# 3.6 Solving Systems of Linear Equations in Three Variables 7.1 Solving Systems of Linear Equations in Three Variables Warm-Up No Solution Infinitely many solutions Here is a system of three linear equations in three variables: x 2 y 3z 3 2 x 5 y 4 z 13 5 x 4 y z 5 The ordered triple (2,-1,1) is a solution to this system since it is a

solution to all three equations. 2 2( 1) 3(1) 2 2 3 3 2(2) 5( 1) 4(1) 4 5 4 13 5(2) 4( 1) 1 10 4 1 5 The graph of a linear equation in three variables is a plane. Three planes in space can intersect in different ways (pg 218). The planes could The planes could intersect in a line. intersect in a The system has single point. infinitely many

The system has solutions exactly one solution The planes could have NO point of intersection. The left figure shows planes that intersect pairwise, but all 3 do not have a common point of intersection. The right figure shows parallel planes. Each system has NO solution. The linear combination method in lesson 3.2 can be extended to solve a system of linear equations in three variables. Solve this system 3x 2 y 4 z 11 2 x y 3 z 4 5 x 3 y 5 z 1

Our strategy will be to use two of the equations to eliminate one of the variables. We will then use two other equations to eliminate the same variable. Once we have two equations with two variables, we can use the technique we learned in lesson 3.2 Solve this system 3x 2 y 4 z 11 2 x y 3 z 4 5 x 3 y 5 z 1

Equation 1 Equation 2 Equation 3 Multiply Eq. 2 by 2 and add it to Eq. 1. Save this result 3 x 2 y 4 z 11 4 x 2 y 6 z 8 7x +10z = 19 Now multiply Eq. 2 by -3 and add it to Eq. 3. Save this result. 6 x 3 y 9 z 12

...5 x 3 y 5 z 1 -x -4z = -13 Solve this new system of linear equation in two variables. Multiply the bottom eq. by 7 and add it to the top eq. 7 x 10 z 19 7 x 28 z 91 -18z=-72 or z = 4 Substituting z=4 into either of the new equations will give x = -3finally substituting these values into any of the original equations give y = 2. Our final solution is (-3,2,4)

Here is a system with No Solution. Watch what happens when we try to solve it. x y z 2 3 x 3 y 3 z 14 x 2 y z 4 3 x 3 y 3 z 6 ..3 x 3 y 3 z 14 0=8 Equation 1 Equation 2 Equation 3 Add -3 times Eq 1 to Eq 2. Since this is a false equation, you can conclude the original system of equations

has no solution. Here is a system with MANY solutions. Watch what happens when we try to solve it. Equation 1 x y z 2 Equation 2 x y z 2 2 x 2 y z 4 Equation 3 x y z 2 x y z 2 2x + 2y =4

Add Eq. 1 to Eq. 2 New EQ. 1 x y z 2 Add Eq 2 to Eq 3 2 x 2 y z 4 3x +3y =6 New EQ 2 Solving this new system of two equation by adding -3 times the first eq. to 2 times the second eq. produces the identity 0 = 0. So, the system has infinitely many solution. You could describe the solution this way: divide New Eq 1 by 2 to get x+y=2, or

y=-x+2. Substituting this into the orignial Equation 1 produces z = 0. So any ordered triple of the form (x, -x+2,0) is a solution. For example (0,2,0) and (2,0,0) are solutions. Substitution Method x y z 24 5 x 3 y z 56 x y z Since x+y=z, substitute this for z in the first two equations x y ( x y ) 24 5 x 3 y ( x y ) 56 Simplify

2 x 2 y 24 6 x 4 y 56 Finally, solve this linear system of two equations and two variables to get x = 4 and y =8 Since z=x+y, z = 12. Our final solution is (4,8,12) www.pleasanton.k12.ca.us