Hey guys! Today, we're diving into the fascinating world of solving systems of equations. You know, those things that look like a puzzle with two or more equations and variables. Specifically, we're going to tackle a problem where we need to figure out how to eliminate one of the variables – in this case, 'x' – so we can solve for the others. It might sound intimidating, but trust me, it's like cracking a code, and once you get the hang of it, it's super satisfying. So, let's jump right in and break down the process step-by-step!
The Challenge: Eliminating 'x' in a System of Equations
So, we've got this system of equations staring back at us, and the mission, should we choose to accept it (and we do!), is to figure out which operation will make those pesky 'x' terms vanish when we add the equations together. This is a classic strategy in algebra, and it's all about manipulating the equations so that when you combine them, some terms cancel out. Think of it like a mathematical magic trick – but instead of pulling a rabbit out of a hat, we're making variables disappear! The system looks like this:
3x + (1/2)y = 3
6x - y = 2
The big question here is: What can we do to one (or both) of these equations so that when we add them, the 'x' terms eliminate each other? To make this happen, we need the coefficients of 'x' in the two equations to be opposites – that is, numbers that add up to zero. For instance, if one equation has '3x', we'd want the other to have '-3x'. This way, when we add them, '3x + (-3x)' becomes zero, and 'x' is gone! This approach is super useful because it simplifies the system, making it much easier to solve for the remaining variables.
To really nail this, we need to understand the fundamental principle at play: we're using the addition property of equality. This basically says that if we add the same thing to both sides of an equation, the equation remains balanced. We can also multiply both sides of an equation by the same number without changing its validity. These are the tools in our algebraic toolbox that allow us to manipulate equations and set them up for variable elimination. We're not just blindly changing things; we're strategically tweaking the equations while maintaining their integrity. Think of it like adjusting ingredients in a recipe – you want to change the flavor, but you still need the cake to bake properly!
Now, let's look closely at our equations again. We have '3x' in the first equation and '6x' in the second. To eliminate 'x' by adding the equations, we need to transform one of these terms into its opposite. The easiest way to do this is to focus on the '3x' in the first equation. What do we need to multiply '3x' by to get '-6x'? Well, we need to multiply by '-2'. And remember, we have to multiply every term in the equation by '-2' to keep it balanced. So, multiplying the entire first equation by '-2' is our key move. This will give us a new first equation that, when added to the second, will make the 'x' terms vanish like a puff of smoke. It's all about setting up the puzzle pieces just right so they fit together perfectly and solve our problem.
The Solution: Multiplying to Eliminate
Okay, let's put our plan into action. We've identified that the magic trick here involves multiplying the first equation by -2. Why -2? Because that's the number that will turn our 3x into -6x, which is the exact opposite of the 6x in the second equation. This is crucial – we're aiming for those 'x' terms to cancel each other out when we add the equations together. It's like a mathematical dance, where we choreograph the steps so that certain elements gracefully disappear.
So, let's break it down. We take the first equation:
3x + (1/2)y = 3
And we multiply every single term by -2. This is super important. We can't just multiply the 'x' term and call it a day. We need to distribute the -2 across the entire equation to keep things balanced. Remember, an equation is like a scale – whatever you do to one side, you have to do to the other to maintain equilibrium. So, we're multiplying both sides of the equation by -2, ensuring that the equality holds.
When we do the multiplication, we get:
-2 * (3x) + -2 * (1/2)y = -2 * 3
Which simplifies to:
-6x - y = -6
Now, look at what we've achieved! We've transformed our first equation into a new form where the 'x' term is -6x. This is exactly what we wanted. Now, we're perfectly set up to eliminate the 'x' terms when we add this modified equation to the second equation in our original system. It's like preparing the ingredients for a recipe – we've prepped our first equation, and now it's ready to combine with the second equation to create our solution.
Let's recap quickly. We started with a system of equations and a mission to eliminate the 'x' variable. We realized that multiplying the first equation by -2 would give us opposite 'x' terms. We carefully performed the multiplication, making sure to distribute the -2 to every term in the equation. And now, we have a new equation that's perfectly poised to help us solve the system. This is a fantastic example of how strategic manipulation can simplify complex problems in algebra. By applying the multiplication property of equality, we've taken a significant step toward finding the solution. It's like we've unlocked a secret passage in a puzzle, and the rest of the solution is just around the corner!
Verifying the Elimination
Alright, now comes the fun part: the moment of truth! We've multiplied the first equation by -2, and we've got our transformed equation. Now, we need to actually add the two equations together to see if our plan worked and those 'x' terms vanish as expected. This is where we get to witness the magic of algebra in action, where careful manipulation leads to a beautiful simplification. It's like watching a perfectly executed move in a game – satisfying and rewarding!
So, let's bring back our modified first equation:
-6x - y = -6
And let's line it up with our second equation from the original system:
6x - y = 2
Now, we're going to add these two equations together, term by term. We add the 'x' terms, the 'y' terms, and the constants. This is like combining like ingredients in a recipe – we're grouping the similar elements together to see what the final result will be. When we add equations, we're essentially saying that if the left sides are equal to each other and the right sides are equal to each other, then the sums of the left sides and the sums of the right sides must also be equal. It's a logical extension of the basic principles of equality.
