2. Objectives  Solve systems of linear equations in three variables using left-to-right elimination  Find solutions of dependent systems  Determine when a system of equations is inconsistent

3. Operations on a System of Equations The following operations result in an equivalent system. 1. Interchange any two equations. 2. Multiply both sides of any equation by the same nonzero number. 3. Multiply any equation by a number, add the result to a second equation, and then replace the second equation with the sum.

4. Left-to-Right Elimination Method of Solving Systems of Linear Equations in Three Variables x, y, and z 1. If necessary, interchange two equations or use multiplication to make the coefficient of x in the first equation a 1. 2. Add a multiple of the first equation to each of the following equations so that the coefficients of x in the second and third equations become 0. 3. Multiply (or divide) both sides of the second equation by a number that makes the coefficient of y in the second equation equal to 1. 4. Add a multiple of the (new) second equation to the (new) third equation so that the coefficient of y in the newest third equation becomes 0.

5. Left-to-Right Elimination Method of Solving Systems of Linear Equations in Three Variables x, y, and z 5. Multiply (or divide) both sides of the third equation by a number that makes the coefficient of z in the third equation equal to 1. This gives the solution for z in the system of equations. 6. Use the solution for z to solve for y in the second equation. Then substitute values for y and z to solve for x in the first equation. (This is called back substitution.)

6. Example Solve the system Solution Interchange equations 1 and 2. Multiply equation 1 by –2 and add it to the second equation. Multiply equation 1 by –3 and add it to the second equation.

7. Example (cont) To get 1 as the coefficient of y of the second equation multiply by  1. Multiply equation 2 by –4 and add it to the third equation. Solve for z.

8. Example (cont) Substituting z =  2 for z in the second equation and solving for y gives y + 3(  2) =  5, so y = 1. Substituting z =  2 and y = 1 in the first equation and solving for x gives x – 1 + 2(  2)) =  3, so x = 2. The solution is (2, 1,  2).

9. Example A manufacturer of furniture has three models of chairs: Anderson, Blake, and Colonial. The numbers of hours required for framing, upholstery, and finishing for each type of chair are given in the table. The company has 1500 hours per week for framing, 2100 hours for upholstery, and 850 hours for finishing. How many of each type of chair can be produced under these conditions?

10. Example (cont) Let x represent Anderson Let y represent Blake Let z represent Colonial Framing: 2x + 3y + z = 1500 Upholstery: x + 2y + 3z = 2100 Finishing: x + 2y + 0.5z= 850 Interchange equations 1 and 2.

11. Example (cont) (  2)Eq1 + Eq2 (  1)Eq1 + Eq3 Solve for z. Solve for y.

12. Example (cont) Solve for x. Thus, 200 Anderson, 200 Blake, and 500 Colonial chairs should be made under the given conditions.

13. Nonunique Solutions It is not always possible to reduce a system of three equations in three variables to a system in which the third equation contains one variable. A system in which 0 = x (any number) is inconsistent. A system in which 0 = 0 is a dependent system.

14. Example Solve the system. (  3) Eq1 + Eq 2 Eq(1) + Eq(3) 0 = 15 is impossible. The system is inconsistent and has no solution.

15. Example Solve the system. (  1) Eq1 (  1)Eq1+ Eq2 (2)Eq1 + Eq3

16. Example (cont) (1/3)Eq2 The equation has infinitely many solutions. z = any number; y = 2 – z and x = 2 + z Eq2 + Eq3

17. Assignment Pg. 505-508 #1-25 odd #27, 31, 34,