Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall

Chapter 4 Systems of Equations

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 4.3 Systems of Linear Equations and Problem Solving

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example Hilton University Drama Club sold 311 tickets for a play. Student tickets cost $0.50 each; non-student tickets cost $1.50 each. If total receipts were $ How many of each ticket were sold? Solution 1. UNDERSTAND Read and reread the problem. Suppose 200 student tickets were sold. (311 – 200) = 111 non-student tickets sold Receipts: 200(0.50) + 111(1.50) = $ Proposed solution is incorrect. Let s = student ticketsLet n = non-student tickets continue

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Hilton University Drama Club sold 311 tickets for a play. Student tickets cost $0.50 each; non-student tickets cost $1.50 each. If total receipts were $ How many of each ticket were sold? 2. TRANSLATE. student tickets + non-student tickets = total tickets s + n = 311 price of student + price of non-students = total receipts $0.50s + $1.50n = $ SOLVE. Multiply the second equation by  2. continue

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Hilton University Drama Club sold 311 tickets for a play. Student tickets cost $0.50 each; non-student tickets cost $1.50 each. If total receipts were $ How many of each ticket were sold? Substitute n = 230 into the first equation to determine s. continue

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Hilton University Drama Club sold 311 tickets for a play. Student tickets cost $0.50 each; non-student tickets cost $1.50 each. If total receipts were $ How many of each ticket were sold? 4. INTERPRET. Check: Substitute s = 81 and n = 230 into both equations. State: There were 81 student tickets and 230 non- student tickets sold for the play.

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 2 One number is 4 more than twice the second number. Their total is 25. Find the numbers. Solution 1. UNDERSTAND Read and reread the problem. If the second number is 5. The first number is 4 more than twice the second number, (4 + 25) = 14. Is the total 25? No: = 19. Our proposed solution is incorrect, but we have a better understanding of the problem. Since we are looking for two numbers, we let x = first numbery = second number continue

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 2. TRANSLATE. Translate the given facts. Words: One number is 4 more than twice the second Translate: x = 4 + 2y Second statement: Words:Their total is 25. Translate: x + y = SOLVE. Solve the system. Since the first equation is solved for x, we will use substitution. continue

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Replace y with 7 in the equation x = 4 + 2y and solve for x. continue The ordered pair solution of the system is (18, 7).

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 4. INTERPRET Since the solution of the system is (18, 7), then the first number we are looking for is 18 and the second number is 7. Check: 18 is 4 more than twice 7, the sum of the two number is 25. State: The numbers are 18 and 7.

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 3 Terry Watkins can row about 10.6 kilometers in 1 hour downstream and 6.8 kilometers upstream in 1 hour. Find how fast he can row in still water, and find the speed of the current. Solution 1. INTERPRET. Read and reread the problem. Use the formula d = rt (distance = rate  time). However, we have the effect of the water current in this problem. The rate when traveling downstream would be r + w and the rate upstream would be r – w, where r is the speed of the rower in still water, and w is the speed of the current. continue

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Terry Watkins can row about 10.6 kilometers in 1 hour downstream and 6.8 kilometers upstream in 1 hour. Find how fast he can row in still water, and find the speed of the current. 1. INTERPRET (continued) Suppose Terry can row 9km/hr in still water, and the water current is 2 km/hr. Since he rows for 1 hour in each direction, downstream would be (r + w)t = d or (9 + 2)(1) = 11 km Upstream would be (r – w)t = d or (9 – 2)(1) = 7 km Proposed solution is incorrect. We are looking for two rates: Let r = rate of the rower in still water Let w = rate of the water current continue

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Terry Watkins can row about 10.6 kilometers in 1 hour downstream and 6.8 kilometers upstream in 1 hour. Find how fast he can row in still water, and find the speed of the current. 2. TRANSLATE. 3. SOLVE. Use the elimination method. continue RateTime (hr)Distance Upstreamr + w110.6 Downstream r  w 16.8

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Terry Watkins can row about 10.6 kilometers in 1 hour downstream and 6.8 kilometers upstream in 1 hour. Find how fast he can row in still water, and find the speed of the current. 3. SOLVE. Since r = 8.7 substitute into equation 1 and determine w. 4. INTERPRET. Check: Check that r + w = 10.6 and r – w = 6.8, both equations check. State: Terry’s rate in still water is 8.7 km/hr and the rate of the water current is 1.9 km/hr.

