Solving Linear Equations: Elimination/Reduction Method

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Hey guys! Today, we're diving into a classic math problem: solving a system of three linear equations using the elimination/reduction method. This method is super useful for tackling problems like the one we have here: x + 2y + z = 7, 3x + y - z = 5, and 2x + 3y - z = 3. If you've ever felt intimidated by these kinds of problems, don't worry! We'll break it down step-by-step so you can conquer them like a math pro.

Understanding the Problem

Before we jump into the solution, let's make sure we understand what we're dealing with. We have a system of three equations, each with three variables: x, y, and z. Our goal is to find the values of x, y, and z that satisfy all three equations simultaneously. Think of it like finding the exact point where three lines intersect in 3D space. The elimination/reduction method helps us do this by strategically adding or subtracting equations to eliminate variables one by one.

Solving systems of linear equations is a fundamental skill in mathematics with applications that extend far beyond the classroom. These skills are essential in various fields, including engineering, economics, computer science, and physics. In engineering, for instance, solving linear equations can help determine the forces acting on a structure or the flow of current in an electrical circuit. Economists use these equations to model supply and demand curves and predict market equilibrium. In computer science, they are crucial for algorithms related to computer graphics and optimization problems. Moreover, in physics, linear equations help in solving problems related to kinematics and dynamics, such as projectile motion and the interactions between forces. Therefore, mastering these methods is not just about solving math problems; it is about gaining tools that are applicable across numerous disciplines, enhancing your ability to tackle complex real-world challenges. Each step we take in understanding this method is a step towards broadening your mathematical toolkit and preparing you for advanced problem-solving scenarios.

Step-by-Step Solution

Okay, let's get our hands dirty and solve this thing! We'll go through each step in detail, so you can follow along easily.

Step 1: Choose a Variable to Eliminate

First, we need to decide which variable we want to eliminate first. Looking at our equations:

  1. x + 2y + z = 7
  2. 3x + y - z = 5
  3. 2x + 3y - z = 3

Notice that the 'z' variable has opposite signs in equations 2 and 3. This makes it a great candidate for elimination! It means we can easily get rid of 'z' by adding those two equations together.

The decision to eliminate a specific variable often hinges on identifying the most straightforward path to simplification. In our case, the 'z' variable presents a unique advantage because it appears with opposite signs in two equations. This strategic positioning allows for immediate cancellation through addition, reducing the complexity of the system. But, let's think about why this is so crucial. Imagine we chose a different variable, like 'x' or 'y,' which don't have such conveniently opposite signs. Eliminating those would require additional steps, such as multiplying one or both equations by a constant to align the coefficients. This not only adds more computational work but also increases the potential for making errors. Therefore, the intelligent selection of a variable to eliminate is a critical skill in efficiently solving linear systems. It reflects a deep understanding of the algebraic structures at play and an ability to strategize for the most streamlined solution process. This is the kind of insight that transforms a complex problem into a manageable task, illustrating why smart choices at the outset can significantly impact the ease and accuracy of the solution.

Step 2: Eliminate the Variable (z)

Let's add equations 2 and 3:

(3x + y - z) + (2x + 3y - z) = 5 + 3

This simplifies to:

5x + 4y = 8. Let's call this equation 4.

Now we have a new equation with only 'x' and 'y'! Awesome, right?

This step embodies the core principle of the elimination method: simplifying a system of equations by reducing the number of variables. When we added equations 2 and 3, the 'z' variable vanished, leaving us with an equation in just 'x' and 'y.' This transformation is a pivotal moment because it decreases the dimensionality of the problem. Think of it as moving from a three-dimensional challenge (three variables) to a two-dimensional one (two variables), which is inherently easier to visualize and solve. However, the elegance of this step might make us overlook some important nuances. For instance, the coefficients of the variables play a crucial role in whether we can directly add or subtract equations. If 'z' had coefficients other than 1 and -1, we would have needed to multiply one or both equations by a constant to ensure that 'z' would cancel out upon addition. This highlights the importance of carefully examining the equations and planning our moves strategically. The goal is not just to eliminate a variable but to do so in the most efficient way, minimizing the number of operations and the chances of making a mistake. This approach showcases how strategic algebraic manipulation can significantly impact the complexity of problem-solving, turning what might seem like a daunting task into a series of manageable steps.

Step 3: Eliminate the Same Variable Again

We need to eliminate 'z' again, but this time using a different pair of equations. Let's use equations 1 and 2. To eliminate 'z', we'll add equation 1 to equation 2:

(x + 2y + z) + (3x + y - z) = 7 + 5

Simplifying gives us:

4x + 3y = 12. Let's call this equation 5.

Now we have two equations (4 and 5) with only 'x' and 'y'. We're getting closer!

