Download Three Variable Systems of Equations Joseph Lee Metropolitan Community College

Three Variable Systems of Equations
Joseph Lee
Metropolitan Community College
Joseph Lee
Three Variable Systems of Equations
Example 1.
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Joseph Lee
Three Variable Systems of Equations
Example 1.
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. First, we will take two equations and eliminate a
variable. Let’s take the first two equations and eliminate z.
Joseph Lee
Three Variable Systems of Equations
Example 1.
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. First, we will take two equations and eliminate a
variable. Let’s take the first two equations and eliminate z.
x +y +z =6
2x − 3y − z = 5
Joseph Lee
Three Variable Systems of Equations
Example 1.
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. First, we will take two equations and eliminate a
variable. Let’s take the first two equations and eliminate z.
x +y +z =6
2x − 3y − z = 5
3x − 2y = 11
Joseph Lee
Three Variable Systems of Equations
Example 1. (Continued)
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. (Continued) Now, we want to take two different
equations but eliminate the same variable: z. We will take the first
and last equations.
Joseph Lee
Three Variable Systems of Equations
Example 1. (Continued)
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. (Continued) Now, we want to take two different
equations but eliminate the same variable: z. We will take the first
and last equations.
2(x + y + z) = (6)2
3x + 2y − 2z = −1
Joseph Lee
Three Variable Systems of Equations
Example 1. (Continued)
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. (Continued) Now, we want to take two different
equations but eliminate the same variable: z. We will take the first
and last equations.
2(x + y + z) = (6)2
3x + 2y − 2z = −1
2x + 2y + 2z = 12
3x + 2y − 2z = −1
Joseph Lee
Three Variable Systems of Equations
Example 1. (Continued)
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. (Continued) Now, we want to take two different
equations but eliminate the same variable: z. We will take the first
and last equations.
2(x + y + z) = (6)2
3x + 2y − 2z = −1
2x + 2y + 2z = 12
3x + 2y − 2z = −1
5x + 4y = 11
Joseph Lee
Three Variable Systems of Equations
Example 1. (Continued)
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. (Continued) We now have two equations with two
variables. We may proceed as usual from here.
3x − 2y = 11
5x + 4y = 11
Joseph Lee
Three Variable Systems of Equations
Example 1. (Continued)
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. (Continued) We now have two equations with two
variables. We may proceed as usual from here.
3x − 2y = 11
5x + 4y = 11
2(3x − 2y ) = (11)2
5x + 4y = 11
Joseph Lee
Three Variable Systems of Equations
Example 1. (Continued)
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. (Continued) We now have two equations with two
variables. We may proceed as usual from here.
3x − 2y = 11
5x + 4y = 11
2(3x − 2y ) = (11)2
5x + 4y = 11
6x − 4y = 22
5x + 4y = 11
Joseph Lee
Three Variable Systems of Equations
Example 1. (Continued)
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. (Continued) We now have two equations with two
variables. We may proceed as usual from here.
3x − 2y = 11
5x + 4y = 11
2(3x − 2y ) = (11)2
5x + 4y = 11
6x − 4y = 22
5x + 4y = 11
11x = 33
Joseph Lee
Three Variable Systems of Equations
Example 1. (Continued)
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. (Continued)
11x = 33
x =3
Joseph Lee
Three Variable Systems of Equations
Example 1. (Continued)
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. (Continued)
11x = 33
x =3
We substute x = 3 into an equation containing only x and y .
3x − 2y = 11
Joseph Lee
Three Variable Systems of Equations
Example 1. (Continued)
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. (Continued)
11x = 33
x =3
We substute x = 3 into an equation containing only x and y .
3x − 2y = 11
3(3) − 2y = 11
Joseph Lee
Three Variable Systems of Equations
Example 1. (Continued)
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. (Continued)
11x = 33
x =3
We substute x = 3 into an equation containing only x and y .
