Unit 14.1 Systems of Equations with Three Variables
Systems of Equations with Three Variables
A system of equations with three variables typically involves three equations and can be solved using various methods such as substitution or elimination. Later in the unit, other methods are discussed. It follows the same processes as solving a systems of equations with 2 variables. Keep in mind that if you have 3 unknowns, you need 3 equations
Solving a systems of equations with 3 unknowns
Consider the following system of equations with three variables
2x – y + z = 4 (eq. 1)
x + 3y – 2z = -6 (eq. 2)
3x – 2y + 4z = 10 (eq. 3)
Solve for x, y and z. We will be using the elimination method
Pick any 2 equations to use the elimination method on first. We will use equation 1 and 2 and eliminate z.
Since we cannot directly get rid of z by adding, multiply equation 1 by 2 first
2 (2x – y + z = 4)
4x – 2y + 2z = 8 (eq. 1a)
Add equation 1a to equation 2 to eliminate z. Label the result as equation 4
4x – 2y + 2z = 8 (eq. 1a)
+ x + 3y – 2z = -6 (eq. 2)
5x + y = 2 (eq. 4)
Pick the remaining equation (in this case equation 3) and any other equation and eliminate the same variable. We will use equation 3 and equation 2
Since we cannot directly get rid of z with out multiplying, multiply equation 2 by 2 first
2 (x + 3y – 2z = -6)
2x + 6y – 4z = -12 (eq. 2a)
Add equation 2a to equation 3 to eliminate z Label the result as equation 5
2x + 6y – 4z = -12 (eq. 2a)
+ 3x – 2y + 4z = 10 (eq. 3)
5x + 4y = -2 (eq. 5)
On equations 4 and 5, use elimination to get rid of one of the variables. Since x has a coefficient of 5 in both, we can multiply one by -1 to get rid of x
-1 (5x + y = 2) (eq. 4)
-5x – y = -2 (eq. 4a)
Add equations 4a and 5 to get rid of x and solve for y
-5x – y = -2 (eq. 4a)
+ 5x + 4y = -2 (eq. 5)
3y = -4
y = –[latex]\frac{4}{3}[/latex]
Now that we have y, we can plug it into equation 4 or 5 to solve for x
5x + y = 2 (eq. 4)
5x + (-[latex]\frac{4}{3}[/latex]) = 2
5x = [latex]\frac{10}{3}[/latex]
x = [latex]\frac{2}{3}[/latex]
Two variables down, one to go. Pick any of the original equations and plug x and y into them to solve for z
2x – y + z = 4 (eq. 1)
2([latex]\frac{2}{3}[/latex]) – (-[latex]\frac{4}{3}[/latex]) + z = 4
z = [latex]\frac{4}{3}[/latex]
Our solutions are
x = [latex]\frac{2}{3}[/latex]
y = –[latex]\frac{4}{3}[/latex]
z = [latex]\frac{4}{3}[/latex]