The repetition of the elimination process using a different pair of equations is a critical step in solidifying our reduction. By applying the same operation to another set of equations, we generate a second equation in the same two variables ('x' and 'y' in our case). This move is not merely redundant; it is essential for creating a solvable system. Think of it as building a foundation – one equation alone can only tell us so much, but two equations provide a framework for a unique solution. This step underscores a fundamental concept in linear algebra: to solve for 'n' variables, we generally need 'n' independent equations. Without this second equation, we would be stuck with an underdetermined system, meaning there are infinitely many solutions or no solutions at all. The act of strategically re-eliminating 'z' transforms the problem from a three-variable conundrum into a manageable two-variable system, setting the stage for the final algebraic maneuvers that will reveal the values of 'x' and 'y.' This methodical approach, characterized by thoughtful repetition, demonstrates the power of breaking down complex problems into smaller, solvable parts, reinforcing the importance of structured problem-solving in mathematics.

Step 4: Solve for x and y

Now we have a system of two equations with two variables:

  1. 5x + 4y = 8 (Equation 4)
  2. 4x + 3y = 12 (Equation 5)

We can use elimination again! Let's multiply equation 4 by 3 and equation 5 by -4 to eliminate 'y':

  • 3 * (5x + 4y) = 3 * 8 -> 15x + 12y = 24
  • -4 * (4x + 3y) = -4 * 12 -> -16x - 12y = -48

Now add these two equations:

(15x + 12y) + (-16x - 12y) = 24 + (-48)

This simplifies to:

-x = -24

So, x = 24! 🎉

Now that we have 'x', we can plug it into either equation 4 or 5 to solve for 'y'. Let's use equation 4:

5 * 24 + 4y = 8

120 + 4y = 8

4y = -112

y = -28! 🎉🎉

The maneuver of solving for 'x' and 'y' is the culmination of our variable reduction efforts, and it showcases the elegance of the elimination method. By strategically multiplying and adding equations, we've managed to isolate 'x,' giving us a definitive numerical value. This moment is particularly satisfying because it marks the transition from a system of interconnected equations to a concrete solution for one variable. However, let's not overlook the mathematical finesse at play here. The choice to multiply each equation by a specific factor (3 and -4 in our case) wasn't arbitrary; it was a calculated move to ensure that the 'y' terms would cancel out upon addition. This highlights the importance of foresight in algebraic problem-solving. Recognizing that we could align the coefficients of 'y' to be opposites allowed us to eliminate this variable efficiently, paving the way for a straightforward determination of 'x.' Furthermore, the subsequent substitution of 'x' back into one of the equations demonstrates a fundamental concept in algebra: once we solve for one variable in a system, we can use that value to find the others. This interconnectedness of variables within a system means that each solution we uncover unlocks further possibilities for discovery, transforming a complex puzzle into a series of manageable steps. This approach not only provides a solution but also reinforces the interconnected and strategic nature of algebraic problem-solving.

Step 5: Solve for z

We have x = 24 and y = -28. Now, we just need to plug these values into any of the original equations to solve for 'z'. Let's use equation 1:

24 + 2 * (-28) + z = 7

24 - 56 + z = 7

-32 + z = 7

z = 39! 🎉🎉🎉

Step 6: Verify the Solution

To make sure we're correct, let's plug x = 24, y = -28, and z = 39 into all three original equations and see if they hold true:

  1. 24 + 2*(-28) + 39 = 24 - 56 + 39 = 7 (Correct!)
  2. 3*24 + (-28) - 39 = 72 - 28 - 39 = 5 (Correct!)
  3. 224 + 3(-28) - 39 = 48 - 84 - 39 = -75 != 3 (Incorrect!)

Uh oh! It looks like there is an error with equation 3. Let’s re-evaluate the original calculation. The mistake lies in the arithmetic of verifying the solution for the third equation. Let’s correct it:

  1. 2 * 24 + 3 * (-28) - 39 = 48 - 84 - 39 = -75

There seems to be a transcription error in the original equation. Equation 3 should be:

2x + 3y - z = -39 instead of 2x + 3y - z = 3

Let's verify the solution again with the corrected equation 3:

  1. 224 + 3(-28) - 39 = 48 - 84 - 39 = -75 != -3 (Still Incorrect!)

It seems there may be other errors. We need to identify a mistake within the process, but we will proceed based on the assumption there were no errors made in copying the equations.

Let's plug x = 24 and y = -28 into equation 3 and solve for z:

2 * 24 + 3 * (-28) - z = 3

48 - 84 - z = 3

-36 - z = 3

-z = 39

z = -39

With z = -39, let's re-verify:

  1. 24 + 2*(-28) + (-39) = 24 - 56 - 39 = -71 != 7 (Incorrect!)

It appears there may be an error with the initial values and process.

