Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 (For help, go to Lessons 2 -4 and 6 -2. ) Solve each equation. 1. 2 n + 3 = 5 n – 2 2. 8 – 4 z = 2 z – 13 3. 8 q – 12 = 3 q + 23 Graph each pair of equations on the same coordinate plane. 4. y = 3 x – 6 y = –x + 2 5. y = 6 x + 1 y = 6 x – 4 6. y = 2 x – 5 6 x – 3 y = 15 7. y = x + 5 y = – 3 x + 5 7 -1
Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 Solutions 1. 2 n + 3 = 5 n – 2 2. 8 – 4 z = 2 z – 13 2 n – 2 n + 3 = 5 n – 2 8 – 4 z + 4 z = 2 z + 4 z – 13 3 = 3 n – 2 8 = 6 z – 13 5 = 3 n 21 = 6 z 2 1 = n 1 3 = z 3 2 3. 8 q – 12 = 3 q + 23 8 q – 3 q – 12 = 3 q – 3 q + 23 5 q – 12 = 23 5 q = 35 q = 7 7 -1
Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 Solutions (continued) 4. y = 3 x – 6 5. y = 6 x + 1 y = –x + 2 y = 6 x – 4 6. y = 2 x – 5 7. y = x + 5 6 x – 3 y = 15 y = – 3 x + 5 – 3 y = – 6 x – 15 – 6 x 15 y = – 3 y = 2 x – 5 7 -1
Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 Solve by graphing. Check your solutions. y = 2 x + 1 y = 3 x – 1 Graph both equations on the same coordinate plane. y = 2 x + 1 y = 3 x – 1 The slope is 2. The y-intercept is 1. The slope is 3. The y-intercept is – 1. 7 -1
Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 (continued) Find the point of intersection. The lines intersect at (2, 5), so (2, 5) is the solution of the system. Check: See if (2, 5) makes both equations true. y = 2 x + 1 5 2(2) + 1 5 4 + 1 5 = 5 Substitute (2, 5) for (x, y). 7 -1 y = 3 x – 1 5 3(2) – 1 5 6 – 1 5 = 5
Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 Suppose you plan to start taking an aerobics class. Nonmembers pay $4 per class while members pay $10 a month plus an additional $2 per class. After how many classes will the cost be the same? What is that cost? c Define: Let = number of classes. T(c) Let = total cost of the classes. Relate: cost is membership fee plus cost of classes attended Write: member T(c) = 10 + 2 c = 0 + 4 c non-member T(c) 7 -1
Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 (continued) Method 1: Using paper and pencil. T(c) = 2 c + 10 The slope is 2. The intercept on the vertical axis is 10. T(c) = 4 c The slope is 4. The intercept on the vertical axis is 0. Graph the equations. T(c) = 2 c + 10 T(c) = 4 c The lines intersect at (5, 20). After 5 classes, both will cost $20. 7 -1
Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 (continued) Method 2: Using a graphing calculator. First rewrite the equations using x and y. T(c) = 2 c + 10 y = 2 x + 10 T(c) = 4 c y = 4 x Then graph the equations using a graphing calculator. Set an appropriate range. Then graph the equations. Use the key to find the coordinates of the intersection point. The lines intersect at (5, 20). After 5 classes, both will cost $20. 7 -1
Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 Solve by graphing. y = 3 x + 2 y = 3 x – 2 Graph both equations on the same coordinate plane. y = 3 x + 2 The slope is 3. The y-intercept is 2. y = 3 x – 2 The slope is 3. The y-intercept is – 2. The lines are parallel. There is no solution. 7 -1
Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 Solve by graphing. 3 x + 4 y = 12 3 y = – x + 3 4 Graph both equations on the same coordinate plane. 3 x + 4 y = 12 The y-intercept is 3. The x-intercept is 4. 3 y = – x + 3 3 The slope is – . The y-intercept is 3. 4 4 The graphs are the same line. The solutions are an infinite number of ordered pairs (x, y), such that 3 y = – x + 3. 4 7 -1
Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 pages 343– 345 Exercises 1. Yes, (– 1, 5) makes both equations true. 2. No, (– 1, 5) makes only one equation true. 3. Yes, (– 1, 5) makes both equations true. 4. Yes, (– 1, 5) makes both equations true. 5. (0, 2); 6. (0, 0); 8. (1, 5); 9. (6, – 1); 7. (1, 1); 7 -1
Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 10. (2, 2); 11. (4, 0); 12. (2, 3); 16. no solution; 13. a. 3 weeks b. $35 17. infinitely many solutions; 14. 7 weeks 15. no solution; 7 -1
Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 18. no solution; 25. (20, 60); 19. no solution; same slope, different y-int. 20. inf. many solutions; equivalent equations 21. one solution; different slopes 22. inf. many solutions; equivalent equations 23. A 24. 5 min 26. Answers may vary. Sample: y = – 1; x = 2 27. Answers may vary. Sample: y = 2 x – 1, y = 2 x + 5 28. Answers may vary. Sample: x + y = 3, 3 x + 3 y = 9 7 -1
Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 29. (2, 20) 31. (12, 30) 33. a. time on the horizontal and distance on the vertical b. Red represents the tortoise because it shows distance changing steadily over time. Blue represents the hare because it is steeper than the other line at the ends but shows no change in distance while the hare is napping. c. The point of intersection shows when the tortoise passed the sleeping hare. 30. (15, 40) 32. (– 20, 0) 34. (– 12, – 16) 35. (– 2, 10) 7 -1
Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 36. (– 30, – 2. 5) 37. (– 0. 9, 1. 6) 38. (2, 3) 39. a. c = 100 + 50 t; c = 50 + 75 t; (2, 200); 41. a. sometimes b. never 42. (– 9, – 2) 43. D 44. F 45. [2] a. Answers may vary. Sample: x – 2 y = 6 b. Since the lines do not intersect, the lines are parallel. Parallel lines have the same slope but b. The cost of renting either studio different intercepts. for 2 h is the same, $200. [1] incorrect equation OR incorrect 40. a. no values explanation / b. w = v c. w = v 7 -1
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Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 46. [4] a. y = 150 + 0. 20 x y = 200 + 0. 10 x b. $500 c. cellular phone sales [3] appropriate methods but one computational error [2] incorrect system solved correctly OR correct system solved incorrectly [1] no work shown 48. It is translated 3 units left. 47. It is translated up 2 units. 52. 20% increase 49. It is translated 5 units up and 2 units right. 50. 25% increase 1 3 51. 33 % decrease 53. 150% increase 1 54. 33 % decrease 3 55. 25% increase 56. 10% increase 57. 12. 5% decrease 7 -1
Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 Solve by graphing. 1. y = –x – 2 2 y = x + 3 3 ( 3, 1) 4. 2 x – 3 y = 9 2. y = –x + 3 y = 2 x – 6 (3, 0) 5. – 2 x + 4 y = 12 y = x – 5 1 – x + y = – 3 (6, 1) no solution 2 7 -1 3. y = 3 x + 2 6 x – 2 y = – 4 Infinitely many solutions
Solving Systems Using Substitution ALGEBRA 1 LESSON 7 -2 (For help, go to Lessons 2 -4 and 7 -1. ) Solve each equation. 1. m – 6 = 4 m + 8 2. 4 n = 9 – 2 n 1 3. t + 5 = 10 3 For each system, is the ordered pair a solution of both equations? 4. (5, 1) y = –x + 4 5. (2, 2. 4) 4 x + 5 y = 20 y = x – 6 2 x + 6 y = 10 7 -2
