# 3.9 Solving Systems of Equations in Three Variables

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3.5 Solving Systems of Equations in Three Variables Mrs. Spitz Fall 2006.

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3.9 Solving Systems of Equations in Three Variables
Algebra II Mrs. Spitz Fall 2006

Objective Solve a system of equations in three variables.

Assignment Pp #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.
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? = 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.

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 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) 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) 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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