Solving Linear Systems 5x + Y + 3z = 76, -x + 2y + 5z = 35, And -4x - 5y + Z = 22

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Hey guys! Let's dive into solving a system of linear equations. This type of problem pops up all the time in math, physics, engineering, and even economics. We're going to tackle the system:

  • 5x + y + 3z = 76
  • -x + 2y + 5z = 35
  • -4x - 5y + z = 22

We'll explore a few different methods to crack this, making sure you understand each step. So, buckle up and let's get started!

Understanding Linear Systems

Before we jump into solving, let's make sure we're all on the same page about what a linear system actually is. At its heart, a linear system is just a set of two or more linear equations that we're trying to solve simultaneously. Each equation represents a line (in 2D) or a plane (in 3D), and the solution to the system is the point (or set of points) where all the lines or planes intersect. Think of it like finding the common ground for all the equations.

In our case, we have three equations with three unknowns (x, y, and z). Each equation represents a plane in 3D space, and we're looking for the point where all three planes intersect. This point, represented by the values of x, y, and z, will satisfy all three equations at the same time. That's the solution to our system.

Now, why do these systems matter? Well, they're incredibly versatile for modeling real-world situations. For instance, you might use a linear system to:

  • Balance chemical equations: The coefficients in a balanced equation must satisfy a system of linear equations.
  • Analyze electrical circuits: Kirchhoff's laws lead to linear systems that describe the flow of current.
  • Optimize resource allocation: Businesses use linear programming (which relies on solving linear systems) to decide how to best use their resources.
  • Predict weather patterns: Complex weather models often involve solving large systems of linear equations.

The beauty of linear systems lies in their predictability and the well-established methods we have for solving them. We can use techniques like substitution, elimination, and matrices to systematically find the solutions, even for very large systems. So, understanding linear systems is a fundamental skill in many fields.

Method 1: Elimination Method

The elimination method, also known as the addition method, is a classic technique for solving systems of linear equations. The core idea behind this method is to strategically manipulate the equations so that when we add them together, one or more variables cancel out. This reduces the system to a simpler one with fewer variables, making it easier to solve. We then repeat this process until we can isolate the value of one variable, and then back-substitute to find the others.

Let's break down how we can apply the elimination method to our system:

  1. Label the equations: It's helpful to label our equations for easy reference. Let's call them:

    • Equation 1: 5x + y + 3z = 76
    • Equation 2: -x + 2y + 5z = 35
    • Equation 3: -4x - 5y + z = 22

  2. Choose a variable to eliminate: We need to pick a variable that we can eliminate by adding multiples of two equations. Looking at our system, the 'x' variable seems like a good candidate because the coefficients have opposite signs in Equation 1 and Equation 2. Let's aim to eliminate 'x' first.

  3. Multiply equations to match coefficients: To eliminate 'x' from Equation 1 and Equation 2, we need the coefficients of 'x' to be opposites. We can achieve this by multiplying Equation 2 by 5. This gives us a new equation:

    • 5 * (Equation 2): -5x + 10y + 25z = 175

  4. Add equations to eliminate the variable: Now, we can add the modified Equation 2 to Equation 1:

    • (5x + y + 3z) + (-5x + 10y + 25z) = 76 + 175
    • This simplifies to: 11y + 28z = 251. Let's call this Equation 4.

  5. Repeat the process to eliminate the same variable again: We need to eliminate 'x' again, but this time using a different pair of equations. Let's use Equation 1 and Equation 3. To eliminate 'x', we can multiply Equation 2 by -4 and Equation 3 by 5, and add the resulting equation into Equation 1. The new equations are:

    • -4 * (Equation 2): 4x - 8y - 20z = -140
    • 5 * (Equation 3): -20x - 25y + 5z = 110

    Adding these together along with Equation 1 (multiplied by 4) does not result in eliminating x. A different approach would be to use Equation 3 and Equation 1. To eliminate x, multiply Equation 2 by -4. This gives us 4x - 8y - 20z = -140. We can then add this to Equation 3:

    • (-4x - 5y + z) + (4x - 8y - 20z) = 22 + (-140)
    • This simplifies to: -13y - 19z = -118. Let's call this Equation 5.

