Transcript Equation - Brookwood High School

Solve Systems of Linear
Equations in 3 Variables
1.7 (M3)
General Steps for Solving Systems
with 3 variables
1.
2.
3.
4.
5.
6.
Combine 2 equations to make a new equation with 2
unknowns (eliminate 1 of the variables)
Do the same with 2 different equations (make sure you
eliminate the same variable)
Solve the system of the 2 new equations from steps #1
and #2
Solve for 1 variable.
Substitute back into 1 of the new equations to find a 2nd
variable.
Substitute both back into one of the original equations to
find the 3rd variable.
Special Situations


If you get a false statement (like 0 = -1) when
you are trying to solve, the original system has
no solution.
If you get 0 = 0 when solving, the system has
infinitely many solutions.
EXAMPLE 1
Use the elimination method
Solve the system.
4x + 2y + 3z = 1
Equation 1
2x – 3y + 5z = –14
Equation 2
6x – y + 4z = –1
Equation 3
SOLUTION
STEP 1 Rewrite the system as a linear system in two
variables.
4x + 2y + 3z = 1
12x – 2y + 8z = –2
16x
+ 11z = –1
Add 2 times Equation 3
to Equation 1.
New Equation 1
EXAMPLE 1
Use the elimination method
2x – 3y + 5z = –14
–18x + 3y –12z = 3
–16x
– 7z = –11
Add – 3 times Equation 3
to Equation 2.
New Equation 2
STEP 2 Solve the new linear system for both of its
variables.
16x + 11z = –1
Add new Equation 1
–16x – 7z = –11
and new Equation 2.
4z = –12
z = –3
Solve for z.
x=2
Substitute into new
Equation 1 or 2 to find x.
EXAMPLE 1
STEP 3
Use the elimination method
Substitute x = 2 and z = – 3 into an original
equation and solve for y.
6x – y + 4z = –1
6(2) – y + 4(–3) = –1
y =1
Write original Equation 3.
Substitute 2 for x and –3
for z.
Solve for y.
EXAMPLE 2
Solve a three-variable system with no solution
Solve the system.
x+y+z=3
4x + 4y + 4z = 7
Equation 1
3x – y + 2z = 5
Equation 3
Equation 2
SOLUTION
When you multiply Equation 1 by – 4 and add the
result to Equation 2, you obtain a false equation.
–4x – 4y – 4z = –12
4x + 4y + 4z = 7
Add – 4 times Equation 1
to Equation 2.
New Equation 1
0 = –5
Because you obtain a false equation, you can conclude
that the original system has no solution.
EXAMPLE 3
Solve a three-variable system with many solutions
Solve the system.
x+y+z=4
x+y–z=4
3x + 3y + z = 12
Equation 1
Equation 2
Equation 3
SOLUTION
STEP 1 Rewrite the system as a linear system in
two variables.
x+y+z=4
Add Equation 1
x +y–z=4
to Equation 2.
2x + 2y = 8
New Equation 1
EXAMPLE 3
Solve a three-variable system with many solutions
x+y–z=4
3x + 3y + z = 12
4x + 4y = 16
Add Equation 2
to Equation 3.
New Equation 2
STEP 2 Solve the new linear system for both of its
variables.
Add –2 times new
–4x – 4y = –16
Equation 1
4x + 4y = 16
to new Equation 2.
Because you obtain the identity 0 = 0, the system
has infinitely many solutions.
EXAMPLE 3
Solve a three-variable system with many solutions
STEP 3 Describe the solutions of the system. One
way to do this is to divide new Equation 1
by 2 to get x + y = 4, or y = –x + 4.
Substituting this into original Equation 1
produces z = 0. So, any ordered triple of the
form (x, –x + 4, 0) is a solution of the system.
GUIDED PRACTICE
for Examples 1, 2 and 3
Solve the system.
1. 3x + y – 2z = 10
6x – 2y + z = –2
x + 4y + 3z = 7
(1, 3, –2)
ANSWER
3.
2.
x+y–z=2
2x + 2y – 2z = 6
5x + y – 3z = 8
ANSWER
x+y+z=3
x+y–z=3
2x + 2y + z = 6
ANSWER
Infinitely many solutions
no solution
EXAMPLE 4
Marketing
Solve a system using substitution
The marketing department of a
company has a budget of $30,000 for
advertising. A television ad costs
$1000, a radio ad costs $200, and a
newspaper ad costs $500. The
department wants to run 60 ads per
month and have as many radio ads as
television and newspaper ads
combined. How many of each type of
ad should the department run each
month?
EXAMPLE 4
Solve a system using substitution
SOLUTION
STEP 1 Write verbal models for the situation.
EXAMPLE 4
Solve a system using substitution
STEP 2 Write a system of equations. Let x be the
number of TV ads, y be the number of
radio ads, and z be the number of
newspaper ads.
x + y + z = 60
1000x + 200y + 500z = 30,000
y=x+z
Equation 1
Equation 2
Equation 3
STEP 3 Rewrite the system in Step 2 as a linear
system in two variables by substituting
x + z for y in Equations 1 and 2.
EXAMPLE 4
Solve a system using substitution
x + y + z = 60
x + (x + z) + z = 60
2x + 2z = 60
Write Equation 1.
Substitute x + z for y.
New Equation 1
1000x + 200y + 500z = 30,000 Write Equation 2.
1000x + 200(x + z) + 500z = 30,000 Substitute x + z for y.
1200x + 700z = 30,000 New Equation 2
EXAMPLE 4
Solve a system using substitution
STEP 4 Solve the linear system in two variables
from Step 3.
–1200x – 1200z = – 36,000
1200x +700z = 30,000
Add –600 times new Equation 1
to new Equation 2.
– 500z = – 6000
z = 12
Solve for z.
x = 18
Substitute into new Equation
1 or 2 to find x.
y = 30
Substitute into an original
equation to find y.
The solution is x = 18, y = 30, and z = 12, or (18, 30, 12). So, the department should run 18
TV ads, 30 radio ads, and 12 newspaper ads each month.
Do #’s 1-3 on p. 35 with a partner.