Solving A Linear System: Find The Solution!

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Hey guys! Let's dive into the fascinating world of linear systems and learn how to crack them! Linear systems pop up everywhere, from simple math problems to complex engineering designs, so mastering them is super useful. In this article, we'll tackle the following linear system as an example:

x + y + z = 7
2x - y + z = 12
x - 2y - z = 0

We'll break down a straightforward method to find the solution. So, grab your thinking caps, and let's get started!

Understanding Linear Systems

Before we jump into solving, let's quickly recap what a linear system actually is. Think of it as a set of equations, each involving the same variables (like x, y, and z in our example), where the highest power of any variable is 1. These systems can represent relationships between different quantities, and finding the 'solution' means finding values for the variables that make all the equations true at the same time.

The solution to a linear system is a set of values for the variables that satisfy all the equations simultaneously. In simpler terms, it’s the point where all the lines (or planes, in 3D) intersect. Linear systems are incredibly important in various fields, from engineering and physics to economics and computer science. They help us model and solve real-world problems involving multiple interconnected variables.

For instance, in economics, you might use a linear system to model the supply and demand of a product. In engineering, it could be used to analyze the forces acting on a structure. The possibilities are endless! Now, let’s get into the nitty-gritty of how to solve these systems.

Method 1: Elimination - Our Primary Weapon

The elimination method is a classic technique for solving linear systems. The basic idea is to strategically add or subtract multiples of the equations to eliminate variables one by one. This simplifies the system until we can easily solve for the remaining variables. Let’s apply this method to our system:

x + y + z = 7   (Equation 1)
2x - y + z = 12  (Equation 2)
x - 2y - z = 0   (Equation 3)

Step 1: Eliminate 'y' from Equations 1 and 2

Notice that the 'y' terms in Equation 1 and Equation 2 have opposite signs (+y and -y). This makes them perfect candidates for elimination! Simply adding Equation 1 and Equation 2 will cancel out the 'y' variable:

(x + y + z) + (2x - y + z) = 7 + 12

This simplifies to:

3x + 2z = 19  (Equation 4)

Step 2: Eliminate 'y' again, this time from Equations 1 and 3

To eliminate 'y' from Equations 1 and 3, we need to make the coefficients of 'y' opposites. Multiply Equation 1 by 2:

2 * (x + y + z) = 2 * 7

Which gives us:

2x + 2y + 2z = 14 (Modified Equation 1)

Now, add Modified Equation 1 to Equation 3:

(2x + 2y + 2z) + (x - 2y - z) = 14 + 0

This simplifies to:

3x + z = 14   (Equation 5)

Step 3: Solve the New System (Equations 4 and 5)

We’ve now reduced our original three-equation system to a simpler two-equation system with two variables (x and z):

3x + 2z = 19  (Equation 4)
3x + z = 14   (Equation 5)

To eliminate 'x', subtract Equation 5 from Equation 4:

(3x + 2z) - (3x + z) = 19 - 14

This gives us:

z = 5

Awesome! We've found the value of z!

Step 4: Back-Substitute to Find 'x'

Now that we know z = 5, we can substitute this value back into either Equation 4 or Equation 5 to solve for 'x'. Let’s use Equation 5:

3x + 5 = 14

Subtract 5 from both sides:

3x = 9

Divide by 3:

x = 3

We've got x! x = 3. Things are really coming together now.

Step 5: Back-Substitute Again to Find 'y'

We have x = 3 and z = 5. Let's substitute these values back into any of the original equations to solve for 'y'. Let’s use Equation 1:

3 + y + 5 = 7

Combine the constants:

y + 8 = 7

Subtract 8 from both sides:

y = -1

And there we have it! We’ve found y = -1.

Step 6: State the Solution

We've found x = 3, y = -1, and z = 5. The solution to the linear system is the ordered triple (3, -1, 5). This means that these values, when plugged into the original equations, will make all three equations true.

Solution

The solution to the system of equations is:

  • x = 3
  • y = -1
  • z = 5

So, the solution set is (3, -1, 5). You can always double-check your answer by plugging these values back into the original equations to make sure they hold true!

Method 2: Substitution

The substitution method is another powerful way to solve linear systems. Instead of eliminating variables, we solve one equation for one variable and then substitute that expression into the other equations. This process reduces the number of variables until we can solve for them one by one.

