4.4 Problem Solving Using Systems of Equations

Transcription

1 4.4 Problem Solving Using Systems of Equations

2 Steps 1) Read, read, read. Draw a diagram, if needed. 2) Assign a variable to each unknown (need 2 variables). 3) Translate so you have 2 equations. 4) Solve the system. 5) State the conclusion clearly. Use units. 6) Check answer in original wording of problem.

3 Motion Motion: distance = rate time

4 Ex. When a crew rows with the current, it travels 16 miles in 2 hours. Against the current, the crew rows 8 miles in 2 hours. Find the rate of rowing in still water and the rate of the current. rate time = distance w/current x + y 2 16 Against current x y 2 8 x = crew s rowing rate in still water y = rate of the current system: 2(x + y) = 16 2(x y) = 8 Elimination Method: 2(x + y) = 16 2x + 2y = 16 2(x y) = 8 2x 2y = 8 4x + 0 = 24 4x = 24 4x = x = 6

5 Find y (x = 6): 2x + 2y = 16 2(6) + 2y = y = y 12 = y = 4 2y = y = 2 ans: rate of crew is 6 mph rate of current is 2 mph

6 Interest Simple Interest: Interest = principal rate time I = Prt

7 Ex. You deposit $9,200 in a savings account, which has a rate of 4%. Find the interest at the end of the first year. I =? P = 9,200 r = 0.04 t = 1 I = Prt I = 9200(0.04)(1) I = 368 $368 in interest after 1 yr.

8 Ex. You invest $20,000 in two accounts paying 7% and 9% annual interest. If the total interest earned for the year was $1550, how much was invested at each rate?

9 Ex. You invest $20,000 in two accounts paying 7% and 9% annual interest. If the total interest earned for the year was $1550, how much was invested at each rate? Principal Rate = Interest

10 Ex. You invest $20,000 in two accounts paying 7% and 9% annual interest. If the total interest earned for the year was $1550, how much was invested at each rate? x + y = 20, x y = 1,550 Principal Rate = Interest 7% investment x x 9% investment y y Substitution: x + y = 20,000 X + y x = 20,000 x y = 20,000 x 0.07x y = 1, x (20,000 x) = x x = x = x = x = x = x = 12,500 7% acct. has x = $12,500 9% acct. has 20,000 x = 20,000 12,500 = $7,500 Ans: $12,500 at 7% $7,500 at 9%

11 Mixture Mixture: amount strength = ingredient

12 Ex. A 70 milliliter solution of acid in water contains 60% acid. How much acid is in the solution? solution = 70 concentration = 0.6 ingredient =? amount strength = ingredient 70(0.6) = ingredient 42 = ingredient Ans: 42 ml of acid

13 Ex. A chemistry experiment calls for a 30% sulfuric acid solution. If the lab supply room has only 50% and 20% sulfuric acid solutions, how much of each should be mixed to obtain 12 liters of a 30% acid solution?

14 Ex. A chemistry experiment calls for a 30% sulfuric acid solution. If the lab supply room has on 50% and 20% sulfuric acid solutions, how much of each should be mixed to obtain 12 liters of a 30% acid solution? amount strength = amount of pure sulfuric acid

15 Ex. A chemistry experiment calls for a 30% sulfuric acid solution. If the lab supply room has on 50% and 20% sulfuric acid solutions, how much of each should be mixed to obtain 12 liters of a 30% acid solution? x = amount of 20% soln. y = amount of 50% soln. System: x + y = x + 0.5y = 3.6 amount strength = amount of pure sulfuric acid weak (20% solution) x x strong (50% solution) y y 30% solution (12) Elimination Method: -0.2(x + y) = -0.2(12) -0.2x 0.2y = x + 0.5y = x + 0.5y = y = y = y = y = 1.2

16 Find x (y = 4): ans: 8 ml of 20% solution and 4 ml of 50% solution x + y = 12 x + 4 = x 4 = 12 4 x = 8

17 Number -Value Value: number value = total value

18 Ex. A community center sells a total of 301 tickets for a basketball game. An adult tickets costs $3. A student ticket costs $1. The sponsors collect $487 in ticket sales. Find the number of each type of ticket sold. x = # adult tickets y = # student tickets total # of tickets = 301 x + y = 301 cost of adult ticket = $3 cost of student ticket = $1 total cost = $487 3x + 1y = 487 system: x + y = 301 3x + y = 487 Elimination Method: x + y = 301-1(x + y) = -1(301) 3x + y = 487 3x + y = 487 -x y = x + y = 487 2x + 0 = 186 2x = 186 2x = x = 93

19 Find y (x = 93): x + y = y = y 93 = y = 208 ans: 93 adult tickets 208 student tickets

20 Geometry Ex. An American football field is a rectangle with a perimeter of 1040 feet. The length is 200 feet more than the width. Find the width and length of the rectangular field.

21 Ex. An American football field is a rectangle with a perimeter of 1040 feet. The length is 200 feet more than the width. Find the width and length of the rectangular field. W = width L = length L = W ,040 = 2L + 2W P = 2L + 2W W L L = W ,040 = 2L + 2W 1,040 = 2(W + 200) + 2W 1,040 = 2W W 1040 = 4W , = 4W = 4W 640 = 4W = W width = W = 160 feet length = W = = 360 feet

22 Ex. The perimeter of a badminton court is 128 feet. After a game of badminton, a player s coach estimates that the athlete has run a total of 444 feet, which is equivalent to six times the court s length plus nine times its width. What are the dimensions of a standard badminton court? Elimination Method: L L = length W = width system: 2L + 2W = 128 6L + 9W = 444 W 2L + 2W = 128-3(2L + 2W) = -3(128) 6L + 9W = 444 6L + 9W = 444-6L 6W = L + 9W = W = 60 3W = 60 3W = W = 20

23 Find L (W = 20): ans: Width is 20 feet Length is 44 feet 2L + 2W = 128 2L + 2(20) = 128 2L + 40 = 128 2L = L = 88 2L = L = 44

24 complementary angles 2 angles whose sum is 90 B A A + B = 90 supplementary angles 2 angles whose sum is 180 B A A + B = 180 angle comp supp

25 Ex. Two angles are supplementary. The measure of one angle is 20 less than 19 times the measure of the other. Find the measure of each angle. x = first angle y = second angle The measure of one angle is 20 less than 19 times the measure of the other. y = 19x 20 x + y = 180 (supplementary angles) y = 19x 20 Substitution Method: x + y = 180 Find y (x = 10) x + (19x 20) = 180 y = 19x 20 20x 20 = 180 y = 19(10) 20 20x = y = x = 200 y = x = x = 10 Ans: 10, 170