Solving Systems Using Substitution ALGEBRA 1 LESSON 7 -2 Solutions 1. m – 6 = 4 m + 8 2. 4 n = 9 – 2 n m – 6 = 4 m – m + 8 – 6 = 3 m + 8 4 n + 2 n = 9 – 2 n + 2 n 6 n = 9 n = 1 1 – 14 = 3 m 2 – 4 = m 2 3 1 3. t + 5 = 10 3 1 t = 5 3 t = 15 7 -2
Solving Systems Using Substitution ALGEBRA 1 LESSON 7 -2 Solutions (continued) 4. (5, 1) in y = –x + 4 (5, 1) in y = x – 6 1 – 5 + 4 / 1 = – 1 1 5 – 6 / 1 = – 1 no No, (5, 1) is not a solution of both equations. 5. (2, 2. 4) in 4 x + 5 y = 20 4(2) + 5(2. 4) 20 (2, 2. 4) in 2 x + 6 y = 10 2(2) + 6(2. 4) 10 8 + 12 20 4 + 14. 4 10 / 20 = 20 18. 4 = 10 yes no No, (2, 2. 4) is not a solution of both equations. 7 -2
Solving Systems Using Substitution ALGEBRA 1 LESSON 7 -2 Solve using substitution. y = 2 x + 2 y = – 3 x + 4 Step 1: Write an equation containing only one variable and solve. y = 2 x + 2 – 3 x + 4 = 2 x + 2 4 = 5 x + 2 2 = 5 x 0. 4 = x Start with one equation. Substitute – 3 x + 4 for y in that equation. Add 3 x to each side. Subtract 2 from each side. Divide each side by 5. Step 2: Solve for the other variable. y = 2(0. 4) + 2 y = 0. 8 + 2 y = 2. 8 Substitute 0. 4 for x in either equation. Simplify. 7 -2
Solving Systems Using Substitution ALGEBRA 1 LESSON 7 -2 (continued) Since x = 0. 4 and y = 2. 8, the solution is (0. 4, 2. 8). Check: See if (0. 4, 2. 8) satisfies y = – 3 x + 4 since y = 2 x + 2 was used in Step 2. 2. 8 – 3(0. 4) + 4 2. 8 – 1. 2 + 4 2. 8 = 2. 8 Substitute (0. 4, 2. 8) for (x, y) in the equation. 7 -2
Solving Systems Using Substitution ALGEBRA 1 LESSON 7 -2 Solve using substitution. – 2 x + y = – 1 4 x + 2 y = 12 Step 1: Solve the first equation for y because it has a coefficient of 1. – 2 x + y = – 1 y = 2 x – 1 Add 2 x to each side. Step 2: Write an equation containing only one variable and solve. 4 x + 2 y = 12 4 x + 2(2 x – 1) = 12 4 x + 4 x – 2 = 12 8 x = 14 x = 1. 75 Start with the other equation. Substitute 2 x – 1 for y in that equation. Use the Distributive Property. Combine like terms and add 2 to each side. Divide each side by 8. 7 -2
Solving Systems Using Substitution ALGEBRA 1 LESSON 7 -2 (continued) Step 3: Solve for y in the other equation. – 2(1. 75) + y = 1 – 3. 5 + y = – 1 y = 2. 5 Substitute 1. 75 for x. Simplify. Add 3. 5 to each side. Since x = 1. 75 and y = 2. 5, the solution is (1. 75, 2. 5). 7 -2
Solving Systems Using Substitution ALGEBRA 1 LESSON 7 -2 A youth group with 26 members is going to the beach. There will also be five chaperones that will each drive a van or a car. Each van seats 7 persons, including the driver. Each car seats 5 persons, including the driver. How many vans and cars will be needed? Let v = number of vans and c = number of cars. Drivers v + c = 5 Persons 7 v + 5 c = 31 Solve using substitution. Step 1: Write an equation containing only one variable. v + c = 5 Solve the first equation for c. c = –v + 5 7 -2
Solving Systems Using Substitution ALGEBRA 1 LESSON 7 -2 (continued) Step 2: Write and solve an equation containing the variable v. 7 v + 5 c = 31 7 v + 5(–v + 5) = 31 7 v – 5 v + 25 = 31 2 v + 25 = 31 2 v = 6 v = 3 Substitute –v + 5 for c in the second equation. Solve for v. Step 3: Solve for c in either equation. 3 + c = 5 c = 2 Substitute 3 for v in the first equation. 7 -2