Let's revisit our system:

x + y + z = 7   (Equation 1)
2x - y + z = 12  (Equation 2)
x - 2y - z = 0   (Equation 3)

Step 1: Solve for One Variable in One Equation

Look for an equation where it's easy to isolate a variable. Equation 1 looks promising. Let's solve it for 'x':

x = 7 - y - z   (Equation 4)

Step 2: Substitute into the Other Equations

Now, substitute this expression for 'x' (7 - y - z) into Equations 2 and 3:

Substitute into Equation 2:

2(7 - y - z) - y + z = 12

Simplify:

14 - 2y - 2z - y + z = 12

Combine like terms:

-3y - z = -2   (Equation 5)

Substitute into Equation 3:

(7 - y - z) - 2y - z = 0

Simplify:

7 - y - z - 2y - z = 0

Combine like terms:

-3y - 2z = -7   (Equation 6)

Step 3: Solve the New System (Equations 5 and 6)

We now have a system of two equations with two variables (y and z):

-3y - z = -2   (Equation 5)
-3y - 2z = -7   (Equation 6)

Let's use elimination here. Subtract Equation 6 from Equation 5:

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

Simplify:

z = 5

Great! We found z = 5 again.

Step 4: Back-Substitute to Find 'y'

Substitute z = 5 into either Equation 5 or 6. Let's use Equation 5:

-3y - 5 = -2

Add 5 to both sides:

-3y = 3

Divide by -3:

y = -1

We found y = -1, just like before!

Step 5: Back-Substitute Again to Find 'x'

Now, substitute y = -1 and z = 5 back into Equation 4 (the equation where we solved for x):

x = 7 - (-1) - 5

Simplify:

x = 7 + 1 - 5
x = 3

We have x = 3, matching our previous result.

Step 6: State the Solution

Using the substitution method, we've also found x = 3, y = -1, and z = 5. The solution to the linear system is (3, -1, 5).

Method 3: Matrices and Gaussian Elimination

For larger systems, matrices provide a compact and efficient way to represent and solve linear equations. Gaussian elimination is a systematic method that uses row operations to transform a matrix into an echelon form, making it easier to solve the system. This method is particularly useful when dealing with systems that have many variables and equations.

Step 1: Represent the System as a Matrix

Let's represent our system as an augmented matrix. The coefficients of the variables and the constants on the right-hand side form the matrix:

[ 1  1  1 | 7 ]
[ 2 -1  1 | 12]
[ 1 -2 -1 | 0 ]

Step 2: Perform Row Operations to Get Echelon Form

The goal is to transform the matrix into row-echelon form. This means we want to get zeros below the main diagonal. We can use the following row operations:

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

Step 2.1: Eliminate the 2 in the second row, first column

Multiply the first row by -2 and add it to the second row:

[ 1  1  1 | 7 ]
[ 0 -3 -1 | -2]
[ 1 -2 -1 | 0 ]

Step 2.2: Eliminate the 1 in the third row, first column

Multiply the first row by -1 and add it to the third row:

[ 1  1  1 | 7 ]
[ 0 -3 -1 | -2]
[ 0 -3 -2 | -7]

Step 2.3: Eliminate the -3 in the third row, second column

Subtract the second row from the third row:

[ 1  1  1 | 7 ]
[ 0 -3 -1 | -2]
[ 0  0 -1 | -5]

Now our matrix is in row-echelon form.

Step 3: Convert Back to Equations and Solve

Convert the matrix back into a system of equations:

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

From the last equation, we get z = 5.

Step 4: Back-Substitute

Substitute z = 5 into the second equation:

-3y - 5 = -2

Solve for y:

-3y = 3
y = -1

Substitute y = -1 and z = 5 into the first equation:

x + (-1) + 5 = 7

Solve for x:

x + 4 = 7
x = 3

Step 5: State the Solution

Using Gaussian elimination, we again find the solution to be x = 3, y = -1, and z = 5. The solution set is (3, -1, 5).

Conclusion

We've explored three different methods for solving linear systems: elimination, substitution, and Gaussian elimination. Each method has its strengths, and the best approach often depends on the specific system you're dealing with.

  • The elimination method is great when you can easily find coefficients that are opposites or multiples of each other.
  • The substitution method works well when you can easily isolate a variable in one of the equations.
  • Gaussian elimination is a systematic approach that's particularly useful for larger systems and can be implemented efficiently using computers.

No matter which method you choose, the key is to be organized and methodical. Double-check your work at each step to avoid errors. With practice, you'll become a pro at solving linear systems! Keep up the great work, and feel free to tackle more examples to sharpen your skills. You've got this!