Solving Linear Systems of Three Equations

In algebra, a system of linear equations is a set of two or more linear equations that share the same variables. Solving a system of equations means finding the values of the variables that satisfy all the equations simultaneously. This means that when you substitute these values into each equation, the equation becomes a true statement.

One method for solving systems of equations is called the elimination method. This method involves manipulating the equations to eliminate one variable at a time. Once you've eliminated a variable, you can solve for the remaining variable and then substitute that value back into one of the original equations to find the value of the eliminated variable.

Let's look at how to solve a system of three linear equations with three unknowns using the elimination method.

Step 1: Choose two equations and eliminate one variable.

To eliminate a variable, you need to make sure the coefficients of that variable in the two equations are opposites. You may need to multiply one or both equations by a constant to achieve this. For example, if you have the equations:

Equation 1: 2x + 3y - z = 5

Equation 2: x - 2y + 3z = 8

You could multiply Equation 2 by -2 to get:

Equation 2: -2x + 4y - 6z = -16

Now, if you add Equation 1 and the modified Equation 2, the x terms will cancel out, leaving you with an equation with only y and z.

Step 2: Choose a different pair of equations and eliminate the same variable.

Now you have a new equation with just y and z. Choose a different pair of the original equations and eliminate the same variable (in this case, x). You may need to multiply one or both equations by a constant to achieve this. For example, you could choose Equation 1 and Equation 3:

Equation 1: 2x + 3y - z = 5

Equation 3: 3x + y + 2z = 10

You could multiply Equation 1 by -3 and Equation 3 by 2 to get:

Equation 1: -6x - 9y + 3z = -15

Equation 3: 6x + 2y + 4z = 20

Adding these equations will eliminate the x terms, leaving you with another equation with only y and z.

Step 3: Solve the resulting system of two equations with two unknowns.

You now have two equations with two unknowns (y and z). You can solve this system using any method you prefer, such as substitution or elimination. Once you solve for y and z, you can substitute these values back into any of the original equations to solve for x.

Example

Let's solve the following system of equations using the elimination method:

Equation 1: 2x + 3y - z = 5

Equation 2: x - 2y + 3z = 8

Equation 3: 3x + y + 2z = 10

**Step 1:** Eliminate x from Equation 1 and Equation 2.

Multiply Equation 2 by -2:

-2x + 4y - 6z = -16

Add this equation to Equation 1:

7y - 7z = -11

**Step 2:** Eliminate x from Equation 1 and Equation 3.

Multiply Equation 1 by -3:

-6x - 9y + 3z = -15

Multiply Equation 3 by 2:

6x + 2y + 4z = 20

Add these equations:

-7y + 7z = 5

**Step 3:** Solve the resulting system of two equations with two unknowns.

We now have the following system of equations:

7y - 7z = -11

-7y + 7z = 5

Notice that if we add these two equations, the y and z terms cancel out, leaving us with 0 = -6. This is a contradiction, indicating that the system of equations has no solution. A system of equations with no solution is considered **inconsistent**.

Summary

The elimination method is a powerful tool for solving systems of linear equations. It involves systematically eliminating variables by manipulating the equations until you have a single equation with one unknown. You can then solve for that unknown and substitute it back into the original equations to find the values of the other unknowns. It's important to remember that not all systems of equations have solutions, and those that do may have infinitely many solutions. By carefully applying the elimination method, you can determine whether a system has a solution and, if so, find the values of the variables that satisfy all the equations.