3x − 2y = 11
3(3) − 2y = 11
9 − 2y = 11
Joseph Lee
Three Variable Systems of Equations
Example 1. (Continued)
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. (Continued)
11x = 33
x =3
We substute x = 3 into an equation containing only x and y .
3x − 2y
3(3) − 2y
9 − 2y
−2y
Joseph Lee
= 11
= 11
= 11
=2
Three Variable Systems of Equations
Example 1. (Continued)
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. (Continued)
11x = 33
x =3
We substute x = 3 into an equation containing only x and y .
3x − 2y
3(3) − 2y
9 − 2y
−2y
y
Joseph Lee
= 11
= 11
= 11
=2
= −1
Three Variable Systems of Equations
Example 1. (Continued)
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. (Continued) Similarly, we substitute x = 3 and y = −1
into an equation containing x, y , and z.
x +y +z =6
Joseph Lee
Three Variable Systems of Equations
Example 1. (Continued)
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. (Continued) Similarly, we substitute x = 3 and y = −1
into an equation containing x, y , and z.
x +y +z =6
(3) + (−1) + z = 6
Joseph Lee
Three Variable Systems of Equations
Example 1. (Continued)
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. (Continued) Similarly, we substitute x = 3 and y = −1
into an equation containing x, y , and z.
x +y +z =6
(3) + (−1) + z = 6
2+z =6
Joseph Lee
Three Variable Systems of Equations
Example 1. (Continued)
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. (Continued) Similarly, we substitute x = 3 and y = −1
into an equation containing x, y , and z.
x +y +z
(3) + (−1) + z
2+z
z
Joseph Lee
=6
=6
=6
=4
Three Variable Systems of Equations
Example 1. (Continued)
Solve the following system of equations.
x +y +z =6
2x − 3y − z = 5
3x + 2y − 2z = −1
Solution. (Continued) Similarly, we substitute x = 3 and y = −1
into an equation containing x, y , and z.
x +y +z
(3) + (−1) + z
2+z
z
=6
=6
=6
=4
Thus, (3, −1, 4) is the solution to this system of equations.
Joseph Lee
Three Variable Systems of Equations
Example 2.
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Joseph Lee
Three Variable Systems of Equations
Example 2.
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. First, we will take two equations and eliminate a
variable. Let’s take the last two equations and eliminate z.
Joseph Lee
Three Variable Systems of Equations
Example 2.
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. First, we will take two equations and eliminate a
variable. Let’s take the last two equations and eliminate z.
5y + 3z = 1
2x − 5z = 7
Joseph Lee
Three Variable Systems of Equations
Example 2.
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. First, we will take two equations and eliminate a
variable. Let’s take the last two equations and eliminate z.
5y + 3z = 1
2x − 5z = 7
5(5y + 3z) = (1)5
3(2x − 5z) = (7)3
Joseph Lee
Three Variable Systems of Equations
Example 2.
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. First, we will take two equations and eliminate a
variable. Let’s take the last two equations and eliminate z.
5y + 3z = 1
2x − 5z = 7
5(5y + 3z) = (1)5
3(2x − 5z) = (7)3
25y + 15z = 5
6x − 15z = 21
Joseph Lee
Three Variable Systems of Equations
Example 2.
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. First, we will take two equations and eliminate a
variable. Let’s take the last two equations and eliminate z.
5y + 3z = 1
2x − 5z = 7
5(5y + 3z) = (1)5
3(2x − 5z) = (7)3
25y + 15z = 5
6x − 15z = 21
6x + 25y = 26
Joseph Lee
Three Variable Systems of Equations
Example 2. (Continued)
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. (Continued) Notice we have two equations with two
variables.
3x + 4y = −4
6x + 25y = 26
Joseph Lee
Three Variable Systems of Equations
Example 2. (Continued)
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. (Continued) Notice we have two equations with two
variables.
3x + 4y = −4
6x + 25y = 26
−2(3x + 4y ) = (−4)(−2)
6x + 25y = 26
Joseph Lee
Three Variable Systems of Equations
Example 2. (Continued)
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. (Continued) Notice we have two equations with two
variables.
