Download Systems of Equations with THREE Variables

Systems of Equations with THREE Variables
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To solve a system of equations with three variables, you must have three equations. The idea
is to solve for the variables one-by-one until you have your ordered triple (x, y, z). Yes, this will
take a lot of space for each problem!
Sometimes you may find it easy to find your first answer:
x + 2y + z = 9
3y – z = -1
3z = 12
 The last equation only has one variable, so solve down and get z = 4.
Now you can replace the z back into the remaining two equations:
x + 2y + (4) = 9
3y – (4) = -1


x + 2y = 5
3y = 3
 Now this equation only has one variable, so
solve and get y = 1.
Now use the remaining equation to solve for x.
x + 2(1) = 5

x+2=5

x=3
Ordered triple is (x, y, z) = (3, 1, 4).
Sometimes you may find substitution to be a good idea (even though this was not one of
the favorite methods when doing two equations and two variables).
x+y+z=0
x+y=3

y+z=1
Notice how both of the first two equations have x + y.
(x + y) + z = 0

(3) + z = 0

z = -3
The two remaining equations are x + y = 3 and y + z = 1. Since x + y = 3 does not have a z in it,
it does us no good, but look at y + z = 1.
y+z=1

y + (-3) = 1

y=4
You still need x, so go back to one of the original equations that had an x in it and substitute in
the y and z values as appropriate. I am going to use the second equation.
x+y=3

x + (4) = 3

x = -1
Ordered triple is (-1, 4, -3).
CHECK your answer into EACH of the equations to make certain it works!
Three equations, three variables, but only two variables in each equation
4x + 9y = 8
8x + 6z = -1
6y + 6z = -1

There are three variables and three equations, but none of the
equations have all three variables. None of the equations have the same
two variables either. Substitution would not be ideal here.
Your first step is to decide which variable you want to make disappear first. You must use the
two equations that contain that particular variable.
I notice that both the second and third equations contain a 6z, so I am going to make z
disappear. (It is not always this easy.)

( -1) 
8x + 6z = -1
6y + 6z = -1
8x + 6z = -1
-6y – 6z = 1 Now “squush.”
8x – 6y = 0
Now use the 8x – 6y = 0 with the first equation which also has x and y. You must choose your
method. (Multiplication is the best method. Make x disappear.)
( 2) 
( -1) 
4x + 9y = 8
8x – 6y = 0
8x + 18y = 16
-8x + 6y = 0 Now “squush.”
24y = 16

y = 16
24
=
2
3
Now that we know what y is, we must find x and z. Go back to the original forms since each
only contained two variables.
4x + 9y = 8

6y + 6z = -1 
4x + 9(2/3) = 8

4x + 6 = 8

4x = 2

x=½
6(2/3) + 6z = -1

4 + 6z = -1

6z = -5

z = -5/6
Ordered triple is (1/2, 2/3, -5/6).
Sometimes substitution is not as obvious, does not work at all, or is too cumbersome to
use. Be creative. Look for a way using two equations to make a variable “disappear.”
x+ y+ z=6
2x – y + 3z = 9
- x + 2y + 2z = 9

One way of solving this one is to use the first two equations and
make y go away. Then use the last two equations and make the
same variable go away. Whatever you do, the same variable has to
go away so you work the system down from 3 equations and 3
variables to 2 equations and 2 variables.
First two equations:
Since the y values have opposite coefficients, you can eliminate by addition.
x+ y+ z=6
2x – y + 3z = 9
3x
+ 4z = 15
Last two equations:
Since the y coefficients are not the same or opposites, you must use multiplication.
2x – y + 3z = 9
- x + 2y + 2z = 9
(2)


4x – 2y + 6z = 18
- x + 2y + 2z = 9
3x
+ 8z = 27
Now use the two equations you found with only x’s and z’s.
Since the x coefficients are the same, we can use the subtraction method.
3x + 4z = 15
3x + 8z = 27

( -1) 
3x + 4z = 15
-3x – 8z = -27
-4z = -12

z = 3 (One down, two to go!)
Since we now have z, use one of the two equations you just used to find x. (Why x and not y?
The equations you just used only have TWO variables, and one of them you just solved for, and
the other is x.)
3x + 4z = 15 
3x + 4(3) = 15

3x + 12 = 15 
3x = 3 
x=1
We have now found x and z, but we still must find y. Go back to one of the original equations in
the system. I am going to use the first equation. Substitute the x and z we have found.
x+y+z=6 
(1) + y + (3) = 6

y+4=6

y=2
Ordered triple is (1, 2, 3).
It is still a good idea to CHECK the other two equations as well!
2x – y + 3z = 9

