1. Copyright © 2010 Pearson Education, Inc. Systems of Linear Equations in Three Variables Learn basic concepts about systems in three variables Learn basic concepts about systems in three variables Solve systems using elimination and substitution Solve systems using elimination and substitution Solve systems with no solution Solve systems with no solution Solve systems with infinitely many solutions Solve systems with infinitely many solutions 9.3

2. Slide 6.3 - 2 Copyright © 2010 Pearson Education, Inc. Basic Concepts The solution to a system of linear equations in three variables is an ordered triple, (x, y, z). A system of linear equations can have zero, one, or infinitely many solutions.

3. Slide 6.3 - 3 Copyright © 2010 Pearson Education, Inc. Solving a System of Equations in Three Variables STEP 1: Eliminate one variable, such as x, from two of the equations. STEP 2: Apply the techniques discussed in Section 6.1 to solve the two resulting equations in two variables from Step 1. If x is eliminated, then solve these equations to find y and z.

4. Slide 6.3 - 4 Copyright © 2010 Pearson Education, Inc. Solving a System of Equations in Three Variables STEP 2:If there are no solutions for y and z, then the given system also has no solutions. If there are infinitely many solutions for y and z, then write y in terms of z and proceed to Step 3. STEP 3:Substitute the values for y and z in one of the given equations to find x. The solution is (x, y, z). Check your answer as in Example 1.

5. Slide 6.3 - 5 Copyright © 2010 Pearson Education, Inc. Example 2 Solve the following system.

6. Slide 6.3 - 6 Copyright © 2010 Pearson Education, Inc. Example 2 Solution STEP 1:Eliminate the variable x from the 2nd and 3rd equations.

7. Slide 6.3 - 7 Copyright © 2010 Pearson Education, Inc. Example 2 Solution continued STEP 2:Use the two resulting equations and eliminate the variable y. Substitute 2 for z into either equation and find y. 3y – 6z = –15

8. Slide 6.3 - 8 Copyright © 2010 Pearson Education, Inc. Example 2 Solution continued STEP 2: STEP 3:Substitute y = –1 and z = 2 in any given equation.

9. Slide 6.3 - 9 Copyright © 2010 Pearson Education, Inc. Example 2 Solution continued STEP 3: The solution is (1, –1, 2).

10. Slide 6.3 - 10 Copyright © 2010 Pearson Education, Inc. Example 4 Three students buy lunch in the cafeteria. One student buys 2 hamburgers, 1 order of fries, and 1 soda for $9. Another student buys 1 hamburger, 2 orders of fries, and 1 soda for $8. The third student buys 3 hamburgers, 3 orders of fries, and 2 sodas for $18. If possible, find the cost of each item. Interpret the results.

11. Slide 6.3 - 11 Copyright © 2010 Pearson Education, Inc. Example 4 Solution Let x be the cost of a hamburger, y be the cost of an order of fries, and z be the cost of a soda. Then the purchases of the three students can be expressed as a system of linear equations.

12. Slide 6.3 - 12 Copyright © 2010 Pearson Education, Inc. Example 4 Solution continued STEP 1:Eliminate z in equations (1) & (3).

13. Slide 6.3 - 13 Copyright © 2010 Pearson Education, Inc. Example 4 Solution continued STEP 2:The equations x – y = 2 and x – y = 1 are inconsistent because the difference between two numbers cannot be both 1 and 2. Step 3 is not necessary– the system of equations has no solutions.

14. Slide 6.3 - 14 Copyright © 2010 Pearson Education, Inc. Example 5 Solve the following system of linear equations.

15. Slide 6.3 - 15 Copyright © 2010 Pearson Education, Inc. Example 5 Solution STEP 1:Because x does not appear in equation (3) eliminate x from equation (1).

16. Slide 6.3 - 16 Copyright © 2010 Pearson Education, Inc. Example 5 Solution continued STEP 2:Adding the resulting equation from Step 1 and the third given equation gives the equation 0 = 0, which indicates that there are infinitely many solutions. The variable y can be written in terms of z as y = z – 1.

17. Slide 6.3 - 17 Copyright © 2010 Pearson Education, Inc. Example 5 Solution continued STEP 3:To find x, substitute the results from Step 2 in the first given equation. Solutions are of the form (–1, z–1, z), where z is any real number. For example, if z = 2, then (–1, 1, 2) is one possible solution.