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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 Courtneys 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 Courtneys score on the first test
- Let s Courtneys score on the second test
- Let t Courtneys 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
- f 85

The first test score is 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.

Courtneys 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

doesnt necessarily have a solution that is a

unique ordered pair. Similarly, a system of

three linear equations in three variables doesnt

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

- 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

- 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 ?

- 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

Ex. 2 Solve this system of equations

- 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.

- 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

Ex. 2 Solve this system of equations

- 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

- 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

Ex. 2 Check your work!!!

- 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

- 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

- 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