3.6 Systems with Three
Variables
There are three strategies for solving
systems with three equations:
1. Solve by elimination
2. Solve an equivalent system
3. Solving by substitution
Solve by elimination
x – 3y +3z = -4
2x + 3y – z = 15
4x – 3y –z = 19
Why pick elimination?
Note that the y values in this
set are inverses. They will go
away once you add the
equations together.
Step 1: Pair the equation to eliminate y, because the y
terms are already additive inverses
x – 3y +3z = -4
2x + 3y – z = 15
3x
+2z = 11
2x + 3y – z = 15
4x – 3y –z = 19
6x
-2z =34
Step 2: Take the two new equations and use elimination to
solve for the remaining two variables (in this case, x and z)
3x +2z = 11
6x – 2z = 34
9x
= 45
x=5
Plug in x into either equation
to solve for z
3 (5) + 2z = 11
15 + 2z = 11
2z = -4
z = -2
Step 3: Substitute the values for x and z back into one of
the original equations (whichever seems easier) and solve
for y
x – 3y +3z = -4
2x + 3y – z = 15
4x – 3y –z = 19
(5, 1, -2) is the solution
x=5, z =-2
x – 3y +3z = -4
5 – 3y + 3(-2)=-4
5 – 3y -6 = -4
-3y = -3
y=1
3x3 Systems #1
2x + y – z = 5
3x - y + 2z = -1
x-y - z =0
3x3 Systems #1
2x + y – z = 5
3x - y + 2z = -1
x-y - z =0
Solving an Equivalent System
What is an equivalent system?
2x + y – z = 5
x + 4y + 2z = 16
15x + 6y - 2z = 12
Very similar to the direct elimination
method, but in this case you may need
to multiply one or more of the equations
by a constant in order for a variable to
drop out
Step 1: Pair the equation to eliminate z, because the z
terms are already additive inverses in equations #2
2x + y - z = 5
x + 4y +2z = 16
x + 4y + 2z = 16
15x + 6y - 2z = 12
Note: the Zs are inverses in system #2 but not
in #1. Multiply the first equation by 2 to get the
Zs so that they will drop out
Step 2: Manipulate the first equation by multiplying by 2.
Now the z variable will drop out of the equations when you
combine them.
(2x + y - z = 5)2
x + 4y +2z = 16
4x + 2y -2 z = 10
x + 4y +2z = 16
5x + 6y
=26
x + 4y + 2z = 16
15x + 6y - 2z = 12
16x + 10y
= 28
These equations are all
ready set up to allow us
to eliminate z. Simply add
together to get your two
variable equation.
Step 3: Take the two new equations and use elimination to
solve for the remaining two variables (in this case, x and y)
5x + 6y =16
16x + 10y = 28
Again, we do not
have a variable
that simply drops
out for us. Get a
common multiple
that you can
eliminate. In this
case the easiest is
to get the Ys to
cancel
(5x + 6y =16) 5
(16x + 10y = 28) -3
25x + 30y = 130
-48x - 30y = -84
-23x
x = -2
= 46
Plug x= -2 into either equation to solve for y.
5x + 6y =26
5(-2) + 6y =26
16x + 10y = 28
-10 + 6y = 26
6y = 36
Y=6
Step 4: Substitute the values for x and y back into one of
the original equations (whichever seems easier) and solve
for y
2x + y – z = 5
2x + y – z = 5
x + 4y + 2z = 16
15x + 6y - 2z = 12
(-2, 6, -3) is the solution
2 (-2) + (6) – z = 5
-4 +6 –z =5
2–z=5
-z = 3
z = -3
3x3 Systems #2
x + 2y + z =10
2x – y + 3z = -5
2x – 3y – 5z = 27
Note: this ALMOST
Looks like we could solve
The system by elimination.
What do we have to do to
make one if the variables
Inverse of one another?
-2(x + 2y + z =10)
2x – y + 3z = -5
2x – 3y – 5z = 27
Now try on your own
-2x - 4y -2z =-20
2x – y + 3z = -5
2x – 3y – 5z = 27
Now try on your own
-2x - 4y -2z =-20
2x – y + 3z = -5
2x – 3y – 5z = 27
Solve by Substitution
X – 2y + z = -4
-4x + y -2z = 1
2x + 2y – z = 10
Why pick substitution?
Note in each equation a
different variable is by itself.
Step 1: Choose one equation to solve for one of its variables.
x - 2y + z = -4
x – 2y = -z -4
x = 2y – z - 4
X = 2y – z - 4
Step 2: Substitute the expression for x into each of the
other two equations.
-4x + y -2z = 1
-4(2y – z – 4) + y -2z = 1
-8y + 4z + 16 + y -2z = 1
-7y + 2z + 16 = 1
-7y+ 2z = -15
2x + 2y –z =10
2(2y – z – 4) + 2y –z =10
4y – 2z – 8 + 2y –z =10
6y – 3z – 8 =10
6y -3z = 18
Step 3: Take the remaining equations and solve the system
to solve for z and y.
-7y + 2z = -15
6y – 3z = 18
We need to
eliminate a
variable.
Let’s get rid
of z since on
of the terms
is already
negative
(-7y + 2z = -15)3
(6y – 3z = 18)2
-21y + 6z = -45
12y – 6z = 36
-9y
y =1
= -9
-7y + 2z = -15
Plug y into
one of the
equations to
solve for z
-7(1) + 2z = -15
2z = -8
z = -4
Step 4: Substitute the values for y and z into one of the original equations
x – 2y + z = -4
-4x + y -2z = 1
2x + 2y – z = 10
(2, 1, -4) is the solution to the
system of equations.
x- 2y + z = -4
x -2(1) + (-4) = -4
x – 2 – 4 = -4
x – 6 = -4
x=2
3x3 Systems #3
x – 3y + z =6
2x – 5y – z = -2
-x + y + 2z = 7
3x3 Systems #3
x – 3y + z = 6
2x – 5y – z = -2
-x + y + 2z = 7
3x3 Systems #4
6a + 12b – 8c = 24
9a + 18b – 12c = 30
4a + 8b – 7c = 26
Hint: can you factor?
3x3 Systems #4
6a + 12b – 8c = 24
9a + 18b – 12c = 30
4a + 8b – 7c = 26
The a, b, and c terms of the first two equations
are the same. They exist as parallel planes,
so there will be no solution to this set of
equations!
3x3 Systems #5
4x – 6y + 4z = 12
6x – 9y + 6z = 18
5x – 8y + 10z =20
3x3 Systems #5
4x – 6y + 4z = 12
6x – 9y + 6z = 18
5x – 8y + 10z =20
The first two equations will be planes that lie on top of each other.
Equations 1 and 3 (or 2 and 3)
are planes that will intersect.
How do planes intersect?
A Line
We are looking for one point where all three planes will
intersect. In this case we say that there is no unique
solution.