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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?. Let C – # of Child tickets. 6C + 15A. = 63. 6C + 15A = 63. A – # of Adult tickets. – 6A. = -36. -6C. = 6. -6. C + A.
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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? LetC – # of Child tickets 6C + 15A = 63 6C + 15A = 63 A – # of Adult tickets – 6A = -36 -6C = 6 -6 C + A 9A 0 + = 27 9 9 3 A = C + A = 6 C + (3) = 6 -3 -3 C = 3 They bought 3 Adult and 3 Child tickets!!
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. Objective: (1) Students will solve word problems using systems of equations. Agenda: 06/06/11 1.) Warm-up 2.) Questions WS Systems Review 3.) Lesson Word Problems – Current (ppt.) 4.) Class/Home Work WS Systems Current 5.) Bring Your Text Book on WEDNESDAY!!
8.4 Word Problems – Current Notes: Word Problems Type: Current With the 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. Against the current LetF – Rate of fish swimming Rate ● Time = Distance C – Rate of current (F + C)5 = 45 F + C 5 45 with 15 (F – C)5 = 15 F – C against 5 5F + 5C = 45 5F = 45 + 5C 5(6) + 5C = 45 = 15 5F – 5C 30 + 5C = 45 -30 -30 + 0 = 60 10F 10 10 5C = 15 C = 3 m.p.h. 5 5 F = 6 m.p.h.
8.4 Word Problems – Current Try this one ! ! ! With the current 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. Against the current LetF – Rate of fish swimming Rate ● Time = Distance C – Rate of current (F + C)4 = 56 F + C 4 56 with 24 (F – C)4 = 24 F – C against 4 4F + 4C = 56 4F = 56 + 4C 4(10) + 4C = 56 = 24 4F – 4C 40 + 4C = 56 -40 -40 + 0 = 80 8F 8 8 4C = 16 C = 4 m.p.h. 4 4 F = 10 m.p.h.
8.4 Word Problems – Current Notes: Word Problems Type: Current 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? LetD – Rate of Duck swimming Rate ● Time = Distance C – Rate of current D + C 2D = 12 + 2C 2 3 12 with 12 D – C = 12 against 6D – 6C 6 + 6C 6D = 36 2D + 2C = 12 2(4) + 2C = 12 6D – 6C = 12 8 + 2C = 12 + 0 = 48 12D -8 -8 12 12 2C = 4 C = 2 m.p.h. 4 m.p.h. D = 2 2
8.4 Word Problems – Current Try it ! ! ! 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? LetS – Rate of Shark swimming Rate ● Time = Distance C – Rate of current S + C + 4C 4S = 80 4 2 80 with 80 S – C = 80 against 8S – 8C 8 + 8C 8S = 160 4S + 4C = 80 4(15) + 4C = 80 8S – 8C = 80 60 + 4C = 80 + 0 = 240 16S -60 -60 16 16 4C = 20 C = 5 m.p.h. 15 m.p.h. S = 4 4
8.4 Word Problems – Current Try it ! ! ! 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? LetT – Rate of Tarzan & Jane Rate ● Time = Distance W – Rate of Wind T + W + 5W 5T = 35 5 2 35 with 10 T – W = 10 against 10T –10W 10 +10W 10T = 70 5T + 5W = 35 5(4) + 5w = 35 10T–10W = 10 20 + 5W = 35 + 0 = 80 20T -20 -20 20 20 5W = 15 W = 3 m.p.h. 4 m.p.h. T = 5 5
