hence the name “linear.”
Now, when we have a system of these equations, we're essentially looking for the point (or points) where the lines intersect. This point of intersection represents the solution to the system, because it's the only point that satisfies all the equations simultaneously. In other words, if you plug the x and y values of the intersection point into each equation in the system, you'll get a true statement. Think of it like finding a secret meeting place that all the equations agree on. This is why ordered pairs are so important – they represent these potential meeting places, and our job is to figure out which one is the correct solution.
There are several methods for solving systems of linear equations, each with its own advantages and disadvantages. Some common methods include graphing, substitution, and elimination (also known as the addition method). Graphing is a visual approach where you plot the lines and find their intersection. Substitution involves solving one equation for one variable and substituting that expression into the other equation. Elimination involves manipulating the equations to eliminate one variable, making it easier to solve for the other. We'll primarily focus on the elimination method in our example, as it's often a very efficient approach.
When we talk about finding a solution to a system, we're looking for an ordered pair (x, y) that makes both equations true. This ordered pair represents a point on the coordinate plane where the lines represented by the equations intersect. If there's no intersection, there's no solution. If the lines are the same, there are infinitely many solutions (all the points on the line).
Problem Statement: Finding the Solution
Alright, let's get down to business! We have the following system of linear equations:
\begin{cases}
3x + y = 1 \\
5x + y = 3
\end{cases}
Our mission, should we choose to accept it (and we do!), is to determine which of the following ordered pairs is a solution to this system:
A. (-2, 1) B. (-0.5, 0.5) C. (0.5, -0.5) D. (1, -2)
So, what's the plan of attack? We could try plugging in each ordered pair into both equations to see which one works. That's a perfectly valid approach, especially for multiple-choice questions. However, let's also demonstrate how to solve the system using the elimination method. This will give us a deeper understanding of the process and provide an alternative way to verify our answer.
Method 1: Solving by Elimination
The elimination method, also known as the addition method, is a powerful technique for solving systems of linear equations. The core idea behind this method is to manipulate the equations in such a way that when you add them together, one of the variables cancels out. This leaves you with a single equation in one variable, which is much easier to solve. Once you've solved for one variable, you can substitute that value back into either of the original equations to solve for the other variable. Let's see how it works in our case.
Looking at our system:
\begin{cases}
3x + y = 1 \\
5x + y = 3
\end{cases}
Notice that both equations have a y term with a coefficient of 1. This is perfect for elimination! To eliminate y, we can multiply one of the equations by -1. Let's multiply the first equation by -1:
-1 * (3x + y = 1) => -3x - y = -1
Now our system looks like this:
\begin{cases}
-3x - y = -1 \\
5x + y = 3
\end{cases}
Now comes the fun part – adding the equations together. When we add the left-hand sides, we get (-3x - y) + (5x + y) = 2x. When we add the right-hand sides, we get -1 + 3 = 2. So, our new equation is:
2x = 2
This is a simple equation to solve! Dividing both sides by 2, we get:
x = 1
Fantastic! We've found the value of x. Now, to find y, we can substitute x = 1 into either of the original equations. Let's use the first equation, 3x + y = 1:
3(1) + y = 1
3 + y = 1
Subtracting 3 from both sides, we get:
y = -2
So, our solution is the ordered pair (1, -2). We've solved the system using the elimination method!
Method 2: Testing the Answer Choices
As mentioned earlier, another way to tackle this problem is to simply test each of the given answer choices. This might seem like a brute-force approach, but it can be surprisingly efficient, especially in a multiple-choice setting. The key is to plug in the x and y values from each ordered pair into both equations in the system. If the ordered pair satisfies both equations, then it's the solution. If it fails even one equation, it's not the solution.
Let's go through the answer choices one by one:
- A. (-2, 1):
- Equation 1: 3(-2) + 1 = -6 + 1 = -5 ≠ 1 (Fails)
- Since it fails the first equation, we don't even need to check the second equation. (-2, 1) is not the solution.
- B. (-0.5, 0.5):
- Equation 1: 3(-0.5) + 0.5 = -1.5 + 0.5 = -1 ≠ 1 (Fails)
- Again, it fails the first equation, so we can eliminate it.
- C. (0.5, -0.5):
- Equation 1: 3(0.5) + (-0.5) = 1.5 - 0.5 = 1 (Passes)
- Equation 2: 5(0.5) + (-0.5) = 2.5 - 0.5 = 2 ≠ 3 (Fails)
- It passes the first equation, but fails the second, so it's not the solution.
- D. (1, -2):
- Equation 1: 3(1) + (-2) = 3 - 2 = 1 (Passes)
- Equation 2: 5(1) + (-2) = 5 - 2 = 3 (Passes)
- It passes both equations! This is our solution.
As you can see, testing the answer choices can be a quick way to find the solution, especially if you're comfortable with arithmetic. However, it's also valuable to understand methods like elimination, as they can be used to solve systems even when answer choices aren't provided.
The Verdict: The Correct Ordered Pair
Drumroll, please! Both the elimination method and the method of testing answer choices lead us to the same conclusion: the correct ordered pair is D. (1, -2). This ordered pair satisfies both equations in the system, making it the solution.
Key Takeaways for Solving Linear Systems
Alright, guys, we've successfully navigated this system of linear equations! Before we wrap up, let's highlight some key takeaways that you can apply to similar problems:
- Understand the Goal: Remember that solving a system of linear equations means finding the ordered pair (or pairs) that make all the equations true simultaneously. This represents the point(s) of intersection between the lines represented by the equations.
- Choose Your Weapon: There are several methods for solving systems, including graphing, substitution, and elimination. The best method depends on the specific problem and your personal preference. In many cases, elimination is a powerful and efficient choice.
- Master Elimination: The elimination method involves manipulating the equations to eliminate one variable. This often involves multiplying one or both equations by a constant so that the coefficients of one variable are opposites. Then, you can add the equations together, eliminating that variable.
- Test the Answer Choices: If you're given multiple-choice options, don't hesitate to test them! Plug each ordered pair into both equations to see if it satisfies the system. This can be a quick and reliable way to find the solution.
- Check Your Work: Always double-check your solution by plugging it back into the original equations. This will help you catch any arithmetic errors and ensure that you've found the correct answer.
Practice Makes Perfect
Solving systems of linear equations is a fundamental skill in algebra, and like any skill, it improves with practice. So, don't be afraid to tackle more problems! The more you practice, the more comfortable and confident you'll become. You'll start to recognize patterns, choose the most efficient methods, and solve these problems like a pro.
So there you have it! We've successfully solved a system of linear equations and explored the techniques involved. Keep practicing, and you'll be a master of linear systems in no time. Good luck, and happy solving!