3x + 4y = −4
6x + 25y = 26
−2(3x + 4y ) = (−4)(−2)
6x + 25y = 26
−6x − 8y = 8
6x + 25y = 26
Joseph Lee
Three Variable Systems of Equations
Example 2. (Continued)
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. (Continued) Notice we have two equations with two
variables.
3x + 4y = −4
6x + 25y = 26
−2(3x + 4y ) = (−4)(−2)
6x + 25y = 26
−6x − 8y = 8
6x + 25y = 26
17y = 34
Joseph Lee
Three Variable Systems of Equations
Example 2. (Continued)
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. (Continued)
17y = 34
y =2
Joseph Lee
Three Variable Systems of Equations
Example 2. (Continued)
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. (Continued)
17y = 34
y =2
We may now substitute y = 2 into an equation only containing x
and y to solve for x or into an equation containing y and z to
solve for z.
Joseph Lee
Three Variable Systems of Equations
Example 2. (Continued)
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. (Continued)
17y = 34
y =2
We may now substitute y = 2 into an equation only containing x
and y to solve for x or into an equation containing y and z to
solve for z.
3x + 4y = −4
Joseph Lee
Three Variable Systems of Equations
Example 2. (Continued)
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. (Continued)
17y = 34
y =2
We may now substitute y = 2 into an equation only containing x
and y to solve for x or into an equation containing y and z to
solve for z.
3x + 4y = −4
3x + 4(2) = −4
Joseph Lee
Three Variable Systems of Equations
Example 2. (Continued)
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. (Continued)
17y = 34
y =2
We may now substitute y = 2 into an equation only containing x
and y to solve for x or into an equation containing y and z to
solve for z.
3x + 4y = −4
3x + 4(2) = −4
3x + 8 = −4
Joseph Lee
Three Variable Systems of Equations
Example 2. (Continued)
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. (Continued)
17y = 34
y =2
We may now substitute y = 2 into an equation only containing x
and y to solve for x or into an equation containing y and z to
solve for z.
3x + 4y = −4
3x + 4(2) = −4
3x + 8 = −4
3x = −12
Joseph Lee
Three Variable Systems of Equations
Example 2. (Continued)
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. (Continued)
17y = 34
y =2
We may now substitute y = 2 into an equation only containing x
and y to solve for x or into an equation containing y and z to
solve for z.
3x + 4y = −4
3x + 4(2) = −4
3x + 8 = −4
3x = −12
x = −4
Joseph Lee
Three Variable Systems of Equations
Example 2. (Continued)
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. (Continued)
5y + 3z = 1
Joseph Lee
Three Variable Systems of Equations
Example 2. (Continued)
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. (Continued)
5y + 3z = 1
5(2) + 3z = 1
Joseph Lee
Three Variable Systems of Equations
Example 2. (Continued)
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. (Continued)
5y + 3z = 1
5(2) + 3z = 1
10 + 3z = 1
Joseph Lee
Three Variable Systems of Equations
Example 2. (Continued)
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. (Continued)
5y + 3z
5(2) + 3z
10 + 3z
3z
Joseph Lee
=1
=1
=1
= −9
Three Variable Systems of Equations
Example 2. (Continued)
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. (Continued)
5y + 3z
5(2) + 3z
10 + 3z
3z
z
Joseph Lee
=1
=1
=1
= −9
= −3
Three Variable Systems of Equations
Example 2. (Continued)
Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. (Continued)
5y + 3z
5(2) + 3z
10 + 3z
3z
z
=1
=1
=1
= −9
= −3
Thus, (−4, 2, −3) is the solution to the system of equations.
Joseph Lee
Three Variable Systems of Equations
Example 3.
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Joseph Lee
Three Variable Systems of Equations
Example 3.
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. First, we will take two equations and eliminate a
variable. Let’s take the first two equations and eliminate z.
Joseph Lee
Three Variable Systems of Equations
Example 3.
