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3.9 Solving Systems of Equations in Three VariablesPowerPoint Presentation

3.9 Solving Systems of Equations in Three Variables

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Objective

- Solve a system of equations in three variables.

Assignment

- Pp. 147-148 #11-23 all

Application

- Courtney has a total of 256 points on three Algebra tests. His score on the first test exceeds his score on the second by 6 points. His 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

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.

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.

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.

GraphsGraphs a unique solution.

3. Each of the diagrams below shows three planes that have no points in common. These systems of equations have no solutions.

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. 1: Solve this system of equationsSet 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 equationsNow 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: Solve this system of equationsEx. 2: Check your work!!! that only has an x and y in it.

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 ✔

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: Solve this system of equations
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