Adding the 'x' terms, we have -6x + 6x. And what does that equal? Zero! This is exactly what we wanted! The 'x' terms have canceled each other out, just like we planned. It's like hitting the bullseye on a target – a clear indication that our strategy was correct. This is why we chose to multiply by -2 – it set us up perfectly for this elimination. It's all about foresight and planning in algebra, thinking ahead to how the operations will affect the equation and lead us toward the solution.
Now, let's move on to the 'y' terms. We have -y + (-y), which simplifies to -2y. So, the 'y' terms combine to give us -2y. Finally, let's add the constants: -6 + 2, which equals -4. So, the constants combine to give us -4.
Putting it all together, the result of adding the two equations is:
-2y = -4
Look at that! We've gone from a system of two equations with two variables to a single equation with just one variable. The 'x' is gone, and we're left with a simple equation that we can easily solve for 'y'. This is the power of elimination – it simplifies the problem by reducing the number of variables we need to deal with. It's like taking a complicated puzzle and breaking it down into smaller, more manageable pieces. By eliminating one variable, we've made the entire system much easier to solve.
Solving for 'y' and Finding 'x'
We've successfully eliminated 'x' from our system of equations, and we're now sitting pretty with a simple equation in terms of 'y': -2y = -4. This is like reaching a checkpoint in a game – we've overcome a major hurdle, and the path to the final solution is now much clearer. Solving for 'y' here is a breeze – it's a straightforward application of basic algebraic principles. It's a moment to appreciate the elegance of mathematics, where a series of logical steps leads us to a clear and concise answer.
To isolate 'y', we need to get rid of the -2 that's multiplying it. The golden rule of algebra is that whatever we do to one side of the equation, we must do to the other. So, to get 'y' by itself, we'll divide both sides of the equation by -2. This is a fundamental operation in algebra, and it's all about maintaining balance and equality. It's like a delicate balancing act, where we carefully adjust both sides of the equation to keep everything in equilibrium.
Dividing both sides by -2, we get:
-2y / -2 = -4 / -2
On the left side, the -2 in the numerator and the -2 in the denominator cancel each other out, leaving us with just 'y'. On the right side, -4 divided by -2 equals 2. So, we have:
y = 2
Boom! We've solved for 'y'. It's like finding the missing piece of a jigsaw puzzle – a satisfying moment of discovery. We now know that the value of 'y' that satisfies both equations in our original system is 2. This is a significant step forward, but we're not quite at the finish line yet. We've found 'y', but we still need to find 'x'. Think of it like a treasure hunt – we've found one treasure, but there's still another one waiting to be discovered.
Now that we know 'y', we can plug this value back into either of our original equations to solve for 'x'. This is the beauty of solving systems of equations – once you find one variable, you can use that information to find the others. It's like a domino effect, where solving for one variable sets off a chain reaction that leads to the solution for the rest. The choice of which equation to use is entirely up to us. We can pick the one that looks simpler or easier to work with. It's like choosing the right tool for the job – we want to use the one that will make the task as efficient and straightforward as possible.
For this example, let's use the first original equation:
3x + (1/2)y = 3
We'll substitute y = 2 into this equation:
3x + (1/2)(2) = 3
Simplifying, we get:
3x + 1 = 3
Now, we need to isolate 'x'. We'll subtract 1 from both sides:
3x = 2
And finally, we'll divide both sides by 3:
x = 2/3
The Grand Finale: The Solution Unveiled
We've reached the summit, guys! After navigating through the world of equations and variables, we've successfully solved our system. It's like completing a challenging quest in a video game – a sense of accomplishment and satisfaction washes over you. We started with a system of two equations, faced the challenge of eliminating 'x', and now we stand victorious, with the values of both 'x' and 'y' in hand. This journey is a testament to the power of algebraic manipulation and strategic problem-solving.
So, let's recap our adventure. We were given the following system of equations:
3x + (1/2)y = 3
6x - y = 2
Our mission was to figure out which operation would eliminate the 'x' terms when the equations were added together. We identified that multiplying the first equation by -2 was the key to success. This transformed the first equation into:
-6x - y = -6
When we added this modified equation to the second equation, the 'x' terms vanished in a puff of mathematical smoke, leaving us with:
-2y = -4
We then solved for 'y', finding that:
y = 2
With 'y' in our grasp, we plugged it back into one of the original equations to solve for 'x'. We chose the first equation and found:
x = 2/3
Therefore, the solution to our system of equations is x = 2/3 and y = 2. This means that the point (2/3, 2) is the intersection of the two lines represented by our equations. It's a geometric interpretation of our algebraic solution – the point where the two lines cross each other on a graph.
To truly appreciate the elegance of this solution, it's helpful to think about what we've accomplished. We've taken a seemingly complex problem and broken it down into manageable steps. We've used the principles of algebra to manipulate equations, eliminate variables, and ultimately arrive at a clear and precise answer. This is the essence of mathematical problem-solving – the ability to transform a challenge into a series of solvable steps.
And there you have it, guys! We've conquered another algebraic mountain. Remember, solving systems of equations is a fundamental skill in mathematics, and the techniques we've explored today – like elimination – are powerful tools in your problem-solving arsenal. Keep practicing, keep exploring, and you'll be solving equations like a pro in no time!