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**3.9 Solving Systems of Equations in Three Variables**

Algebra II Mrs. Spitz Fall 2006

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Objective Solve a system of equations in three variables.

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Assignment Pp #11-23 all

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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?

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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.

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**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..

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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.

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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.

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**Solve Next substitute 85 for f and 79 for s in f + s + t = 256.**

The third test score is 92. Courtney’s test scores were 85, 79, and 92.

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**Examine Now check your results against the original problem.**

Is the total number of points on the three tests 256 points? = 256 ✔ Is one test score 6 more than another test score? = 85 ✔ Do two of the tests total 164 points? =164 ✔ Our answers are correct.

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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.

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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.

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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.

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

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**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

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**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

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**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) 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

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**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 ✔

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**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) 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

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