1. Warm-Up A child’s ticket costs $6 for admission into the movie theater. Adult tickets cost $15. How many of each did a family get if they spent $63 for 6 tickets? A = 3 Let C – # of Child tickets A – # of Adult tickets C = 3 They bought 3 Adult and 3 Child tickets

2. Objective: (1) Students will solve word problems using systems of equations. CA Standards 9.0: Students solve a system of two linear equations in two variables algebraically and are able to interpret the answer graphically. Students are able to solve a system of two linear inequalities in two variables and to sketch the solution sets. Agenda: 12/09/2011 1.) Warm-up 2.) Answers to Homework 3.) Lesson Word Problems – Current (ppt.) 4.) Assignment 5.) Stay on Task!!

3. 8.4 Word Problems – Current Notes: Word Problems Type: Current Ex 1) A fish swims 45 miles down stream in 5 hours. It takes the same amount of time to travel 15 miles upstream. How fast is the current. Let F – Rate of fish swimming C – Rate of current Rate ● Time = Distance 45 15 5 5 F + C F – C 5F + 5C = 45 5F– 5C = 15 10F + 0 = 60 10 F = 6 m.p.h. 5F + 5C = 45 5(6) + 5C = 45 30 + 5C = 45 -30 5C = 15 5 5 C = 3 m.p.h. with against With the current (F + C)5 = 45 (F – C)5 = 15 Against the current

4. 8.4 Word Problems – Current Try this one ! ! ! Ex 2) A fish swims 56 miles down stream in 4 hours. It takes the same amount of time to travel 24 miles upstream. How fast is the current. Let F – Rate of fish swimming C – Rate of current Rate ● Time = Distance 56 24 4 4 F + C F – C 4F + 4C = 56 4F– 4C = 24 8F + 0 = 80 8 8 F = 10 m.p.h. 4F + 4C = 56 4(10) + 4C = 56 40 + 4C = 56 -40 4C = 16 4 4 C = 4 m.p.h. with against With the current (F + C)4 = 56 (F – C)4 = 24 Against the current

5. Ex 3) A duck swims 12 miles in 2 hours with the current. The same duck swims the same distance in 6 hours against the current. How fast is the duck in still water? 8.4 Word Problems – Current Notes: Word Problems Type: Current Let D – Rate of Duck swimming C – Rate of current Rate ● Time = Distance 12 2 6 D + C D – C 2D + 2C = 12 6D– 6C = 12 12D + 0 = 48 12 D = 4 m.p.h. 2D + 2C = 12 2(4) + 2C = 12 8 + 2C = 12 -8 2C = 4 2 2 C = 2 m.p.h. with against 3 6D + 6C = 36 6D – 6C = 12

6. 8S – 8C = 80 Ex 4) A shark swims 80 miles in 4 hours with the current. The same shark swims the same distance in 8 hours against the current. How fast is the shark in still water? 8.4 Word Problems – Current Let S – Rate of Shark swimming C – Rate of current Rate ● Time = Distance 80 4 8 S + C S – C 4S + 4C = 80 8S– 8C = 80 16S + 0 = 240 16 S = 15 m.p.h. 4S + 4C = 80 4(15) + 4C = 80 60 + 4C = 80 -60 4C = 20 4 4 C = 5 m.p.h. with against 2 8S + 8C = 160 Try it ! ! !

7. 10T Ex 5) Tarzan and Jane swung threw the rain forest on a windy day. If they were able to travel 35 miles in 5 hours with the wind and only 10 miles in 10 hours against the wind, what was the rate of speed of the wind? 10T–10W = 10 8.4 Word Problems – Current Let T – Rate of Tarzan & Jane W – Rate of Wind Rate ● Time = Distance 35 10 5 T + W T – W 5T + 5W = 35 10T–10W = 10 20T + 0 = 80 20 T = 4 m.p.h. 5T + 5W = 35 5(4) + 5w = 35 20 + 5W = 35 -20 5W = 15 5 5 W = 3 m.p.h. with against 2 +10W = 70 Try it ! ! !