  6. Solve the new system of two equations: Now we have two equations (Equation 4 and Equation 5) with two unknowns (y and z):

    • Equation 4: 11y + 28z = 251
    • Equation 5: -13y - 19z = -118

    We can use elimination again to solve this smaller system. To eliminate 'y', we can multiply Equation 4 by 13 and Equation 5 by 11, and add the resulting equations:

    • 13 * (Equation 4): 143y + 364z = 3263
    • 11 * (Equation 5): -143y - 209z = -1298
    • Adding these gives us: 155z = 1965
    • Solving for z: z = 1965 / 155 = 12.6774 (approximately 13)

  7. Back-substitute to find the other variables: Now that we have the value of z, we can substitute it back into either Equation 4 or Equation 5 to find 'y'. Let's use Equation 4:

    • 11y + 28 * (13) = 251
    • 11y + 364 = 251
    • 11y = -113
    • y = -113 / 11 = -10.2727 (approximately -10)

  8. Back-substitute again to find the last variable: Finally, we can substitute the values of 'y' and 'z' back into any of the original equations (Equation 1, Equation 2, or Equation 3) to find 'x'. Let's use Equation 1:

    • 5x + (-10) + 3 * (13) = 76
    • 5x - 10 + 39 = 76
    • 5x + 29 = 76
    • 5x = 47
    • x = 47 / 5 = 9.4 (approximately 9)

  9. Write the solution as an ordered triple: The solution to the system is approximately (9, -10, 13). We can check our solution by plugging these values back into the original equations to make sure they hold true. This method emphasizes manipulating equations to systematically eliminate variables, leading us to a solution in a step-by-step manner. The approximate solution we found may not be precise due to rounding, but it gives us a good idea of the solution.

Method 2: Substitution Method

The substitution method is another powerful technique for tackling systems of linear equations. The core idea here is to solve one equation for one variable, and then substitute that expression into the other equations. This effectively reduces the number of variables in the remaining equations, making the system simpler to solve. We repeat this process until we have an equation with only one variable, which we can easily solve. Then, we back-substitute to find the values of the other variables.

Let's see how the substitution method works with our system:

  1. Label the equations: As before, let's label our equations:

    • Equation 1: 5x + y + 3z = 76
    • Equation 2: -x + 2y + 5z = 35
    • Equation 3: -4x - 5y + z = 22

  2. Choose an equation and solve for one variable: We want to pick an equation and a variable that are easy to isolate. Looking at our system, the 'y' variable in Equation 1 seems like a good choice because it has a coefficient of 1. Let's solve Equation 1 for 'y':

    • 5x + y + 3z = 76
    • y = 76 - 5x - 3z. Let's call this Equation 4.

  3. Substitute the expression into the other equations: Now we substitute the expression for 'y' (Equation 4) into the other two equations (Equation 2 and Equation 3). This will eliminate 'y' from these equations:

    • Substituting into Equation 2:
      • -x + 2(76 - 5x - 3z) + 5z = 35
      • -x + 152 - 10x - 6z + 5z = 35
      • -11x - z = -117. Let's call this Equation 5.
    • Substituting into Equation 3:
      • -4x - 5(76 - 5x - 3z) + z = 22
      • -4x - 380 + 25x + 15z + z = 22
      • 21x + 16z = 402. Let's call this Equation 6.

  4. Solve the new system of two equations: Now we have two equations (Equation 5 and Equation 6) with two unknowns (x and z):

    • Equation 5: -11x - z = -117
    • Equation 6: 21x + 16z = 402

    We can use substitution or elimination to solve this smaller system. Let's use substitution again. Solve Equation 5 for 'z':

    • z = 117 - 11x. Let's call this Equation 7.

