3-5 Systems With Three Variables
Quick Review
To solve a system of three equations, either pair two equations and eliminate the same variable from both equations, using one equation twice, or choose an equation, solve for one variable, and substitute the expression for that variable into the other two equations. Then, solve the remaining system.
Example
Solve by elimination.
1
2
3
{
x
+
y
+
z
=
10
2
x
−
y
+
z
=
9
−
3
x
+
2
y
+
2
z
=
5
table with 1 row and 2 columns , row1 column 1 , table with 3 rows and 1 column , row1 column 1 , begin circle , 1 , end circle , row2 column 1 , begin circle , 2 , end circle , row3 column 1 , begin circle , 3 , end circle , end table , column 2 left brace . table with 3 rows and 4 columns , row1 column 1 , x plus , column 2 y plus , column 3 z equals , column 4 10 , row2 column 1 , 2 x minus , column 2 y plus , column 3 z equals , column 4 9 , row3 column 1 , negative 3 x plus , column 2 2 y plus , column 3 2 z equals , column 4 5 , end table , end table
| Add equations
1
begin circle , 1 , end circle and
2
begin circle , 2 , end circle to eliminate y. |
4
3
x
+
2
z
=
19
begin circle , 4 , end circle 3 x plus 2 z equals 19
|
| Add 2 times
2
begin circle , 2 , end circle to
3
begin circle , 3 , end circle to eliminate y. |
5
x
+
4
z
=
23
begin circle , 5 , end circle x plus 4 z equals 23
|
| Add
−
3
negative 3 times
5
begin circle , 5 , end circle to
4
begin circle , 4 , end circle to eliminate x. |
z
=
5
z equals 5
|
| Substitute z = 5 into
5
.
begin circle , 5 , end circle .
|
x
=
3
x equals 3
|
| Substitute z = 5 and x = 3 into
1
begin circle , 1 , end circle or
2
.
begin circle , 2 , end circle .
|
y
=
2
y equals 2
|
| The solution to the system is (3, 2, 5). |
|
Exercises
Solve each system by elimination.
-
{
x
+
y
−
2
z
=
8
5
x
−
3
y
+
z
=
−
6
−
2
x
−
y
+
4
z
=
−
13
left brace . table with 3 rows and 4 columns , row1 column 1 , x plus , column 2 y minus , column 3 2 z equals , column 4 8 , row2 column 1 , 5 x minus , column 2 3 y plus , column 3 z equals , column 4 negative 6 , row3 column 1 , negative 2 x minus , column 2 y plus , column 3 4 z equals , column 4 negative 13 , end table
-
{
−
x
+
y
+
2
z
=
−
5
5
x
+
4
y
−
4
z
=
4
x
−
3
y
−
2
z
=
3
left brace . table with 3 rows and 4 columns , row1 column 1 , negative x plus , column 2 y plus , column 3 2 z equals , column 4 negative 5 , row2 column 1 , 5 x plus , column 2 4 y minus , column 3 4 z equals , column 4 4 , row3 column 1 , x minus , column 2 3 y minus , column 3 2 z equals , column 4 3 , end table
Solve each system by substitution.
-
{
3
x
+
y
−
2
z
=
22
x
+
5
y
+
z
=
4
x
=
−
3
z
left brace . table with 3 rows and 1 column , row1 column 1 , 3 x plus y minus 2 z equals 22 , row2 column 1 , x plus 5 y plus z equals 4 , row3 column 1 , x equals negative 3 z , end table
-
{
x
+
2
y
+
z
=
14
y
=
z
+
1
x
=
−
3
z
+
6
left brace . table with 3 rows and 1 column , row1 column 1 , x plus 2 y plus z equals 14 , row2 column 1 , y equals z plus 1 , row3 column 1 , x equals negative 3 z plus 6 , end table
3-6 Solving Systems Using Matrices
Quick Review
A matrix can represent a system of equations where each row stands for a different equation. The columns contain the coefficients of the variables and the constants.
Example
Solve using a matrix.
{
6
x
+
3
y
=
−
15
2
x
+
4
y
=
10
left brace . table with 2 rows and 1 column , row1 column 1 , 6 x plus 3 y equals negative 15 , row2 column 1 , 2 x plus 4 y equals 10 , end table
Enter coefficients as matrix elements
[
6
3
−
15
2
4
10
]
.
. matrix with 2 rows and 3 columns , row1 column 1 , 6 , column 2 3 , column 3 negative 15 , row2 column 1 , 2 , column 2 4 , column 3 10 , end matrix . .
Divide the first row by 3 to get
[
2
1
−
5
2
4
10
]
.
. matrix with 2 rows and 3 columns , row1 column 1 , 2 , column 2 1 , column 3 negative 5 , row2 column 1 , 2 , column 2 4 , column 3 10 , end matrix . . Subtract the first row from the second row to get
[
2
1
−
5
0
3
15
]
.
. matrix with 2 rows and 3 columns , row1 column 1 , 2 , column 2 1 , column 3 negative 5 , row2 column 1 , 0 , column 2 3 , column 3 15 , end matrix . . Multiply the second row by
1
3
1 third to get
[
2
1
−
5
0
1
5
]
.
. matrix with 2 rows and 3 columns , row1 column 1 , 2 , column 2 1 , column 3 negative 5 , row2 column 1 , 0 , column 2 1 , column 3 5 , end matrix . . Subtract the second row from the first row to get
[
2
1
−
10
0
1
5
]
.
. matrix with 2 rows and 3 columns , row1 column 1 , 2 , column 2 1 , column 3 negative 10 , row2 column 1 , 0 , column 2 1 , column 3 5 , end matrix . . Divide the first row by 2 to get
[
1
0
−
5
0
1
5
]
.
. matrix with 2 rows and 3 columns , row1 column 1 , 1 , column 2 0 , column 3 negative 5 , row2 column 1 , 0 , column 2 1 , column 3 5 , end matrix . . The solution to the system is
(
−
5
,
5
)
.
open negative 5 comma 5 close .
Exercises
Solve each system using a matrix.
-
{
4
x
−
12
y
=
−
1
6
x
+
4
y
=
4
left brace . table with 2 rows and 4 columns , row1 column 1 , 4 x minus , column 2 12 y , column 3 equals , column 4 negative 1 , row2 column 1 , 6 x plus , column 2 4 y , column 3 equals , column 4 4 , end table
-
{
7
x
+
2
y
=
5
13
x
+
14
y
=
−
1
left brace . table with 2 rows and 4 columns , row1 column 1 , 7 x plus , column 2 2 y , column 3 equals , column 4 5 , row2 column 1 , 13 x plus , column 2 14 y , column 3 equals , column 4 negative 1 , end table
-
{
−
5
x
+
3
y
+
4
z
=
2
3
x
−
y
−
z
=
4
x
−
6
y
−
5
z
=
−
4
left brace . table with 3 rows and 4 columns , row1 column 1 , negative 5 x plus , column 2 3 y plus , column 3 4 z equals , column 4 2 , row2 column 1 , 3 x minus , column 2 y minus , column 3 z equals , column 4 4 , row3 column 1 , x minus , column 2 6 y minus , column 3 5 z equals , column 4 negative 4 , end table
-
{
x
+
y
+
z
=
4
2
x
−
y
+
z
=
5
x
+
y
−
2
z
=
13
left brace . table with 3 rows and 4 columns , row1 column 1 , x plus , column 2 y plus , column 3 z equals , column 4 4 , row2 column 1 , 2 x minus , column 2 y plus , column 3 z equals , column 4 5 , row3 column 1 , x plus , column 2 y minus , column 3 2 z equals , column 4 13 , end table