Let's go through step 3 again using equations 1 and 3:

Eliminate 'z' by subtracting equation 3 from equation 1:

(x + 2y + z) - (2x + 3y - z) = 7 - 3

Simplifying gives us:

-x - y + 2z = 4

This equation does not eliminate z, so we have to try something else. We should add equation 1 to equation 3 instead:

(x + 2y + z) + (2x + 3y - z) = 7 + 3

Simplifying gives us:

3x + 5y = 10. Let's call this equation 5. (Previous equation 5: 4x + 3y = 12)

We need to re-evaluate using the new equations 4 and 5:

  1. 5x + 4y = 8 (Equation 4)
  2. 3x + 5y = 10 (Equation 5)

Let's multiply equation 4 by 5 and equation 5 by -4 to eliminate 'y':

  • 5 * (5x + 4y) = 5 * 8 -> 25x + 20y = 40
  • -4 * (3x + 5y) = -4 * 10 -> -12x - 20y = -40

Now add these two equations:

(25x + 20y) + (-12x - 20y) = 40 + (-40)

This simplifies to:

13x = 0

So, x = 0!

Now that we have 'x', we can plug it into either equation 4 or 5 to solve for 'y'. Let's use equation 4:

5 * 0 + 4y = 8

4y = 8

y = 2!

Now, we plug x = 0 and y = 2 into any of the original equations to solve for 'z'. Let's use equation 1:

0 + 2 * 2 + z = 7

4 + z = 7

z = 3!

Now we have: x = 0, y = 2, and z = 3

Now we need to verify our solution. If we plug in x = 0, y = 2, and z = 3 into the original equations, we get:

  1. 0 + 2(2) + 3 = 7 (Correct!)
  2. 3(0) + 2 - 3 = -1 != 5 (Incorrect!)

It appears the solution may be incorrect based on these calculations. Let’s double-check our work and see where we might have made a mistake.

After reviewing and recalculating, here's the correct verification process:

With the correct solution x=0, y=2, and z=3:

  1. 0 + 2(2) + 3 = 0 + 4 + 3 = 7 (Correct)
  2. 3(0) + 2 - 3 = 0 + 2 - 3 = -1 (Incorrect!)

Recalculating using equations 2 and 3:

Add 3x + y - z = 5 to 2x + 3y - z = 3 to get: 5x + 4y - 2z = 8

Multiply this new equation by 2 and we get: 10x + 8y - 4z = 16. Let's name this new equation equation 4.

Multiply x + 2y + z = 7 by 4 to get: 4x + 8y + 4z = 28. Let's name this new equation equation 5.

If we add equation 4 to equation 5, we get: 14x + 16y = 44. Let's name this equation 6.

Let's use equation 1 and equation 2 instead.

Multiply equation 2 by 1 to get 3x + y - z = 5 (equation 2).

Multiply equation 1 by 1 to get x + 2y + z = 7 (equation 1).

Let's add equation 1 to equation 2.

4x + 3y = 12 (new equation 4).

Let's use equations 2 and 3.

Multiply equation 2 by -1 to get -3x - y + z = -5.

Multiply equation 3 by 1 to get 2x + 3y - z = 3.

Let's add the modified equation 2 to equation 3.

-x + 2y = -2 (new equation 5).

New System:

4x + 3y = 12 (equation 4). -x + 2y = -2 (equation 5).

Let's multiply the new equation 5 by 4 to eliminate x.

4(-x + 2y) = 4(-2) = -4x + 8y = -8.

Adding equation 4 to modified equation 5:

4x + 3y + (-4x + 8y) = 12 + (-8)

11y = 4 y = 4/11

Plugging y = 4/11 into equation 5:

-x + 2(4/11) = -2 -x + 8/11 = -2 -x = -2 - 8/11 -x = -30/11 x = 30/11

Let's plug these numbers into the first equation to solve for z:

30/11 + 2(4/11) + z = 7 30/11 + 8/11 + z = 7 38/11 + z = 7 z = 7 - 38/11 z = (77 - 38)/11 z = 39/11

The solution is x = 30/11, y = 4/11, and z = 39/11.

With the solution x = 30/11, y = 4/11, and z = 39/11, let's re-verify against the original set of equations. This will provide us a true indication of whether our calculated values satisfy all conditions:

  1. Equation 1 Verification: x + 2y + z = 7 (30/11) + 2*(4/11) + (39/11) = 7 (30/11) + (8/11) + (39/11) = 7 77/11 = 7 (Correct!)

  2. Equation 2 Verification: 3x + y - z = 5 3*(30/11) + (4/11) - (39/11) = 5 (90/11) + (4/11) - (39/11) = 5 55/11 = 5 (Correct!)

  3. Equation 3 Verification: 2x + 3y - z = 3 2*(30/11) + 3*(4/11) - (39/11) = 3 (60/11) + (12/11) - (39/11) = 3 33/11 = 3 (Correct!)

We have found an error in the original solution and have re-verified our final solution, which satisfies the set of equations.

Final Answer:

x = 30/11 y = 4/11 z = 39/11

Conclusion

And there you have it! We've successfully solved the system of linear equations using the elimination/reduction method. It might seem like a lot of steps, but with practice, you'll become a pro at this! Remember, the key is to be organized and take it one step at a time. Solving linear equations is a foundational skill in math, and mastering it will open doors to more advanced topics. Keep practicing, and you'll be crushing these problems in no time! This detailed walkthrough should provide you with a clear understanding of the process. Don’t hesitate to try similar problems to further hone your skills. You got this!