    Substitute Equation 7 into Equation 6:

    • 21x + 16(117 - 11x) = 402
    • 21x + 1872 - 176x = 402
    • -155x = -1470
    • x = -1470 / -155 = 9.4838 (approximately 9.48)

  5. Back-substitute to find the other variables: Now that we have the value of 'x', we can substitute it back into Equation 7 to find 'z':

    • z = 117 - 11 * (9.4838)
    • z = 117 - 104.3218
    • z = 12.6782 (approximately 12.68)

  6. Back-substitute again to find the last variable: Finally, we can substitute the values of 'x' and 'z' back into Equation 4 to find 'y':

    • y = 76 - 5 * (9.4838) - 3 * (12.6782)
    • y = 76 - 47.419 - 38.0346
    • y = -9.4536 (approximately -9.45)

  7. Write the solution as an ordered triple: The solution to the system is approximately (9.48, -9.45, 12.68). Again, this is an approximate solution due to rounding. The substitution method shines when one variable can be easily isolated, allowing us to systematically reduce the complexity of the system. This approach is particularly useful when you have a variable with a coefficient of 1 or -1.

Method 3: Using Matrices (Gaussian Elimination)

Now, let's crank things up a notch and use matrices to solve our linear system. This method, often called Gaussian elimination, provides a very organized and efficient way to solve even larger systems of equations. The basic idea is to represent the system as a matrix, and then use row operations to transform the matrix into a simpler form (specifically, row-echelon form or reduced row-echelon form) from which we can easily read off the solution.

Here's the breakdown of how to use Gaussian elimination:

  1. Write the augmented matrix: First, we represent our system of equations as an augmented matrix. This matrix consists of the coefficients of the variables and the constants on the right-hand side of the equations:

    [ 5  1  3 | 76 ]
[ -1 2  5 | 35 ]
[ -4 -5 1 | 22 ]

    The vertical line separates the coefficient matrix from the constant terms.

  2. Perform row operations to get to row-echelon form: Our goal is to transform the matrix into row-echelon form. This means we want to get leading 1s (called pivots) in each row, with zeros below the pivots. The allowed row operations are:

    • Swapping two rows.
    • Multiplying a row by a non-zero constant.
    • Adding a multiple of one row to another row.

    Let's apply these operations to our matrix:

    • Step 1: Get a 1 in the top-left position. We can swap Row 1 and Row 2, and multiply Row 2 by -1:

      [ -1  2  5 | 35 ]  // Row 2
[ 5  1  3 | 76 ]  // Row 1
[ -4 -5 1 | 22 ]

      [ 1 -2 -5 | -35 ]  // Row 1 * -1
[ 5  1  3 | 76 ]
[ -4 -5 1 | 22 ]

    • Step 2: Get zeros below the leading 1 in the first column. We can achieve this by:

      • Replacing Row 2 with Row 2 - 5 * Row 1
      • Replacing Row 3 with Row 3 + 4 * Row 1
      [ 1 -2 -5 | -35 ]
[ 0 11 28 | 251 ]
[ 0 -13 -19 | -118 ]

    • Step 3: Get a leading 1 in the second row. Divide Row 2 by 11:

      [ 1 -2 -5 | -35 ]
[ 0 1 28/11 | 251/11 ]
[ 0 -13 -19 | -118 ]

    • Step 4: Get a zero below the leading 1 in the second column. Replace Row 3 with Row 3 + 13 * Row 2:

      [ 1 -2 -5 | -35 ]
[ 0 1 28/11 | 251/11 ]
[ 0 0 155/11 | 1965/11 ]

    • Step 5: Get a leading 1 in the third row. Multiply Row 3 by 11/155:

      [ 1 -2 -5 | -35 ]
[ 0 1 28/11 | 251/11 ]
[ 0 0 1 | 12.6774 ] // Approximately

    Now the matrix is in row-echelon form.

