Pre-Calculus – Chapter 7 Section 3 Solving Systems of Equations with 3 Variables
The book is confusing on this subject. Please read these notes carefully. 
Recall how to solve a system of equations with 2 variables by substitution or elimination:
Example 1: Solve the system: 5x + 4y = 10
-3x – 5y = 7
(answer: (6, -5))
work: By elimination:
1. Decide which variable to eliminate. I am choosing x for this example.
2. Multiply each equation by a number so the x coefficients are opposites. I will multiply the first equation by 3
and the second equation by 5, therefore each equation will have a coefficient of -15.
3(5x + 4y = 10)  15x + 12y = 30
5(-3x – 5y = 7)  -15x – 25y = 35
3. Add the 2 equations together to eliminate out the x.
15x + 12y = 30
-15x – 25y = 35
-13y = 65
4. Solve the equation for y.
-13y = 65
y = -5
5. Substitute - 5 in for y into either original equation and solve for x. I am choosing the top equation.
5x + 4(-5) = 10
5x – 20 = 10
5x = 30
x=6
6. Solution is (6, -5)
Now let’s look at systems with 3 equations and 3 unknowns.
Example 2: Solve the system: 4x – 3y – 2z = 21
6y – 5z = -10
z = -4
answer (-1/2, -5, -4)
work:
1. Notice the z is already solved for in this system. Therefore, substitute it into the second equation to
solve for y.
6y – 5(-4) = -10
6y + 20 = -10
6y = -30
y = -5
2. Now, substitute y and z into the first equation to find x.
4x – 3(-5) – 2(-4) = 21
4x + 23 = 21
4x = -2
x = -1/2
3. Solution is (-1/2, -5, -4)
Example 3: 4x + y – 3z = 11
2x – 3y + 2z = 9
x + y + z = -3
answer (2, -3,- 2)
work: 1. Decide which variable to eliminate. I am choosing y since two of the equations have a coefficient of 1.
2. Take two equations and eliminate y. I am going to multiply the top equation by 3 and add it to the 2 nd
equation.
3(4x + y – 3z = 11)  12x + 3y – 9z = 33
2nd equation
2x – 3y + 2z = 9
14x – 7z = 42
add equations
3. Take the bottom equation and one of the others and eliminate the same variable (y). I am going to multiply
the bottom equation by -1 and add it to the 1st equation.
-1(x + y + z = -3)  -x – y – z = 3
1st equation
4x + y – 3z = 11
3x – 4z = 14
4. Take the 2 new equations and solve for x and z. Just like in a 2 variable system. (example 1)
14x – 7z = 32
3x – 4z = 14
5. I am going to multiply top equation by 4 and bottom by -7 to eliminate z.
4(14x – 7z = 42)  56x – 28z = 168
-7(3x – 4z = 14)  -21x + 28z = -98
35x = 70
x=2
6. Substitute x = 2 into one of the equations to find z. I substituted into 3x – 4z = 14.
3(2) – 4z = 14
6 – 4z = 14
-4z = 8
z = -2
7. Substitute x and z into one of the original equations to find y. I am substituting into the bottom equation.
x + y + z = -3
2 + y +(– 2) = -3
y = -3
8. Solution: (2, -3, -2)
Example 4: Try this:
x–y+z=4
x + 3y – 2z = -3
3x + 2y + z = 5
(answer: (2, -1, 1))
work: 1. Decide which variable to eliminate from all 3 equations. I chose x. I multiply the 1st equation by -1
and added it to the 2nd equation.
-1(x – y + z = 4)  -x + y – z = -4
2nd equation
x + 3y – 2z = -3
4y – 3z = -7
2. Multiply 1st equation by -3 and add to the bottom (3rd ) equation .
-3(x – y + z = 4)  -3x + 3y – 3z = - 12
3rd equation
3x + 2y + z = 5
5y – 2z = -7
3. Solve the 2 new equations for y and z.
4y – 3z = -7
5y – 2z = -7
I choose to eliminate the z by multiplying the top equation by 2 and the bottom by -3.
2(4y – 3z = -7)  8y – 6z = -14
-3(5y – 2z = -7) -15y + 6z = 21
-7y = 7
y = -1
Substitute y in to find z. I am substituting into the top equation.
4(-1) – 3z = -7
-4 – 3z = -7
-3z = -3
z=1
4. Substitute y and z into one of the original equations to find x. I am substituting into the 1st one.
x –(-1) + 1 = 4
x+2=4
x=2
5. Solution: (2, -1, 1)
Example 5: Try this: 5x – 3y + 2z = 3
2x + 4y – z = 7
x – 11y + 4Z = 3
(answer: no solution)
work: eliminate z - 2nd equation times 2, add to 1st equation
5x – 3y + 2z = 3
4x + 8y – 2z = 14
9x + 5y = 17
2nd equation times 4, add to 3rd equation
8x + 16y – 4Z = 28
x – 11y + 4 = 3
9x + 5y = 31
Solve these 2 equations;
nd
multiply 2 equation by -1
add
Therefore: no solution
9x + 5y = 17
-9x – 5y = -32
0 = -15