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5.2 Systems of Linear Equations in Three Variables.

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Presentation on theme: "5.2 Systems of Linear Equations in Three Variables."— Presentation transcript:

1 5.2 Systems of Linear Equations in Three Variables

2 Example: Solve x + 2y + z = 4 4y – 3z = 1 5z = 5 Solve for one of the variables, somewhere 5z = 5 z = 1 Now substitute into the other equations x + 2y + 1 = 4 4y – 3(1) = 1 This system is in Triangular Form (its equations follow a stair-step pattern).

3 Solve the 2nd equation for y 4y – 3 = 1 4y = 4y = 1 Now, solve the 1st equation for x x + 2(1) + 1 = 4 x + 3 = 4 x = 1 (1, 1, 1)

4 Example: Solve: x + y + z = 0 2x + 2y – 3z = 5 x – 4y – 3z = –11 We will use linear combinations in the 1st and 2nd equations to eliminate the z terms : 3 (x + y + z = 0) 3x + 3y + 3z = 0 2x + 2y – 3z = 5 2x + 2y – 3z = 5 5x + 5y = 5 This system is not in triangular form. Use elimination to get it in triangular form, or linear combinations to produce a system of 2 equations in 2 variables...

5 We will do the same for the 2nd and 3rd equations: 2x + 2y – 3z = 5 2x + 2y – 3z = 5 –1(x – 4y – 3z = –11) –x + 4y + 3z = 11 x + 6y = 16 We now have now produced a system of 2 equations in 2 variables… 5x + 5y = 5 x + 6y = 16

6 x + 6y = 16 5x + 5y = 5 –5x – 30y = –80 5x + 5y = 5 –25y = –75 y = 3

7 Now, find x: x + 6(3) = 16 x + 18 = 16 x = –2 Finally, go back to one of the original equations to find z: x + y + z = 0 – z = z = 0 z = –1(–2, 3, –1)


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