Solving Systems Of Equations By Elimination A Step By Step Guide

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Hey guys! Today, we're diving deep into the fascinating world of solving systems of equations using the elimination method. It might sound intimidating, but trust me, it's a super useful tool in your math arsenal. We'll break it down step-by-step, using a real-world example to make things crystal clear. So, buckle up and let's get started!

Understanding the Elimination Method

The elimination method, also known as the addition method, is a technique used to solve systems of linear equations. The basic idea is to manipulate the equations so that when you add them together, one of the variables cancels out, leaving you with a single equation in one variable. This makes it much easier to solve for that variable. Once you've found the value of one variable, you can substitute it back into either of the original equations to find the value of the other variable. Think of it like a mathematical magic trick – we're strategically eliminating a variable to reveal the solution!

Why Use Elimination?

You might be wondering, why bother with elimination when we have other methods like substitution? Well, elimination shines when the equations are set up in a way that makes it easy to cancel out a variable. For example, if you have 2x + y = 5 and x - y = 1, you can immediately see that adding the equations will eliminate y. In other cases, you might need to do a little bit of algebraic maneuvering first, but the payoff is often worth it. Elimination can be faster and less prone to errors than substitution in certain scenarios, especially when dealing with fractions or more complex equations.

Step-by-Step Guide to Elimination

Let's break down the elimination method into a series of clear steps. We'll then apply these steps to a specific example to solidify your understanding. Trust me, once you've mastered these steps, you'll be solving systems of equations like a pro!

  1. Align the Equations: First, make sure the equations are lined up vertically, with the variables and constants in the same columns (x-terms over x-terms, y-terms over y-terms, constants over constants). This makes it easier to see which terms might cancel out.
  2. Multiply (if necessary): If neither variable has coefficients that are opposites (like 3 and -3), you'll need to multiply one or both equations by a constant. The goal is to create coefficients that are opposites for one of the variables. For example, if you have 2x + y = 5 and x + 3y = 8, you could multiply the first equation by -3 to get -6x - 3y = -15. Now the y-terms have opposite coefficients.
  3. Add the Equations: Once you have opposite coefficients for one variable, add the equations together vertically. This will eliminate one variable, leaving you with a single equation in one variable.
  4. Solve for the Remaining Variable: Solve the resulting equation for the remaining variable. This is usually a straightforward algebraic step.
  5. Substitute and Solve: Substitute the value you just found back into either of the original equations (or any equation in the process). Solve for the other variable.
  6. Check Your Solution: Finally, check your solution by substituting the values of both variables into both original equations. If both equations hold true, you've found the correct solution!

Solving a System of Equations: A Detailed Example

Okay, let's put these steps into action. We're going to tackle the system of equations you provided, which involves fractions – don't worry, we'll handle them like champs!

4/9 x + 8/3 y = 88/3
5/81 x - 4/9 y = -88/27

This system might look a bit intimidating with all those fractions, but we're going to break it down step-by-step and conquer it. Remember, the key to success is to stay organized and follow the elimination method principles.

Step 1: Eliminate the Fractions (Optional but Recommended)

Fractions can sometimes make things look messier, so let's get rid of them first. To do this, we'll find the least common multiple (LCM) of the denominators in each equation and multiply the entire equation by that LCM.

  • Equation 1: The denominators are 9 and 3. The LCM of 9 and 3 is 9. So, we'll multiply the entire first equation by 9.
    9 * (4/9 x + 8/3 y) = 9 * (88/3)
4x + 24y = 264
  • Equation 2: The denominators are 81, 9, and 27. The LCM of 81, 9, and 27 is 81. So, we'll multiply the entire second equation by 81.
    81 * (5/81 x - 4/9 y) = 81 * (-88/27)
5x - 36y = -264

Now our system looks much cleaner:

4x + 24y = 264
5x - 36y = -264

Step 2: Prepare for Elimination

Now we need to decide which variable to eliminate. Looking at the coefficients, it might seem easier to eliminate y because 24 and 36 have a smaller LCM than 4 and 5. To eliminate y, we need to make the coefficients of y opposites. The LCM of 24 and 36 is 72. So, we'll multiply the first equation by 3 and the second equation by 2 to get 72y and -72y.

  • Multiply Equation 1 by 3:
    3 * (4x + 24y) = 3 * 264
12x + 72y = 792
  • Multiply Equation 2 by 2:
    2 * (5x - 36y) = 2 * (-264)
10x - 72y = -528

Our system now looks like this:

12x + 72y = 792
10x - 72y = -528

Step 3: Eliminate and Solve for x

Now we can add the equations together. Notice that the y terms will cancel out:

(12x + 72y) + (10x - 72y) = 792 + (-528)
22x = 264

Now, solve for x:

x = 264 / 22
x = 12

We've found the value of x! Great job!

Step 4: Substitute and Solve for y

Now we need to find the value of y. We can substitute x = 12 into any of the equations we've used so far. Let's use the simplified first equation (4x + 24y = 264) because it looks relatively easy to work with.

4 * 12 + 24y = 264
48 + 24y = 264
24y = 264 - 48
24y = 216
y = 216 / 24
y = 9

We've found the value of y! Our solution is x = 12 and y = 9.

Step 5: Check the Solution

It's always a good idea to check our solution to make sure we haven't made any mistakes. Let's substitute x = 12 and y = 9 into both of the original equations:

  • Original Equation 1:
    4/9 * 12 + 8/3 * 9 = 88/3
16/3 + 24 = 88/3
16/3 + 72/3 = 88/3
88/3 = 88/3  (True)
  • Original Equation 2:
    5/81 * 12 - 4/9 * 9 = -88/27
20/27 - 4 = -88/27
20/27 - 108/27 = -88/27
-88/27 = -88/27  (True)

Our solution checks out! Both equations are true when x = 12 and y = 9.

The Solution

The solution to the system of equations is x = 12 and y = 9. We can write this as an ordered pair: (12, 9).

Key Takeaways and Tips for Success

  • Organization is Key: Keep your work organized and write down each step clearly. This will help you avoid mistakes and make it easier to check your work.
  • Eliminate Fractions: If you're dealing with fractions, get rid of them early on by multiplying by the LCM of the denominators. This will simplify the equations and make them easier to work with.
  • Choose Wisely: When deciding which variable to eliminate, look for the easiest path. Sometimes multiplying one equation is enough, while other times you'll need to multiply both.
  • Check Your Work: Always check your solution by substituting the values back into the original equations. This is the best way to catch any errors.
  • Practice, Practice, Practice: The more you practice, the more comfortable you'll become with the elimination method. Work through various examples and challenge yourself with different types of systems of equations.

Conclusion

So there you have it! We've successfully solved a system of equations using the elimination method. Remember, the key is to understand the steps, stay organized, and practice regularly. With a little bit of effort, you'll be mastering this technique in no time. Keep up the great work, guys, and happy solving!