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 4 The Candy Barrel shop manager mixes candy worth $2 per pound with trail mix worth $1.50 per pound. Find how many pounds of each she should use to make 50 pounds of a party mix worth $1.80 per pound. Solution 1. INTERPRET. Read and reread the problem. To find the cost of any quantity of items we need to use the formula price per unit  number of units = price of all units Suppose the manager decides to mix 20 pounds of candy. Since the total mixture will be 50 pounds, we need 30 pounds of trail mix. continue

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall The Candy Barrel shop manager mixes candy worth $2 per pound with trail mix worth $1.50 per pound. Find how many pounds of each she should use to make 50 pounds of a party mix worth $1.80 per pound. candy$2  20 pounds=$40 trail mix$1.50  30 pounds=$45 mixture$2.80  50 pounds=$90 Since $40 + $45 ≠ $90 our proposed solution is incorrect. We are looking for two quantities: Let x = the amount of candy Let y = the amount of trail mix continue

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall The Candy Barrel shop manager mixes candy worth $2 per pound with trail mix worth $1.50 per pound. Find how many pounds of each she should use to make 50 pounds of a party mix worth $1.80 per pound. Use a table to organize the given data. 2. Translate. Words: amount of +amount of = 50 candy trail mix x+ y = 50 Words: price of candy+price of mix= (1.8)(50) 2x+ 1.5y= 90 continue Number of pounds Cost per pound Total cost candyx$22x2x trail mixy$ y total50$1.80$90

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall The Candy Barrel shop manager mixes candy worth $2 per pound with trail mix worth $1.50 per pound. Find how many pounds of each she should use to make 50 pounds of a party mix worth $1.80 per pound. 3. SOLVE. Use elimination. Multiply equation (1) by 3 and equation (2) by  2. Substitute x = 30 into equation (1). continue

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall The Candy Barrel shop manager mixes candy worth $2 per pound with trail mix worth $1.50 per pound. Find how many pounds of each she should use to make 50 pounds of a party mix worth $1.80 per pound. 4. INTERPRET. Check: State: The store manager needs to mix 30 pounds of candy with 20 pounds of trail mix to create 50 pounds of party mix worth $1.80 per pound.

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example 6 The measure of the largest angle of a triangle is 90  more than the measure of the smallest angle, and the measure of the remaining angle is 30  more than the measure of the smallest angle. Find the measure of each angle. Solution 1. INTERPRET. Read and reread the problem. The sum of the measures of the angles of a triangle is 180 . Guess a solution. Smallest angle = 30, largest angle = or 120 remaining angle = = 60 The sum of the three angles = = 210, too large. continue

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall The measure of the largest angle of a triangle is 90  more than the measure of the smallest angle, and the measure of the remaining angle is 30  more than the measure of the smallest angle. Find the measure of each angle. Let x = smallest angle Let y = remaining angle Let z = largest angle 2. TRANSLATE. x + y + z = 180 (sum of angles in triangle) z = 90 + x (largest angle is 90 more than smallest) y = 30 + x (remaining angle is 30 more than smallest angle) 3. SOLVE. continue

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall The measure of the largest angle of a triangle is 90  more than the measure of the smallest angle, and the measure of the remaining angle is 30  more than the measure of the smallest angle. Find the measure of each angle. 3. SOLVE. Substitute the last two equations into the first equation. continue The ordered triple is (20, 50, 110).

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall The measure of the largest angle of a triangle is 90  more than the measure of the smallest angle, and the measure of the remaining angle is 30  more than the measure of the smallest angle. Find the measure of each angle. 4. INTERPRET. Check: 20  + 50   = 180  The measure of the largest angle is 90 more than the smallest. The measure of the remaining angle is 30 more than the smallest. State: The measure of the smallest angle is 20 , the measure of the largest angle is 110  and the measure of the remaining angle is 50 .