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. First, we will take two equations and eliminate a
variable. Let’s take the first two equations and eliminate z.
2x + 3y − z = 7
−3x + 2y − 2z = 7
Joseph Lee
Three Variable Systems of Equations
Example 3.
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. First, we will take two equations and eliminate a
variable. Let’s take the first two equations and eliminate z.
2x + 3y − z = 7
−3x + 2y − 2z = 7
−2(2x + 3y − z) = (7)(−2)
−3x + 2y − 2z = 7
Joseph Lee
Three Variable Systems of Equations
Example 3.
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. First, we will take two equations and eliminate a
variable. Let’s take the first two equations and eliminate z.
2x + 3y − z = 7
−3x + 2y − 2z = 7
−2(2x + 3y − z) = (7)(−2)
−3x + 2y − 2z = 7
−4x − 6y + 2z = −14
−3x + 2y − 2z = 7
Joseph Lee
Three Variable Systems of Equations
Example 3.
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. First, we will take two equations and eliminate a
variable. Let’s take the first two equations and eliminate z.
2x + 3y − z = 7
−3x + 2y − 2z = 7
−2(2x + 3y − z) = (7)(−2)
−3x + 2y − 2z = 7
−4x − 6y + 2z = −14
−3x + 2y − 2z = 7
−7x − 4y = −7
Joseph Lee
Three Variable Systems of Equations
Example 3. (Continued)
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. (Continued) Now, we want to take two different
equations but eliminate the same variable: z. We will take the first
and last equations.
Joseph Lee
Three Variable Systems of Equations
Example 3. (Continued)
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. (Continued) Now, we want to take two different
equations but eliminate the same variable: z. We will take the first
and last equations.
3(2x + 3y − z) = (7)3
5x − 4y + 3z = −10
Joseph Lee
Three Variable Systems of Equations
Example 3. (Continued)
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. (Continued) Now, we want to take two different
equations but eliminate the same variable: z. We will take the first
and last equations.
3(2x + 3y − z) = (7)3
5x − 4y + 3z = −10
6x + 9y − 3z = 21
5x − 4y + 3z = −10
Joseph Lee
Three Variable Systems of Equations
Example 3. (Continued)
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. (Continued) Now, we want to take two different
equations but eliminate the same variable: z. We will take the first
and last equations.
3(2x + 3y − z) = (7)3
5x − 4y + 3z = −10
6x + 9y − 3z = 21
5x − 4y + 3z = −10
11x + 5y = 11
Joseph Lee
Three Variable Systems of Equations
Example 3. (Continued)
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. (Continued) We now have two equations with two
variables. We may proceed as usual from here.
−7x − 4y = −7
11x + 5y = 11
Joseph Lee
Three Variable Systems of Equations
Example 3. (Continued)
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. (Continued) We now have two equations with two
variables. We may proceed as usual from here.
−7x − 4y = −7
11x + 5y = 11
5(−7x − 4y ) = (−7)5
4(11x + 5y ) = (11)4
Joseph Lee
Three Variable Systems of Equations
Example 3. (Continued)
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. (Continued) We now have two equations with two
variables. We may proceed as usual from here.
−7x − 4y = −7
11x + 5y = 11
5(−7x − 4y ) = (−7)5
4(11x + 5y ) = (11)4
−35x − 20y = −35
44x + 20y = 44
Joseph Lee
Three Variable Systems of Equations
Example 3. (Continued)
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. (Continued) We now have two equations with two
variables. We may proceed as usual from here.
−7x − 4y = −7
11x + 5y = 11
5(−7x − 4y ) = (−7)5
4(11x + 5y ) = (11)4
−35x − 20y = −35
44x + 20y = 44
9x = 9
Joseph Lee
Three Variable Systems of Equations
Example 3. (Continued)
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. (Continued)
9x = 9
x =1
Joseph Lee
Three Variable Systems of Equations
Example 3. (Continued)
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. (Continued)
9x = 9
x =1
We substute x = 1 into an equation containing only x and y .
