Warm-UpWarm-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 ticketsThey bought 3 Adult and 3 Child tickets

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: 12/09/2011Agenda: 12/09/2011

1.) Warm-up

2.) Answers to Homework

3.) Lesson

Word Problems – Current (ppt.)

4.) Assignment

5.) Stay on Task!!

8.4 Word Problems – 8.4 Word Problems – CurrentCurrentNotes: Word ProblemsType: 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 currentRate ● Time = Distance

4515

5

5

F + C

F – C

5F + 5C = 45

5F – 5C = 15

10F + 0 = 601010

F = 6 m.p.h.

5F + 5C = 455(6) + 5C = 45

30 + 5C = 45-30 -30

5C = 1555

C = 3 m.p.h.

with

against

With the current

(F + C)5 = 45

(F – C)5 = 15

Against the current

8.4 Word Problems – 8.4 Word Problems – CurrentCurrentTry 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 currentRate ● Time = Distance

5624

4

4

F + C

F – C

4F + 4C = 56

4F – 4C = 24

8F + 0 = 8088

F = 10 m.p.h.

4F + 4C = 564(10) + 4C = 56

40 + 4C = 56-40 -40

4C = 1644

C = 4 m.p.h.

with

against

With the current

(F + C)4 = 56

(F – C)4 = 24

Against the 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?

8.4 Word Problems – 8.4 Word Problems – CurrentCurrentNotes: Word ProblemsType: Current

Let D – Rate of Duck swimming

C – Rate of currentRate ● Time = Distance

1212

2

6

D + C

D – C

2D + 2C = 12

6D – 6C = 12

12D + 0 = 481212

D = 4 m.p.h.

2D + 2C = 122(4) + 2C = 12

8 + 2C = 12-8 -8

2C = 422

C = 2 m.p.h.

with

against

3

6D + 6C = 36

6D – 6C = 12

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 – 8.4 Word Problems – CurrentCurrent

Let S – Rate of Shark swimming

C – Rate of currentRate ● Time = Distance

8080

4

8

S + C

S – C

4S + 4C = 80

8S – 8C = 80

16S + 0 = 2401616

S =15 m.p.h.

4S + 4C = 804(15) + 4C = 80

60 + 4C = 80-60 -60

4C = 2044

C = 5 m.p.h.

with

against

2

8S + 8C = 160

Try it ! ! !

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 – 8.4 Word Problems – CurrentCurrent

Let T – Rate of Tarzan & Jane

W – Rate of WindRate ● Time = Distance

3510

5

10

T + W

T – W

5T + 5W = 35

10T–10W = 10

20T + 0 = 802020

T = 4 m.p.h.

5T + 5W = 355(4) + 5w = 3520 + 5W = 35

-20 -20

5W = 1555

W = 3 m.p.h.

with

against

2

+10W = 70

Try it ! ! !