1. 3.6 Solving Systems of Linear Equations in Three Variables ©2001 by R. Villar All Rights Reserved

2. Solving Systems of Linear Equations in Three Variables 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 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) This system is in Triangular Form (its equations follow a stair-step pattern).

3. 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 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 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...

4. x + 6y = 16 5x + 5y = 5 –5x – 30y = –80 5x + 5y = 5 –25y = –75 y = 3 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 –2 + 3 + z = 0 1 + z = 0 z = –1(–2, 3, –1) We can solve this sytem for x and y by using linear combinations. –5( )