#### Transcript x + y + z = 0

```Solving with Three Variables
Section 3-6 pages 148-155
Solving Using Elimination
Solve the system.
2x + y - z = 5
3x - y + 2z = -1
x - y - z = 0
Solving the system
Step 1:
Pick two of the equations and eliminate the
Let’s pick  & 
2x + y - z = 5
3x - y + 2z = -1
Let’s eliminate z:
Multiply first line by 2
4x + 2y - 2z = 10
3x - y + 2z = -1
7x + 1y = 9
Step 2:
Pick two other equations and eliminate the
same variable you eliminated in step 1.
Let’s pick  & 
2x + y - z = 5
x - y - z = 0
Multiply  by -1.
-2x - y + z = -5
x-y-z=0
-1x - 2y = -5
Step 3:
Take the two equations that you just
eliminated z from:
7x + 1y = 9
-1x - 2y = -5
Pick one of the variables to eliminate.
Let’s pick x.
7x + 1y = 9
-7x - 14y = -35
-13y = -26
Solve for y:
y=2
Step 4:
Plug the y value into one of the equations from
step 3 and solve for x.
1x + 1(2) = 9
x=7
Step 5:
Plug the x and y into one of the original
equations and solve for z.
2(7) + 2 - z = 5
16 - z = 5
-z = -11
z = 11
Final Step
Plug the answers into a (x, y, z)
(7, 2, 11)
Try Some
x – 3y + 2z = 11
-x + 4y +3z = 5
2x – 2y – 4z = 2
x – 3y + z = 6
2x – 5y – z = -2
-x + y + 2z = 7
3x + 2y – z = 12
-4x + y – 2z = 4
x – 3y + z + -4
x + y + 2z = 3
2x + y + 3z = 7
-x – 2y + z = 10
Using to solve word problems
A stadium has 49,000 seats. Seats sell for
\$25 in Section A, \$20 in Section B, and
\$15 in Section C. The number of seats
in Section A equals the total number of
seats in Sections B and C. Suppose the
stadium takes in \$1,052,000 from each
sold out event. How many seats does
each section hold?
x = Section A, y = Section B, z = Section C
x + y + z = 49,000
25x + 20y + 15z = 1,052,000
x=y+z
Make sure all variables are on one side:
x + y + z = 49,000
25x + 20y + 15z = 1,052,000
x-y-z=0
Pick two equations to eliminate one of the
variables:
-25(x + y + z = 49,000)
25x + 20y + 15z = 1,052,000
-25x -25y - 25z = -1,225,000
-5y - 10z = -173,000
Pick two different equations and eliminate same
variable:
x + y + z = 49,000
-1(x - y - z = 0)
-x + y + z = 0
2y + 2z = 49,000
-5y - 10z = -173,000
5(2y + 2z = 49,000)
10y + 10z = 245,000
5y = 72,000
y = 14,400
Plug into one of the equations above and solve
for z.
10(14,400) + 10z = 245,000
144,000 + 10z = 245,000
10z = 101,000
z = 10,100
Plug the y and z into one of the original
equations to find x:
x + 14,400 + 10,100 = 49,000
x + 24,500 = 49,000
x = 24,500
Section A has 24,500 seats
Section B has 14,400 seats
Section C has 10,100 seats
```