−7x − 4y = −7
Joseph Lee
Three Variable Systems of Equations
Example 3. (Continued)
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. (Continued)
9x = 9
x =1
We substute x = 1 into an equation containing only x and y .
−7x − 4y = −7
−7(1) − 4y = −7
Joseph Lee
Three Variable Systems of Equations
Example 3. (Continued)
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. (Continued)
9x = 9
x =1
We substute x = 1 into an equation containing only x and y .
−7x − 4y = −7
−7(1) − 4y = −7
−7 − 4y = −7
Joseph Lee
Three Variable Systems of Equations
Example 3. (Continued)
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. (Continued)
9x = 9
x =1
We substute x = 1 into an equation containing only x and y .
−7x − 4y
−7(1) − 4y
−7 − 4y
−4y
Joseph Lee
= −7
= −7
= −7
=0
Three Variable Systems of Equations
Example 3. (Continued)
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. (Continued)
9x = 9
x =1
We substute x = 1 into an equation containing only x and y .
−7x − 4y
−7(1) − 4y
−7 − 4y
−4y
y
Joseph Lee
= −7
= −7
= −7
=0
=0
Three Variable Systems of Equations
Example 3. (Continued)
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. (Continued) Similarly, we substitute x = 1 and y = 0
into an equation containing x, y , and z.
2x + 3y − z = 7
Joseph Lee
Three Variable Systems of Equations
Example 3. (Continued)
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. (Continued) Similarly, we substitute x = 1 and y = 0
into an equation containing x, y , and z.
2x + 3y − z = 7
2(1) + 3(0) − z = 7
Joseph Lee
Three Variable Systems of Equations
Example 3. (Continued)
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. (Continued) Similarly, we substitute x = 1 and y = 0
into an equation containing x, y , and z.
2x + 3y − z = 7
2(1) + 3(0) − z = 7
2−z =7
Joseph Lee
Three Variable Systems of Equations
Example 3. (Continued)
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. (Continued) Similarly, we substitute x = 1 and y = 0
into an equation containing x, y , and z.
2x + 3y − z
2(1) + 3(0) − z
2−z
−z
Joseph Lee
=7
=7
=7
=5
Three Variable Systems of Equations
Example 3. (Continued)
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. (Continued) Similarly, we substitute x = 1 and y = 0
into an equation containing x, y , and z.
2x + 3y − z
2(1) + 3(0) − z
2−z
−z
z
Joseph Lee
=7
=7
=7
=5
= −5
Three Variable Systems of Equations
Example 3. (Continued)
Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. (Continued) Similarly, we substitute x = 1 and y = 0
into an equation containing x, y , and z.
2x + 3y − z
2(1) + 3(0) − z
2−z
−z
z
=7
=7
=7
=5
= −5
Thus, (1, 0, −5) is the solution to this system of equations.
Joseph Lee
Three Variable Systems of Equations
Example 4.
Solve the following system of equations.
x + 2y − z = 3
2x + 3y − 5z = 3
5x + 8y − 11z = 9
Joseph Lee
Three Variable Systems of Equations
Example 4.
Solve the following system of equations.
x + 2y − z = 3
2x + 3y − 5z = 3
5x + 8y − 11z = 9
Solution. First, we will take two equations and eliminate a
variable. Let’s take the first two equations and eliminate x.
Joseph Lee
Three Variable Systems of Equations
Example 4.
Solve the following system of equations.
x + 2y − z = 3
2x + 3y − 5z = 3
5x + 8y − 11z = 9
Solution. First, we will take two equations and eliminate a
variable. Let’s take the first two equations and eliminate x.
x + 2y − z = 3
2x + 3y − 5z = 3
Joseph Lee
Three Variable Systems of Equations
Example 4.
