Solving Equations: Real Number Solutions Explained!
Hey guys! Let's dive into solving some equations where we're looking for real number solutions. We'll tackle this step by step, making sure everything's crystal clear. This topic falls under the exciting world of mathematics, so buckle up and let's get started!
Understanding the Basics
Before we jump into the specific problems, let's quickly refresh our understanding of what it means to find the solution set of an equation. In simple terms, the solution set is the collection of all possible values for the variables (in our case, x and y) that make the equation true. We're working with real numbers, which means we're dealing with all the numbers you can find on a number line – including positive and negative numbers, fractions, decimals, and even those funky irrational numbers like pi and the square root of 2.
When we have equations with two variables (like x and y), we're often looking for pairs of values that satisfy the equation. These pairs can be represented as coordinates (x, y) on a graph. The solution set might be a single point, a set of discrete points, or even a whole line or curve on the graph. The key is to find all the combinations of x and y that make the equation balance perfectly. To begin with, it's super important to understand the equation given. What kind of equation is it? Is it a linear equation, a quadratic equation, or something else? The type of equation will often guide our approach to solving it. For instance, linear equations are generally easier to handle, while quadratic equations might require techniques like factoring or using the quadratic formula.
Next up, let's consider the constraints or conditions that are provided. In our case, we're told that x and y belong to the set of real numbers. This is a crucial piece of information because it narrows down the possibilities. If we were dealing with, say, integers, our solution set would be different. Also, if we are given specific values for one of the variables, like x = -3 or x = 3, that significantly simplifies the problem. We can simply substitute those values into the equation and solve for the other variable. Remember, mathematics is like a puzzle, and each piece of information is a clue to help us find the solution! So, keep your eyes peeled for any hints the problem gives you. The goal is to isolate the variables we're trying to solve for. This often involves using algebraic manipulations, like adding or subtracting the same value from both sides of the equation, or multiplying or dividing both sides by the same non-zero value. The important thing is to maintain the balance of the equation. Whatever you do to one side, you must do to the other. It's like a seesaw – if you add weight to one side, you need to add the same weight to the other side to keep it level.
Solving Equations with a Given x Value
Okay, let's tackle the first part of the problem. We're given specific values for x: x = -3 and x = 3. This makes our job easier because we can substitute these values directly into the given equations and solve for y.
Case 1: x = -3
Let's consider the equation 2x - 3y - 12 = 0. We'll replace x with -3:
2(-3) - 3y - 12 = 0
Now, simplify:
-6 - 3y - 12 = 0
Combine the constants:
-3y - 18 = 0
Add 18 to both sides:
-3y = 18
Divide both sides by -3:
y = -6
So, when x = -3, y = -6. This gives us one solution: (-3, -6). This is a crucial step in finding the overall solution set. Remember, the solution set is a collection of all possible pairs (x, y) that satisfy the equation. When we substitute x = -3, we're essentially fixing one piece of the puzzle and then solving for the other. The same principle applies to other equations and values of x. Think of it like filling in the blanks – you're given one blank (the value of x) and you need to find the value that goes in the other blank (the value of y). And just like in a puzzle, there might be multiple ways to approach it, but the key is to be systematic and careful. Let's delve a bit deeper into the algebraic manipulations we used here. Notice how we carefully isolated y by performing the same operations on both sides of the equation. This is a fundamental technique in algebra, and it's essential for solving a wide range of equations. We started by simplifying the equation, then we combined like terms, and finally, we used inverse operations (addition and division) to get y by itself. This process is like peeling an onion – you remove the outer layers one by one until you get to the core.
Now, what about the other equation, 2x + y = 4? Again, substitute x = -3:
2(-3) + y = 4
Simplify:
-6 + y = 4
Add 6 to both sides:
y = 10
So, when x = -3, y = 10. This gives us another solution: (-3, 10).
Case 2: x = 3
Now, let's do the same for x = 3. First, the equation 2x - 3y - 12 = 0:
2(3) - 3y - 12 = 0
Simplify:
6 - 3y - 12 = 0
Combine constants:
-3y - 6 = 0
Add 6 to both sides:
-3y = 6
Divide both sides by -3:
y = -2
So, when x = 3, y = -2. This gives us the solution (3, -2). Just like before, we've taken a specific value of x and used it to find the corresponding value of y. This is a very powerful technique, especially when you have multiple equations and variables. It allows you to break down a complex problem into smaller, more manageable steps. Imagine you're building a house – you wouldn't try to put up all the walls at once. Instead, you'd start with one wall, make sure it's solid, and then move on to the next. Solving equations is similar – you focus on one variable at a time, make sure you've solved for it correctly, and then move on to the next variable.