  3. Perform back-substitution or further row operations to get to reduced row-echelon form: We can either back-substitute from the row-echelon form, or continue with row operations to get to reduced row-echelon form (where we also have zeros above the leading 1s). Let's continue with row operations to get to reduced row-echelon form.

    • Step 6: Get zeros above the leading 1 in the third column.

      • Replace Row 2 with Row 2 - (28/11) * Row 3
      • Replace Row 1 with Row 1 + 5 * Row 3
      [ 1 -2 0 | 28.3871 ] // Approximately
[ 0 1 0 | -10.2727 ] // Approximately
[ 0 0 1 | 12.6774 ] // Approximately

    • Step 7: Get a zero above the leading 1 in the second column. Replace Row 1 with Row 1 + 2 * Row 2:

      [ 1 0 0 | 9.4 ] // Approximately
[ 0 1 0 | -10.2727 ] // Approximately
[ 0 0 1 | 12.6774 ] // Approximately

    Now the matrix is in reduced row-echelon form.

  4. Read off the solution: The matrix in reduced row-echelon form directly gives us the solution:

    • x ≈ 9.4
    • y ≈ -10.27
    • z ≈ 12.68

    So, the solution is approximately (9.4, -10.27, 12.68).

The matrix method, especially Gaussian elimination, is a powerful and systematic way to solve linear systems. It's particularly well-suited for larger systems and can be easily implemented in computer programs. The key is to master the row operations and apply them strategically to get the matrix into the desired form.

Comparing the Methods

Okay, we've tackled this linear system using three different methods: elimination, substitution, and matrices (Gaussian elimination). Each method has its strengths and weaknesses, and the best choice often depends on the specific system you're dealing with and your personal preference.

  • Elimination Method:
    • Strengths: Conceptually straightforward, especially for smaller systems. It's good for eliminating variables in a systematic way. It can be particularly efficient when coefficients line up nicely for elimination.
    • Weaknesses: Can become cumbersome with larger systems or systems with fractions. It requires careful bookkeeping to avoid errors.
  • Substitution Method:
    • Strengths: Effective when one variable can be easily isolated. It reduces the system's complexity by decreasing the number of variables in each step.
    • Weaknesses: Can lead to messy expressions if substitutions involve fractions or complex terms. It may not be the best choice if no variable is easily isolated.
  • Matrices (Gaussian Elimination):
    • Strengths: Very systematic and efficient, especially for larger systems. It's well-suited for computer implementation. The process is very structured, which reduces the chance of errors.
    • Weaknesses: Can feel abstract initially. It requires understanding matrix operations. It might be overkill for very small systems.

In our example, all three methods led us to the same approximate solution (with slight variations due to rounding). For this particular system, the elimination method might have been slightly quicker due to the presence of some simple coefficients. However, for a much larger system, Gaussian elimination would likely be the most efficient choice.

Ultimately, the best method is the one you're most comfortable with and can apply accurately. It's a good idea to practice all three methods so you can choose the most appropriate one for any given problem. And remember, checking your solution by plugging it back into the original equations is always a good idea!

Conclusion

Alright, guys, we've journeyed through the world of linear systems and explored three powerful methods for solving them: elimination, substitution, and Gaussian elimination using matrices. We cracked the system 5x + y + 3z = 76, -x + 2y + 5z = 35, and -4x - 5y + z = 22 using each approach, highlighting the strengths and nuances of each technique.

Solving linear systems is a fundamental skill in mathematics and its applications. These systems pop up everywhere, from balancing chemical equations to optimizing resource allocation in business. By mastering these methods, you've armed yourselves with valuable tools for tackling a wide range of problems.

Remember, each method has its sweet spot. Elimination is great for smaller systems with easily matched coefficients. Substitution shines when you can isolate a variable quickly. And Gaussian elimination using matrices is the champ for larger, more complex systems. The key is to practice and get comfortable with each approach so you can choose the best one for the job.

So, keep practicing, keep exploring, and keep those linear systems solved! You've got this!