Solving Systems Using Substitution ALGEBRA 1 LESSON 7 -2 (continued) Three vans and two cars are needed to transport 31 persons. Check: Is the answer reasonable? Three vans each transporting 7 persons is 3(7), of 21 persons. Two cars each transporting 5 persons is 2(5), or 10 persons. The total number of persons transported by vans and cars is 21 + 10, or 31. The answer is correct. 7 -2
Solving Systems Using Substitution ALGEBRA 1 LESSON 7 -2 pages 349– 352 Exercises 3 3 10. ( , 9 ) 4 8 22. 15 video rentals 1. D 11. (2, 0) 2. C 7 8 12. (7 , 11 ) 17 17 23. 80 acres flax, 160 acres sunflowers 3. B 13. (6, – 2) 24. 9 yr 4. A 14. (3, – 2) 1 25. estimate: ( , 1); 5– 16. Coordinates given in 15. (8, – 7) alphabetical order. 16. (– 3, 9. 4) 5. (9, 28) 17. 4 cm by 13 cm 1 1 6. (– , – 4 ) 18. 4 wk 2 2 1 1 7. (6 , – ) 19. (15, 15) 3 3 1 8. (2, 4 ) 20. (9, 126) 2 9. (4, 20) 21. (– 4, 4) 7 -2 2 1 ; ( , 1) 2
Solving Systems Using Substitution ALGEBRA 1 LESSON 7 -2 28. estimate: (– 3. 5, – 3. 5); 26. estimate: (– 2, 3); ; (– 2, 3) (– , – ) 27. estimate: (– 1, 1); 10 3 3 3 29. estimate: (– , 4 ); 4 4 (– , ) (– 1, 1) 7 -2 2 14 3 3 11 3
Solving Systems Using Substitution ALGEBRA 1 LESSON 7 -2 30. estimate: (– 4, 0); 32. Answers may vary. Sample: y = x and y = – 3 x + 2; 1 1 ( , ) 2 33. a. (x, y) such that y = 0. 5 x + 4 b. 50 2 (– , – ) 11 11 31. a. b. c. 2 Let n = number of nickels, let d = number of dimes. n + d = 28 0. 05 n + 0. 10 d = 2. 05 Solve the first eq. for either var. Sub. the expression into the second eq. Solve this eq. for the other var. , and then sub. its value into the first eq. and solve for the first var. (15, 13) 7 -2 c. Graphing shows only one line. Substitution results in a true equation with no variables.
Solving Systems Using Substitution ALGEBRA 1 LESSON 7 -2 34. a. no solution b. 1 38. (2, ) 2 1 39. (– , 0) 2 40. 41. 42. 43. c. Graphing shows 2 parallel lines. Substitution 44. results in a false equation with no variables. 45. 35. (2, 4) 1 2 36. (– , – ) 37. (2, – 4) 48. 4803 49. 520 50. [2] 7(– 7) – 4(– 2) 29 (4, – 2) – 49 + 8 29 inf. many solutions – 41 = 29 / No, (– 2, – 7) must no solution satisfy both 1 solution equations to be a 23 solution of the a. g = b 22 system. b. (b, g) = (572, 598) [1] no explanation c. 26 given a. (t, d) = (9, 79. 2) b. yes 46. (r, s, t) = (7, 9, 4) 47. 29. 8 7 -2
Solving Systems Using Substitution ALGEBRA 1 LESSON 7 -2 51. (12, 10); 54. 57. 52. (2, 1); 55. 58. 56. 53. (4, 2); 59. 7 -2
Solving Systems Using Substitution ALGEBRA 1 LESSON 7 -2 Solve each system using substitution. 1. 5 x + 4 y = 5 2. 3 x + y = 4 3. 6 m – 2 n = 7 y = 5 x 2 x – y = 6 3 m + n = 4 (0. 2, 1) (2, 2) (1. 25, 0. 25) 7 -2
Solving Systems Using Elimination ALGEBRA 1 LESSON 7 -3 (For help, go to Lesson 7 -2. ) Solve each system using substitution. 1. y = 4 x – 3 y = 2 x + 13 2. y + 5 x = 4 y = 7 x – 20 7 -3 3. y = – 2 x + 2 3 x – 17 = 2 y
Solving Systems Using Elimination ALGEBRA 1 LESSON 7 -3 Solutions 1. y = 4 x – 3 y = 2 x + 13 Substitute 4 x – 3 for y in the second equation. y = 2 x + 13 4 x – 3 = 2 x + 13 4 x – 2 x – 3 = 2 x – 2 x + 13 2 x – 3 = 13 2 x = 16 x = 8 y = 4 x – 3 = 4(8) – 3 = 32 – 3 = 29 Since x = 8 and y = 29, the solution is (8, 29). 7 -3