For the equation 2x + y = 4, substitute x = 3:
2(3) + y = 4
Simplify:
6 + y = 4
Subtract 6 from both sides:
y = -2
So, when x = 3, y = -2. This gives us the solution (3, -2).
Solving the System of Equations
Now, let's tackle the heart of the problem: finding the solution set for the system of equations:
2x - 3y - 12 = 0
2x + y = 4
We need to find the values of x and y that satisfy both equations simultaneously. There are a couple of ways we can do this: substitution and elimination.
Method 1: Substitution
Let's use the second equation (2x + y = 4) to solve for y:
y = 4 - 2x
Now, substitute this expression for y into the first equation:
2x - 3(4 - 2x) - 12 = 0
Distribute the -3:
2x - 12 + 6x - 12 = 0
Combine like terms:
8x - 24 = 0
Add 24 to both sides:
8x = 24
Divide both sides by 8:
x = 3
Great! We found x = 3. Now, substitute this value back into the equation y = 4 - 2x:
y = 4 - 2(3)
y = 4 - 6
y = -2
So, the solution to the system of equations is (3, -2). Think of the substitution method like a relay race. You start with one equation, isolate a variable, and then pass that information on to the next equation. The key is to choose the equation that makes the substitution easiest. In this case, the second equation was a good choice because it was relatively simple to solve for y. However, don't be afraid to experiment – sometimes a different substitution might lead to a quicker solution. The most important thing is to be systematic and organized. Keep track of your steps, and double-check your work as you go. A small mistake early on can throw off the entire solution, so it's worth taking the time to be careful.
Method 2: Elimination
Another way to solve this system is by elimination. The goal here is to eliminate one of the variables by adding or subtracting the equations. Notice that the coefficients of x in both equations are the same (2). So, we can subtract the second equation from the first:
(2x - 3y - 12) - (2x + y) = 0 - 4
Simplify:
2x - 3y - 12 - 2x - y = -4
-4y - 12 = -4
Add 12 to both sides:
-4y = 8
Divide both sides by -4:
y = -2
Now, substitute y = -2 into either of the original equations. Let's use 2x + y = 4:
2x + (-2) = 4
Add 2 to both sides:
2x = 6
Divide both sides by 2:
x = 3
Again, we find the solution (3, -2). Elimination is like a strategic demolition – you carefully subtract the equations in a way that cancels out one of the variables, leaving you with a simpler equation to solve. The trick is to look for variables with the same or opposite coefficients. If you don't see any, you can multiply one or both equations by a constant to create matching coefficients. The beauty of the elimination method is that it can often be more efficient than substitution, especially when the equations are already set up nicely for it. It's like choosing the right tool for the job – sometimes a hammer is better than a screwdriver, and vice versa. And just like with any mathematical technique, practice makes perfect. The more you use the elimination method, the better you'll become at recognizing when it's the most efficient approach.
Final Solution Set
Okay, let's wrap things up. Based on our calculations:
- When x = -3, the solutions are (-3, -6) and (-3, 10).
- When x = 3, the solution is (3, -2).
- Solving the system of equations gives us the solution (3, -2).
Therefore, the solution set for the given equations, considering the specified conditions, is {(-3, -6), (-3, 10), (3, -2)}. This is the grand finale! We've taken a challenging problem and broken it down into manageable steps. We've used a variety of techniques, from simple substitution to the more powerful elimination method. And we've arrived at the solution set – the collection of all pairs (x, y) that make the equations true. But the journey doesn't end here. Mathematics is a vast and fascinating world, and there's always more to explore. So, keep practicing, keep asking questions, and keep pushing your boundaries. The more you engage with mathematics, the more you'll discover its beauty and power. And remember, the key to success is perseverance. Don't get discouraged if you encounter a difficult problem. Take a break, try a different approach, and most importantly, never give up. You've got this!
Conclusion
Solving equations, especially systems of equations, is a fundamental skill in mathematics. By understanding the different methods available and practicing regularly, you can confidently tackle these problems and build a strong foundation for more advanced topics. Keep up the great work, guys!
Remember, the solution set represents all the possible answers to the problem. It's like a treasure map that leads you to all the hidden solutions. And in mathematics, as in life, the journey of discovery is often just as rewarding as the destination.