Eliminating Y: Multiplying Equations For Solution
Hey guys! Let's dive into a common math problem: eliminating variables in a system of equations. Specifically, we're tackling the question of how to strategically multiply one equation so that when we add it to another, a certain variable disappears. This is a super useful technique in algebra, and we're going to break it down with a clear example.
Understanding the Problem
So, the problem gives us two equations:
- Equation A: x + 3y = 10
- Equation B: 5x - 12y = 25
The core question we're trying to answer is: What number should we multiply Equation A by so that when we add the modified Equation A to Equation B, the 'y' terms cancel each other out? In other words, we want the 'y' terms to have opposite coefficients (like -12 and +12) so they become zero when added. This process is a key part of solving systems of equations, and it's a technique you'll use a lot in algebra and beyond. This method, often referred to as the elimination method, helps us simplify complex problems into more manageable steps. Understanding the underlying principle here is crucial: we're manipulating equations in a way that preserves their equality, allowing us to isolate variables and find solutions more easily. Think of it like balancing a scale; whatever we do to one side of the equation, we must do to the other to maintain the balance. By strategically choosing what to multiply by, we can create opportunities for elimination and move closer to solving for the unknowns.
The Elimination Strategy
To effectively eliminate the 'y' terms, we need to focus on their coefficients. Currently, we have +3y in Equation A and -12y in Equation B. Our goal is to make the 'y' coefficient in Equation A the opposite of the 'y' coefficient in Equation B, but with the same magnitude. This means we want to transform the +3y into a +12y. So, what do we need to multiply 3 by to get 12? That's right, it's 4!
This is the crucial step. We've identified the magic number that will help us eliminate 'y'. But remember, guys, in algebra, what we do to one term in an equation, we must do to all terms to maintain the equality. This means we'll be multiplying the entire Equation A by 4, not just the 'y' term. This principle is fundamental to solving equations. Think of an equation as a balanced scale; if you add or multiply something on one side, you must do the same on the other to keep the scale balanced. Ignoring this principle can lead to incorrect solutions and a lot of frustration! By applying this principle consistently, we ensure that we're not changing the solution set of the equation but simply transforming it into a more useful form. This might seem like a small detail, but it's a cornerstone of algebraic manipulation and problem-solving.
Multiplying Equation A
Let's multiply Equation A (x + 3y = 10) by 4. Remember to distribute the 4 to each term:
4 * (x + 3y) = 4 * (10)
This gives us:
4x + 12y = 40
We'll call this new equation Equation C. Notice how the 'y' term is now +12y, which is the opposite of the -12y in Equation B. This is exactly what we wanted! Now, we have a pathway to eliminate the 'y' variable, which is a major step forward in solving our system of equations. We've essentially set the stage for the next act in our algebraic drama. The beauty of this step is that we've transformed the equation without changing its fundamental truth. The original equation and the multiplied equation represent the same line on a graph, just expressed in different forms. This is a key concept in algebra: manipulating equations to make them more useful while preserving their underlying meaning. It's like reshaping a clay sculpture; you're changing the form, but the amount of clay remains the same.
Adding the Equations
Now, we add Equation C (4x + 12y = 40) to Equation B (5x - 12y = 25):
(4x + 12y) + (5x - 12y) = 40 + 25
Notice what happens to the 'y' terms: +12y and -12y cancel each other out, leaving us with:
9x = 65
Awesome! We've successfully eliminated 'y', and now we have a much simpler equation with only one variable, 'x'. This simplification is the power of the elimination method. By carefully choosing our multiplier, we transformed a two-variable problem into a single-variable one, making it significantly easier to solve. This step highlights the elegance of algebraic manipulation. We're not just blindly following rules; we're strategically using mathematical principles to simplify complex situations. It's like using a lever to lift a heavy object; we're applying a small amount of force in the right way to achieve a significant result. The elimination of 'y' is a critical milestone in our journey to solve the system of equations.
Solving for x
To solve for x, we simply divide both sides of the equation 9x = 65 by 9:
x = 65/9
So, we've found the value of x! This is a significant accomplishment. We've navigated through the complexities of the system of equations and successfully isolated one variable. This process underscores the step-by-step nature of algebraic problem-solving. Each step builds upon the previous one, leading us closer to the solution. Finding 'x' is like discovering a hidden clue in a mystery novel; it unlocks further insights and allows us to move closer to unraveling the entire puzzle. However, our job isn't quite done yet. We've only found the value of one variable. To fully solve the system, we need to find the value of 'y' as well. But don't worry, we're well on our way!
Finding the Value of y (Optional, but Recommended)
While the original question only asked for the value to multiply Equation A by, let's go the extra mile and find the value of 'y' to fully solve the system. We can substitute the value of x (65/9) into either Equation A or Equation B. Let's use Equation A (x + 3y = 10) for simplicity:
(65/9) + 3y = 10
Now, we solve for y:
3y = 10 - (65/9)
3y = (90/9) - (65/9)
3y = 25/9
y = (25/9) / 3
y = 25/27
Excellent! We've found the value of y. Now we have both x and y, which means we've completely solved the system of equations. This demonstrates the interconnectedness of the variables in a system. Finding the value of one variable often paves the way for finding the others. It's like assembling a jigsaw puzzle; each piece fits together to create a complete picture. Solving for 'y' provides a satisfying conclusion to our algebraic journey, showcasing the power and elegance of the elimination method. We've not only answered the original question but also gained a deeper understanding of how to solve systems of equations.
The Answer
So, to answer the original question: We must multiply Equation A by 4 to eliminate the 'y' terms when adding it to Equation B.
Key Takeaways
- The Elimination Method: This is a powerful technique for solving systems of equations by strategically eliminating variables.
- Finding the Right Multiplier: Look at the coefficients of the variable you want to eliminate and determine what number you need to multiply one equation by to create opposite coefficients.
- Multiplying the Entire Equation: Remember to multiply every term in the equation by the chosen value to maintain equality.
- Checking Your Work: If you have time, substitute your solutions for x and y back into the original equations to make sure they hold true. This is a great way to catch any errors.
Keep practicing, and you'll become a pro at eliminating variables! You've got this!