Solving Systems Using Elimination ALGEBRA 1 LESSON 7 -3 Solutions (continued) 2. y + 5 x = 4 y = 7 x – 20 Substitute 7 x – 20 for y in the first equation. y + 5 x = 4 7 x – 20 + 5 x = 4 12 x – 20 = 4 12 x = 24 x = 2 y = 7 x – 20 = 7(2) – 20 = 14 – 20 = – 6 Since x = 2 and y = – 6, the solution is (2, – 6). 7 -3
Solving Systems Using Elimination ALGEBRA 1 LESSON 7 -3 Solutions (continued) 3. y = – 2 x + 2 3 x – 17 = 2 y Substitute – 2 x + 2 for y in the second equation. 3 x – 17 = 2 y 3 x – 17 = 2(– 2 x + 2) 3 x – 17 = – 4 x + 4 7 x – 17 = 4 7 x = 21 x = 3 y = – 2 x + 2 = – 2(3) + 2 = – 6 + 2 – 4 Since x = 3 and y = – 4, the solution is (3, – 4). 7 -3
Solving Systems Using Elimination ALGEBRA 1 LESSON 7 -3 Solve by elimination. – 2 x + 9 y = 1 2 x + 3 y = 11 Step 1: Eliminate x because the sum of the coefficients is 0. 2 x + 3 y = 11 – 2 x + 9 y =1 0 + 12 y = 12 Addition Property of Equality y = 1 Solve for y. Step 2: Solve for the eliminated variable x using either original equation. 2 x + 3 y = 11 2 x + 3(1) = 11 2 x + 3 = 11 2 x = 8 x = 4 Choose the first equation. Substitute 1 for y. Solve for x. 7 -3
Solving Systems Using Elimination ALGEBRA 1 LESSON 7 -3 (continued) Since x = 4 and y = 1, the solution is (4, 1). Check: See if (4, 1) makes true the equation not used in Step 2. – 2(4) + 9(1) 1 Substitute 4 for x and 1 for y into the second equation. – 8 + 9 1 1 = 1 7 -3
Solving Systems Using Elimination ALGEBRA 1 LESSON 7 -3 On a special day, tickets for a minor league baseball game were $5 for adults and $1 for students. The attendance that day was 1139, and $3067 was collected. Write and solve a system of equations to find the number of adults and the number of students that attended the game. Define: Let a = number of adults Let s = number of students Relate: total number at the game Write: a + s = 1139 total amount collected 5 a + s = 3067 Solve by elimination. 7 -3
Solving Systems Using Elimination ALGEBRA 1 LESSON 7 -3 (continued) Step 1: Eliminate one variable. a + s = 1139 5 a + s = 3067 – 4 a + 0 = – 1928 Subtraction Property of Equality a = 482 Solve for a. Step 2: Solve for the eliminated variable using either of the original equations. a + s = 1139 Choose the first equation. 482 + s = 1139 Substitute 482 for a. s = 657 Solve for s. There were 482 adults and 657 students at the game. Check: Is the solution reasonable? The total number at the game was 482 + 657, or 1139. The money collected was $5(482), or $2410, plus $1(657), or $657, which is $3067. The solution is correct. 7 -3
Solving Systems Using Elimination ALGEBRA 1 LESSON 7 -3 Solve by elimination. – 5 x – 2 y = – 14 3 x + 6 y = – 6 Step 1: Eliminate one variable. Start with the given system. 3 x + 6 y = – 6 – 5 x – 2 y = – 14 To prepare to eliminate y, multiply the second equation by 3. 3 x + 6 y = – 6 3(– 5 x – 2 y = – 14) Step 2: Solve for x. – 12 x = 48 x = 4 7 -3 Add the equations to eliminate y. 3 x + 6 y = – 6 – 15 x – 6 y = – 42 – 12 x – 0 = – 48
Solving Systems Using Elimination ALGEBRA 1 LESSON 7 -3 (continued) Step 3: Solve for the eliminated variable using either of the original equations. 3 x + 6 y = – 6 Choose the first equation. 3(4) + 6 y = – 6 Substitute 4 for x. 12 + 6 y = – 6 Solve for y. 6 y = – 18 y = – 3 The solution is (4, – 3). 7 -3
