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: The ordered triple (2,-1,1) is a solution to this system since it is a solution to all three equations.

The graph of a linear equation in three variables is a plane The graph of a linear equation in three variables is a plane. Three planes in space can intersect in different ways The planes could intersect in a line. The system has infinitely many solutions The planes could intersect in a single point. The system has 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.

Substitution Method - Example 1 Since x+y=z, substitute this for z in the first two equations Simplify 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)

Example 2 -3(2z+10) + y – 4z = 20 -4(2z+10) + 2y + 3z = -15 Solving for x in the first equation, we get 2z + 10 =x. Substitute this for x in the last two equations -3(2z+10) + y – 4z = 20 -4(2z+10) + 2y + 3z = -15 Simplify Finally, solve this linear system of two equations and two variables to get y= 0 and y = -5 y – 10z = 50 2y -5z = 25 Since x=2z + 10, x = 0. Our final solution is (0, 0, -5)

Example 3 5r +4(t+2) – 6t = -24 -2(t+2) + 2t = 0 -2t - 4 + 2t = 0 Since t + 2 = s, substitute this for s in the first two equations 5r +4(t+2) – 6t = -24 -2(t+2) + 2t = 0 Simplify Since we have a false statement, this system is inconsistent. -2t - 4 + 2t = 0 -4 = 0 Our final solution is No Solution

Try it on your own! B. A.