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

Objective Solve a system of equations in three variables.

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 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: 1 st 3 + 2(1) +4 = 9 2 nd 3(1) -4 = 1 3 rd 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 = 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: 1 st 2x – y + z = 2(3) – 2 – 1 = 3 2 nd x + 3y – 2z = (2) -2(-1) = 11 3 rd 3x – 2y + 4z 3(3) – 2(2) + 4(-1) = 1

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 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 2 nd 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 = 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 = t = 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? = 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.