Solve a Linear System in Three Variables Objectives: 1.To geometrically interpret the solution to a linear system in three variables 2.To solve a linear system in three variables using substitution and elimination

Activity 1: Graphing in 3D linear equation in three variables A linear equation in three variables x, y, and z can be written ax + by + cz = d, where a, b, c, and d are real numbers, all of which are not zero.

Activity 1: Graphing in 3D We graph this equation in 3-D, on a coordinate system with an x -, y -, and a z -axis, dividing space into eight octants. ordered triple Points in space are located with an ordered triple ( x, y, z ).

Activity 1: Graphing in 3D The solution to a linear equation in three variables is the set of all points ( x, y, z ) that satisfy the equation. In this activity, we will discover the shape of the graph of a linear equation in 3 variables.

Activity 1: Graphing in 3D We are going to use a three-dimensional coordinate system to graph the equation 3 x + 4 y + 6 z = 12. Step 1: Start by finding the x - intercept. Substitute 0 in for y and z and solve for x. Plot this point.

Activity 1: Graphing in 3D Step 2: Next find the y - intercept by substituting 0 in for x and z and solving for y. Plot this point. Step 3: Finally find the z -intercept by substituting 0 in for x and y and solving for z. Plot this point.

Activity 1: Graphing in 3D Step 4: Connect your three points: x - intercept to y -intercept, y -intercept to z - intercept, and z - intercept to x -intercept. What shape is the graph of a linear equation in 3 variables?

Activity 1: Graphing in 3D Recall a postulate from geometry which states: Through any 3 noncollinear points, there exists exactly one plane. Thus, we can conclude that the graph of a linear equation in 3 variables is a plane.

Activity I: Graphing in 3D Microsoft Mathematics 4.0: Macs Click the Graphing tab Choose 3D from the drop down menu Type in the equation and click Graph

Free Graph Paper users/smatheson/gr aphpaper/

Exercise 1 Sketch the graph of the equation. 3 x + 9 y – 3 z = -18

Linear System in 3 Variables linear equation in three variables A linear equation in three variables x, y, and z can be written ax + by + cz = d, where a, b, c, and d are real numbers, all of which are not zero. A linear system of equations in three variables has 3 such equations.

Linear System in 3 Variables linear equation in three variables A linear equation in three variables x, y, and z can be written ax + by + cz = d, where a, b, c, and d are real numbers, all of which are not zero.

Linear System in 3 Variables linear equation in three variables A linear equation in three variables x, y, and z can be written ax + by + cz = d, where a, b, c, and d are real numbers, all of which are not zero. solution ordered triple The solution to such a system is the ordered triple ( x, y, z ) that satisfies all the equations.

Graphs of 3D Systems consistent inconsistent independent dependent Recall that a system of linear equations in two variables can be either consistent or inconsistent, and that consistent systems can be either independent or dependent.

Possible Solutions Geometrically, the solution to any system of equations is the point or points of intersection.

Possible Solutions Geometrically, the solution to any system of equations is the point or points of intersection.

Solving Algebraically We’d probably not want to solve a linear system in 3 variables by graphing. Instead, there would probably be far less bloodshed if we solved such a system algebraically, using either elimination or substitution. For the elimination method, you first eliminate one of your variables so that you have 2 equations with 2 variables. Easy.

Exercise 2 Solve the system. 2 x – y + 6 z = -4 6 x + 4 y – 5 z = x – 2 y + 5 z = 9

Elimination Method In Step 1, you’ll have to eliminate the same variable from 2 different sets of the equations.

Elimination Method no solution The system has no solution if you obtain a contradiction (ex. 0 = 1) while solving the system.

Elimination Method infinitely many solutions The system has infinitely many solutions if you obtain an identity (ex. 0 = 0) while solving the system.

Protip #1: Letter Equations To help you through the often labyrinthine process of solving a 3-variable system, letter each of your equations. 2x – y + 6z = -4 6x + 4y – 5z = -7 -4x – 2y + 5z = 9 A B C

Protip #1: Letter Equations In terms of these letters, write a simple expression that tells you how to add/subtract multiples of each equation. Label the new equation with a new letter. 2x – y + 6z = -4 6x + 4y – 5z = -7 -4x – 2y + 5z = 9 A B C BC + 6x + 4y – 5z = -7 -4x – 2y + 5z = 9 + 2x + 2y = 2 D D x + y = 1 E

Protip #1: Letter Equations Continue this process until the system is solved. 2x – y + 6z = -4 6x + 4y – 5z = -7 -4x – 2y + 5z = 9 A B C + 46x + 19y = -62 F + AB x – 5y + 30z = x + 24y – 30z = -42

Protip #1: Letter Equations Continue this process until the system is solved. 2x – y + 6z = -4 6x + 4y – 5z = -7 -4x – 2y + 5z = 9 A B C + 27x = x – 19y = x + 19y = E F x = -3

Exercise 3 Solve the system. x + y – z = 2 3 x + 3 y – 3 z = 8 2 x – y + 4 z = 7

Exercise 4 Solve the system. x + y + z = 6 x – y + z = 6 4 x + y + 4 z = 24

Protip #2: Multiple Solutions When you discover that you have a consistent, dependent system of equations, how do you write your answer? Graphically, the equations in this system intersect in a line, so you could just write the equation of that line. But what if you want specific solutions, in the form of ordered triples?

Protip #2: Multiple Solutions To write your answers as a set of ordered pairs, set one of the variables in your equation equal to a. Now re-write the other variable in terms of a. x + y + z = 6 x + z = 6 Let x = a Then by substitution in the 2 nd equation: a + z = 6 z = 6 – a Then by substitution in the 1 st equation: a + y + (6 – a) = 6 y = 0

Protip #2: Multiple Solutions Finally, use your new expressions to write an ordered triple. Substitute values in for a to get a specific solution points. x + y + z = 6 x + z = 6 x = a z = 6 – a y = 0 (a, 0, 6 – a) Let a = 0: (0, 0, 6) Let a = 1: (1, 0, 5) Let a = -1: (-1, 0, 7)

Exercise 5 Solve each system. 1.3 x + y – 2 z = 10 6 x – 2 y + z = -2 x + 4 y + 3 z = 7 2. x + y – z = 2 2 x + 2 y – 2 z = 6 5 x + y – 3 z = 8 3. x + y + z = 3 x + y – z = 3 2 x + 2 y + z = 6

Exercise 6 At a carry-out pizza restaurant, an order of 3 slices of pizza, 4 breadsticks, and 2 soft drinks cost $ A second order of 5 slices of pizza, 2 breadsticks, and 3 soft drinks cost $ If four bread sticks and a can of soda cost $.30 more than a slice of pizza, what is the cost of each item?

Substitution Method substitution If it is convenient, you could use substitution to help solve a linear system in three variables. 1.Solve one of the equations for one of the variables. 2.Substitute the expression from Step 1 into both of the other equations. 3.Solve the remaining 2 variable system.

Exercise 7: SAT If 5 sips + 4 gulps = 1 glass and 13 sips + 7 gulps = 2 glasses, how many sips equal a gulp?