Solve the following system of equations.
x + 2y − z = 3
2x + 3y − 5z = 3
5x + 8y − 11z = 9
Solution. First, we will take two equations and eliminate a
variable. Let’s take the first two equations and eliminate x.
x + 2y − z = 3
2x + 3y − 5z = 3
−2(x + 2y − z) = (3)(−2)
2x + 3y − 5z = 3
Joseph Lee
Three Variable Systems of Equations
Example 4.
Solve the following system of equations.
x + 2y − z = 3
2x + 3y − 5z = 3
5x + 8y − 11z = 9
Solution. First, we will take two equations and eliminate a
variable. Let’s take the first two equations and eliminate x.
x + 2y − z = 3
2x + 3y − 5z = 3
−2(x + 2y − z) = (3)(−2)
2x + 3y − 5z = 3
−2x − 4y + 2z = −6
2x + 3y − 5z = 3
Joseph Lee
Three Variable Systems of Equations
Example 4.
Solve the following system of equations.
x + 2y − z = 3
2x + 3y − 5z = 3
5x + 8y − 11z = 9
Solution. First, we will take two equations and eliminate a
variable. Let’s take the first two equations and eliminate x.
x + 2y − z = 3
2x + 3y − 5z = 3
−2(x + 2y − z) = (3)(−2)
2x + 3y − 5z = 3
−2x − 4y + 2z = −6
2x + 3y − 5z = 3
−y − 3z = −3
Joseph Lee
Three Variable Systems of Equations
Example 4. (Continued)
Solve the following system of equations.
x + 2y − z = 3
2x + 3y − 5z = 3
5x + 8y − 11z = 9
Solution. (Continued) Now, we want to take two different
equations but eliminate the same variable: x. We will take the first
and last equations.
Joseph Lee
Three Variable Systems of Equations
Example 4. (Continued)
Solve the following system of equations.
x + 2y − z = 3
2x + 3y − 5z = 3
5x + 8y − 11z = 9
Solution. (Continued) Now, we want to take two different
equations but eliminate the same variable: x. We will take the first
and last equations.
−5(x + 2y − z) = (3)(−5)
5x + 8y − 11z = 9
Joseph Lee
Three Variable Systems of Equations
Example 4. (Continued)
Solve the following system of equations.
x + 2y − z = 3
2x + 3y − 5z = 3
5x + 8y − 11z = 9
Solution. (Continued) Now, we want to take two different
equations but eliminate the same variable: x. We will take the first
and last equations.
−5(x + 2y − z) = (3)(−5)
5x + 8y − 11z = 9
−5x − 10y + 5z = −15
5x + 8y − 11z = 9
Joseph Lee
Three Variable Systems of Equations
Example 4. (Continued)
Solve the following system of equations.
x + 2y − z = 3
2x + 3y − 5z = 3
5x + 8y − 11z = 9
Solution. (Continued) Now, we want to take two different
equations but eliminate the same variable: x. We will take the first
and last equations.
−5(x + 2y − z) = (3)(−5)
5x + 8y − 11z = 9
−5x − 10y + 5z = −15
5x + 8y − 11z = 9
−2y − 6z = −6
Joseph Lee
Three Variable Systems of Equations
Example 4. (Continued)
Solve the following system of equations.
x + 2y − z = 3
2x + 3y − 5z = 3
5x + 8y − 11z = 9
Solution. (Continued) We now have two equations with two
variables. We may proceed as usual from here.
−y − 3z = −3
−2y − 6z = −6
Joseph Lee
Three Variable Systems of Equations
Example 4. (Continued)
Solve the following system of equations.
x + 2y − z = 3
2x + 3y − 5z = 3
5x + 8y − 11z = 9
Solution. (Continued) We now have two equations with two
variables. We may proceed as usual from here.
−y − 3z = −3
−2y − 6z = −6
−2(−y − 3z) = (−3)(−2)
−2y − 6z = −6
Joseph Lee
Three Variable Systems of Equations
Example 4. (Continued)
Solve the following system of equations.
x + 2y − z = 3
2x + 3y − 5z = 3
5x + 8y − 11z = 9
Solution. (Continued) We now have two equations with two
variables. We may proceed as usual from here.
