3.4 Solving Systems of Linear Equations in 3 Variables

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Presentation transcript:

3.4 Solving Systems of Linear Equations in 3 Variables Algebra II

A system of lin. eqns. in 3 variables Looks something like: x+3y-z=-11 2x+y+z=1 5x-2y+3z=21 A solution is an ordered triple (x,y,z) that makes all 3 equations true.

Steps for solving in 3 variables Using the 1st 2 equations, cancel one of the variables. Using the last 2 equations, cancel the same variable from step 1. Use the results of steps 1 & 2 to solve for the 2 remaining variables. Plug the results from step 3 into one of the original 3 equations and solve for the 3rd remaining variable. Write the solution as an ordered triple (x,y,z).

1)Solve the system. x+3y-z=-11 2x+y+z=1 5x-2y+3z=21 x+3y-z=-11 z’s are easy to cancel! 3x+4y=-10 2. 2x+y+z=1 5x-2y+3z=21 Must cancel z’s again! -6x-3y-3z=-3 -x-5y=18 2(2)+(-4)+z=1 4-4+z=1 3. 3x+4y=-10 -x-5y=18 Solve for x & y. 3x+4y=-10 -3x-15y=54 -11y=44 y=- 4 3x+4(-4)=-10 x=2 (2, - 4, 1) z=1

2)Solve the system. -x+2y+z=3 2x+2y+z=5 4x+4y+2z=6 -2(2x+2y+z=5) Cancel z’s again. -4x-4y-2z=-10 0=- 4 Doesn’t make sense! No solution -x+2y+z=3 -(2x+2y+z=5) z’s are easy to cancel! -2x-2y-z=-5 -3x=-2 x=2/3

3)Solve the system. -2x+4y+z=1 3x-3y-z=2 5x-y-z=8 -2x+4y+z=1 3x-3y-z=2 z’s are easy to cancel! x+y=3 5x-y-z=8 Cancel z’s again. -3x+3y+z=-2 2x+2y=6 3. x+y=3 2x+2y=6 Cancel the x’s. -2x-2y=-6 0=0 This is true. IMS

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