Download 3.9 Solving Systems of Equations in Three Variables

3.4 Solving Systems of
Equations in Three Variables
Algebra II
Mrs. Aguirre
Fall 2013
Objective
• Solve a system of equations in three
variables.
Application
• Courtney has a total of 256 points on three
Algebra tests. Her score on the first test
exceeds his score on the second by 6
points. Her total score before taking the
third test was 164 points. What were
Courtney’s test scores on the three tests?
Explore
• Problems like this one can be solved using
a system of equations in three variables.
Solving these systems is very similar to
solving systems of equations in two
variables. Try solving the problem
– Let f = Courtney’s score on the first test
– Let s = Courtney’s score on the second test
– Let t = Courtney’s score on the third test.
Plan
• Write the system of equations from the
information given.
f + s + t = 256
f–s=6
f + s = 164
The total of the scores is 256.
The difference between the 1st and 2nd is 6 points.
The total before taking the third test is the sum of
the first and second tests..
Solve
• Now solve. First use elimination on the
last two equations to solve for f.
f–s=6
f + s = 164
2f
= 170
The first test score is 85.
f = 85
Solve
• Then substitute 85 for f in one of the
original equations to solve for s.
f + s = 164
85 + s = 164
s = 79 The second test score is 79.
Solve
• Next substitute 85 for f and 79 for s in f + s
+ t = 256.
f + s + t = 256
85 + 79 + t = 256
164 + t = 256
t = 92 The third test score is 92.
Courtney’s test scores were 85, 79, and 92.
Examine
• Now check your results against the original
problem.
• Is the total number of points on the three tests
256 points?
85 + 79 + 92 = 256 ✔
• Is one test score 6 more than another test
score?
79 + 6 = 85 ✔
• Do two of the tests total 164 points?
85 + 79 =164 ✔
• Our answers are correct.
Solutions?
• You know that a system of two linear
equations doesn’t necessarily have a
solution that is a unique ordered pair.
Similarly, a system of three linear
equations in three variables doesn’t
always have a solution that is a unique
ordered triple.
Graphs
•
The graph of each equation in a system
of three linear equations in three
variables is a plane. Depending on the
constraints involved, one of the following
possibilities occurs.
Graphs
1.
The three planes
intersect at one point.
So the system has a
unique solution.
2. The three planes
intersect in a line. There
are an infinite number of
solutions to the system.
Graphs
3. Each of the diagrams below shows three
planes that have no points in common. These
systems of equations have no solutions.
Ex. 1: Solve this system of
equations
x  2y  z  9
3 y  z  1
3z  12
• Solve the third equation, 3z =
12
3z = 12
z=4
• Substitute 4 for z in the second
equation 3y – z = -1 to find y.
3y – (4) = -1
3y = 3
y=1
Substitute 4 for z and 1 for y in
the first equation, x + 2y + z =
9 to find x.
x + 2y + z = 9
x + 2(1) + 4 = 9
x+6=9
x = 3 Solution is (3, 1, 4)
Check:
1st 3 + 2(1) +4 = 9 ✔
2nd 3(1) -4 = 1 ✔
3rd 3(4) = 12 ✔
Ex. 2: Solve this system of
equations
2x  y  z  3
x  3 y  2 z  11
3x  2 y  4 z  1
• Set the first two equations
together and multiply the first
times 2.
2(2x – y + z = 3)
4x – 2y +2z = 6
x + 3y -2z = 11
5x + y = 17
Set the next two equations
together and multiply the first
times 2.
2(x + 3y – 2z = 11)
2x + 6y – 4z = 22
3x - 2y + 4z = 1
5x + 4y = 23
Next take the two equations that
only have x and y in them and
put them together. Multiply the
first times -1 to change the
signs.
Ex. 2: Solve this system of
equations
2x  y  z  3
x  3 y  2 z  11
3x  2 y  4 z  1
Next take the two equations that only
have x and y in them and put them
together. Multiply the first times -1
to change the signs.
-1(5x + y = 17)
-5x - y = -17
5x + 4y = 23
3y = 6
y=2
Now you have y = 2. Substitute y into
one of the equations that only has
an x and y in it.
5x + y = 17
5x + 2 = 17
5x = 15
x=3
Now you have x and y. Substitute
values back into one of the
equations that you started with.
2x – y + z = 3
2(3) - 2 + z = 3
6–2+z=3
4+z=3
z = -1
Ex. 2: Check your work!!!
2x  y  z  3
x  3 y  2 z  11
3x  2 y  4 z  1
Solution is (3, 2, -1)
Check:
1st 2x – y + z =
2(3) – 2 – 1 = 3 ✔
2nd x + 3y – 2z = 11
3 + 3(2) -2(-1) = 11 ✔
3rd 3x – 2y + 4z
3(3) – 2(2) + 4(-1) = 1 ✔
Ex. 2: Solve this system of
equations
2x  y  z  3
x  3 y  2 z  11
3x  2 y  4 z  1
Next take the two equations that only
have x and y in them and put them
together. Multiply the first times -1
to change the signs.
-1(5x + y = 17)
-5x - y = -17
5x + 4y = 23
3y = 6
y=2
Now you have y = 2. Substitute y into
one of the equations that only has
an x and y in it.
5x + y = 17
5x + 2 = 17
5x = 15
x=3
Now you have x and y. Substitute
values back into one of the
equations that you started with.
2x – y + z = 3
2(3) - 2 + z = 3
6–2+z=3
4+z=3
z = -1