−y − 3z = −3
−2y − 6z = −6
−2(−y − 3z) = (−3)(−2)
−2y − 6z = −6
2y + 6z = 6
−2y − 6z = −6
Joseph Lee
Three Variable Systems of Equations
Example 4. (Continued)
Solve the following system of equations.
x + 2y − z = 3
2x + 3y − 5z = 3
5x + 8y − 11z = 9
Solution. (Continued) We now have two equations with two
variables. We may proceed as usual from here.
−y − 3z = −3
−2y − 6z = −6
−2(−y − 3z) = (−3)(−2)
−2y − 6z = −6
2y + 6z = 6
−2y − 6z = −6
0 =0
Joseph Lee
Three Variable Systems of Equations
Example 4. (Continued)
Solve the following system of equations.
x + 2y − z = 3
2x + 3y − 5z = 3
5x + 8y − 11z = 9
Solution. (Continued)
0 =0
Arriving at this identity, we conclude that this is a dependent
system of equations, and thus it has infinitely many solutions.
Unlike in the previous section, we will not describe these
solutions.1
Joseph Lee
Three Variable Systems of Equations
Example 5.
Solve the following system of equations.
3x − y + 2z = 4
x − 5y + 4z = 3
6x − 2y + 4z = −8
Joseph Lee
Three Variable Systems of Equations
Example 5.
Solve the following system of equations.
3x − y + 2z = 4
x − 5y + 4z = 3
6x − 2y + 4z = −8
Solution. First, we will take two equations and eliminate a
variable. Let’s take the first and last equations and eliminate x.
Joseph Lee
Three Variable Systems of Equations
Example 5.
Solve the following system of equations.
3x − y + 2z = 4
x − 5y + 4z = 3
6x − 2y + 4z = −8
Solution. First, we will take two equations and eliminate a
variable. Let’s take the first and last equations and eliminate x.
3x − y + 2z = 4
6x − 2y + 4z = −8
Joseph Lee
Three Variable Systems of Equations
Example 5.
Solve the following system of equations.
3x − y + 2z = 4
x − 5y + 4z = 3
6x − 2y + 4z = −8
Solution. First, we will take two equations and eliminate a
variable. Let’s take the first and last equations and eliminate x.
3x − y + 2z = 4
6x − 2y + 4z = −8
−2(3x − y + 2z) = (4)(−2)
6x − 2y + 4z = −8
Joseph Lee
Three Variable Systems of Equations
Example 5.
Solve the following system of equations.
3x − y + 2z = 4
x − 5y + 4z = 3
6x − 2y + 4z = −8
Solution. First, we will take two equations and eliminate a
variable. Let’s take the first and last equations and eliminate x.
3x − y + 2z = 4
6x − 2y + 4z = −8
−2(3x − y + 2z) = (4)(−2)
6x − 2y + 4z = −8
−6x + 2y − 4z = −8
6x − 2y + 4z = −8
Joseph Lee
Three Variable Systems of Equations
Example 5.
Solve the following system of equations.
3x − y + 2z = 4
x − 5y + 4z = 3
6x − 2y + 4z = −8
Solution. First, we will take two equations and eliminate a
variable. Let’s take the first and last equations and eliminate x.
3x − y + 2z = 4
6x − 2y + 4z = −8
−2(3x − y + 2z) = (4)(−2)
6x − 2y + 4z = −8
−6x + 2y − 4z = −8
6x − 2y + 4z = −8
0 = −16
Joseph Lee
Three Variable Systems of Equations
Example 5. (Continued)
Solve the following system of equations.
3x − y + 2z = 4
x − 5y + 4z = 3
6x − 2y + 4z = −8
Solution. (Continued)
0 = −16
Arriving at this contradiction, we conclude that this is an
inconsistent system of equations, and thus it has no solution.
Joseph Lee
Three Variable Systems of Equations