2(1) – (2) + 3(3) = 9 
2–2+9=9 
-x +2y + 2z = 9

-(1) + 2(2) + 2(3) = 9

-1 + 4 + 6 = 9
9=9 

9=9
Three equations, some with three variables each, some with only two variables
4x – 3y + 5z = 43
2x + y = 9
3y – 2z = -9

One has x, y, and z. One has x and y. One has y and z. Where do I
start?!!!
 Trying to eliminate a variable from the equation with x, y, and z
could be a pain, but decide which one you want to eliminate and
use the equation with two variables that has that same variable
present with the three-variable equation to do elimination.
 Substitution could be done (for example: solve down the 2x + y =
9 to get y = 9 – 2x), but it would be a lot of extra work because
you would have to do the substitution into BOTH of the other two
equations.
I have decided to eliminate the x, so I will need to use the two equations that contain an x.
4x – 3y + 5z = 43
2x + y
= 9

( -2) 
4x – 3y + 5z = 43
-4x – 2y
= -18
-5y + 5z = 25
 This gives an equation with y and z,
just like the third equation from the
original problem.
Use this new equation with the third equation from the original problem. You must choose who
to make disappear. I am going to eliminate the y.
-5y + 5z = 25
3y – 2z = -9
(3)
(5)


-15y + 15z = 75
15y – 10z = -45
5z = 30

z = 6 (One down, two to go!)
One of the original two-variable equations contained z, so use it to find another variable value.
3y – 2z = -9 
3y – 2(6) = -9

3y – 12 = -9 
3y = 3 
y=1
Now find the only remaining variable: x. Use any of the original equations containing an x. The
easiest to use is 2x + y = 9.
2x + y = 9

2x + (1) = 9 
2x = 8 
x=4
The ordered triple is (4, 1, 6).
CHECK!!!!
4x – 3y + 5z = 43
4(4) – 3(1) + 5(6) = 43
16 – 3 + 30 = 43
13 + 30 = 43
43 = 43 
2x + y = 9
2(4) + (1) = 9
8+1=9
9=9
3y – 2z = -9
3(1) – 2(6) = -9
3 – 12 = -9
-9 = -9 
Three equations, some with three variables each, some with only two variables
(a second example)
x–y–z=3
y+z=5

The second equation only has y and z. Use the first and last
equations to eliminate x.
x+y–z=8
Since the x coefficients are the same, subtract the second equation from the first. In this case,
not only will the x disappear, but the z will also.
x–y–z=3

x–y–z=3
x + y – z = 8 ( -1) 
-x – y + z = -8
- 2y
= -5

y = 5/2
Since the original second equation had only two variables to begin with, and one of them was y,
use it to solve for z at this point.
y+z=5

(5/2) + z = 5 
z = 5 – (5/2) 
z = 5/2
Use one of the other two original equations to find x. I am going to use the first one.
x–y–z=3 
x – (5/2) – (5/2) = 3 
Ordered triple is (8, 5/2, 5/2).
x–5=3

x=8
There are always special cases to deal with!
2x – y + z = 3
x – 3y + z = 4
-5x – 2z = -5

2x – y + z = 3 ( -3) 
x – 3y + z = 4

The last equation only has x and z, but the first two both have all three
variables. Since the last equation does not have a y value, make it
disappear from the first two equations. You will have to use the
multiplication method.
-6x + 3y – 3z = -9
x – 3y + z = 4
-5x
- 2z = -5
Now “squush”
As it turns out, the equation we derive from the first two equations is the same thing as the original third
equation! This means we are dealing with an infinite number of solutions.
Another special case problem:
x+z=5
y – z = -4
x+y=1

There are three variables and three equations, but none of the equations
have all three variables. None of the equations have the same two variables
either.
Choose one equation and solve that equation for one of its variables.
I am going to choose the first equation and solve it for x.
x+z=5

x = (5 – z)
The second equation does not have an x, but the third one does. Substitute for the x in the third
equation.
x+y=1

(5 – z) + y = 1 
Special case rears its ugly head again!!!!!
We have an infinite number of solutions.
y – z = -4
(Fix order and -5 from both sides.)
This equation is the same as the original second equation.
One more special case:
x+y+z=2
x – y + 2z = 3
3x + 5y + 2z = 6
Use first two equations and eliminate y.
x+y+z=2
x – y + 2z = 3
2x + 3z = 5
Use last two equations and eliminate y.
x – y + 2z = 3 (5) 
5x – 5y + 10z = 15
3x + 5y + 2z = 6
3x + 5y + 2z = 6
8x + 12z = 21
Use these two equations to continue solving. I am going to eliminate x.
2x + 3z = 5
( -4) 
-8x –12z = -20
8x + 12z = 21

8x + 12z = 21
0 = 1 FALSE!!!!!! No solution
or 