1. Slide 4-1 Copyright © 2004 Pearson Education, Inc. Systems of Linear Equations Chapter 4

2. Slide 4-2 Copyright © 2004 Pearson Education, Inc. Page 224, Figure 4.1

3. Slide 4-3 Copyright © 2004 Pearson Education, Inc. Page 225, Figure 4.2

4. Slide 4-4 Copyright © 2004 Pearson Education, Inc. Page 226

5. Slide 4-5 Copyright © 2004 Pearson Education, Inc. Page 226-227

6. Slide 4-6 Copyright © 2004 Pearson Education, Inc. Page 229, Figure 4.4

7. Slide 4-7 Copyright © 2004 Pearson Education, Inc. Page 229, Figure 4.5

8. Slide 4-8 Copyright © 2004 Pearson Education, Inc. Page 229

9. Slide 4-9 Copyright © 2004 Pearson Education, Inc. Page 230

10. Slide 4-10 Copyright © 2004 Pearson Education, Inc. Page 232, Figure 4.6

11. Slide 4-11 Copyright © 2004 Pearson Education, Inc. Page 239, Figure 4.7

12. Slide 4-12 Copyright © 2004 Pearson Education, Inc. Page 239-240

13. Slide 4-13 Copyright © 2004 Pearson Education, Inc. Page 240

14. Slide 4-14 Copyright © 2004 Pearson Education, Inc. Page 247

15. Slide 4-15 Copyright © 2004 Pearson Education, Inc. Page 250, Figure 4.9 Mixture Problems

16. Slide 4-16 Copyright © 2004 Pearson Education, Inc. The illustration above is a bit deceiving if we are adding volume of the same measurement. For example, if the first container has 3 ounces and is mixed with the second container that has 7 ounces, the resulting container must have 10 ounces. Remember that the volume is added in the result. The set-up: For the illustration above, the equation we need to solve becomes:.05x+.20y=.125*10 Since x+y = 10, then y=10-x.05x+.20y=.125*10.05x+.20(10-x)=.125*10 Solving yields: x=5. Since the total liquid is known to be 10 liters, then y must be 5 ounces in this case. Slide Edited by Ted Koukounas