Solve Systems of Three Linear Equations
When graphed, an equation that involves three variables in the form \(Ax+By+Cz=D\) is a plane that extends in all directions. Just like with a system of two linear equations, we can end up with graphs that have one solution, no solution, or infinite solutions.
When graphed, an equation that involves three variables in the form \(Ax+By+Cz=D\) is a plane that extends in all directions. Just like with a system of two linear equations, we can end up with graphs that have one solution, no solution, or infinite solutions.
In order to solve a system of three linear equations algebraically, we can still use substitution and elimination, but we will focus on elimination in this lesson (substitution often gets very messy to solve by hand). When solving, we will take two sets of two equations and eliminate the same variable in both. Let’s look at the following example.
\(\begin{cases} 2x-4y+7z=24 \\ -x+5y+3z=-2 \\ 3x+6y-5z=-7 \end{cases}\)
It doesn’t matter which variable we choose to eliminate first, but if we look at the coefficients, it will probably be easiest to eliminate \(x\) first. We need to choose two different sets of two equations (1st and 2nd, 2nd and 3rd, or 1st and 3rd) and use elimination to eliminate \(x\). For this example, let’s use the 1st and 2nd equations and then the 2nd and 3rd equations.
1st and 2nd Equations
\(\begin{cases} 2x-4y+7z=24 \\ -x+5y+3z=-2 \end{cases}\)
\(\begin{cases} 2x-4y+7z=24 \\ 2(-x+5y+3z=-2) \end{cases}\ \text{(Multiply second equation by 2.)}\)
\(\begin{cases} 2x-4y+7z=24 \\ -2x+10y+6z=-4 \end{cases}\)
\(6y+13z=20\ \text{(Add equations together to eliminate x.)}\)
2nd and 3rd Equations
\(\begin{cases} -x+5y+3z=-2 \\ 3x+6y-5z=-7 \end{cases}\)
\(\begin{cases} 3(-x+5y+3z=-2) \\ 3x+6y-5z=-7 \end{cases}\ \text{(Multiply first equation by 3.)}\)
\(\begin{cases} -3x+15y+9z=-6 \\ 3x+6y-5z=-7 \end{cases}\)
\(21y+4z=-13\ \text{(Add equations together to eliminate x.)}\)
We now have two equations that only involve two variables. This is why it’s very important to choose the same variable to eliminate from each set of two equations. We can solve this system of two equations using the methods covered in the section on Solving Systems of Two Linear Equations. We will then have two of the three variables we are solving for.
\(\begin{cases} 6y+13z=20 \\ 21y+4z=-13 \end{cases}\)
\(\begin{cases} 7(6y+13z=20) \\ -2(21y+4z=-13) \end{cases}\)
\(\begin{cases} 42y+91z=140 \\ -42y-8z=26 \end{cases}\)
\(83z=166\)
\(z=2\)
\(6y+13(2)=20\)
\(6y+26=20\)
\(6y=-6\)
\(y=-1\)
Now that we know what \(y\) and \(z\) are, we can substitute those two values back into one of the three original equations to solve for \(x\).
\(2x-4y+7z=24\)
\(2x-4(-1)+7(2)=24\)
\(2x+4+14=24\)
\(2x+18=24\)
\(2x=6\)
\(x=3\)
We can write our answer in the form \((x,y,z)\) which is an ordered triple. So the solution to this system is \((3,-1,2)\).
If at any point in the process, all of the variables are eliminated and you have a false statement, then you can say there is no solution to the system. Proving that there are infinite solutions requires several steps of true statements, and that will be saved for a future course.
Example 1: Solve the system of three equations: \(\begin{cases} 3x+y-4z=0 \\ -2x-3y+5z=-21 \\ 5x+4y-3z=9 \end{cases}\)
Quick Check
Solve the system of three equations.
\(\begin{cases} -x+2y+8z=4 \\ 2x-5y-7z=13 \\ 7x+4y-9z=12 \end{cases}\)
Quick Check Solutions
Solve the system of three equations.
\(\begin{cases} -x+2y+8z=4 \\ 2x-5y-7z=13 \\ 7x+4y-9z=12 \end{cases}\)
Quick Check Solutions