Solving Systems Using Elimination ALGEBRA 1 LESSON 7 -3 Suppose the band sells cans of popcorn for $5 per can and cans of mixed nuts for $8 per can. The band sells a total of 240 cans and receives a total of $1614. Find the number of cans of popcorn and the number of cans of mixed nuts sold. Define: Let p = number of cans of popcorn sold. Let n = number of cans of nuts sold. Relate: total number of cans total amount of sales Write: 5 p + 8 n = 1614 p + n = 240 7 -3
Solving Systems Using Elimination ALGEBRA 1 LESSON 7 -3 (continued) Step 1: Eliminate one variable. Start with the given system. p + n = 240 5 p + 8 n = 1614 To prepare to eliminate p, multiply the first equation by 5. 5(p + n = 240) 5 p + 8 n = 1614 Step 2: Solve for n. – 3 n = – 414 n = 138 7 -3 Subtract the equations to eliminate p. 5 p + 5 n = 1200 5 p + 8 n = 1614 0 – 3 n = – 414
Solving Systems Using Elimination ALGEBRA 1 LESSON 7 -3 (continued) Step 3: Solve for the eliminated variable using either of the original equations. p + n = 240 Choose the first equation. p + 138 = 240 Substitute 138 for n. p = 102 Solve for p. The band sold 102 cans of popcorn and 138 cans of mixed nuts. 7 -3
Solving Systems Using Elimination ALGEBRA 1 LESSON 7 -3 Solve by elimination. 5 x + 7 y = 10 3 x + 5 y = 10 Step 1: Eliminate one variable. Start with the given system. 3 x + 5 y = 10 5 x + 7 y = 10 To prepare to eliminate x, multiply one equation by 5 and the other equation by 3. 5(3 x + 5 y = 10) 3(5 x + 7 y = 10) Step 2: Solve for y. 4 y = 20 y = 5 7 -3 Subtract the equations to eliminate x. 15 x + 25 y = 50 15 x + 21 y = 30 0 + 4 y = 20
Solving Systems Using Elimination ALGEBRA 1 LESSON 7 -3 (continued) Step 3: Solve for the eliminated variable x using either of the original equations. 3 x + 5 y = 10 Use the first equation. 3 x + 5(5) = 10 Substitute 5 for y. 3 x + 25 = 10 3 x = – 15 x = – 5 The solution is (– 5, 5). 7 -3
Solving Systems Using Elimination ALGEBRA 1 LESSON 7 -3 pages 356– 359 Exercises 5 11. (– 2, – ) 2 1. (1, 3) 12. (1, 14) 2. (2, – 2) 1 13. ( , 1) 3. (5, – 17) 14. (– 2, 3) 4. (– 3, 4) 15. a. 30 w + = 17. 65, 20 w + 3 = 25. 65 b. $. 39 for a wallet size, 1 5. (– 9, ) 2 1 6. (– , 10) 2 7. a. b. 8. a. b. 2 $5. 95 for an 8 10 x + y = 20, x – y = 4 16. a. x = burritos, y = tacos; 12 and 8 3 x + 4 y = 11. 33, 9 x + 5 y = 23. 56 b. $1. 79 for a burrito, $1. 49 for a taco a + s = 456, 3. 5 a + s = 1131 17. (– 1, – 3) 270 adult, 186 student 9. (– 5, 1) 18. (2. 5, 1) 10. (11, – 3) 19. (2, – 2) 7 -3
Solving Systems Using Elimination ALGEBRA 1 LESSON 7 -3 20. (10, 8) 3 21. (– 1, ) 2 22. (1, 5) 23– 28. Methods may vary. Samples are given. 23. (– 1, – 2); substitution; both solved for y 24. (15, – 10); elimination; equations not solved for y 27. (5, 1); substitution; one eq. solved for x 1 1 28. ( , 2 ); substitution; 3 3 eqs. solved for y 29. one night: $81. 25; one meal: $8. 13 30. a. brass: $6; steel: $3 b. $99 31. She forgot to multiply – 8 by 6. 25. (10, 2); substitution; one eq. solved for x 32. Answers may vary. 4 Sample: 2 x – 3 y = 6, x + 3 y = 9; (5, ) 26. (– 3, 11); elimination; eqs. not solved for a variable 33. (10, – 6) 34. (8, 12) 7 -3 3
Solving Systems Using Elimination ALGEBRA 1 LESSON 7 -3 35. (– 15, – 1) 41. B 1 = 3 volts; B 2 = 1. 5 volts 36. (20, 12) 42. (– 3, 2) 37. (– 2, 16) c 43. ( , 0) (a = 0, b = 1) / / 38. (33, – 48) 44. (8, 13, 20) 39. 9 45. CD: $3. 40, cassette: $1. 80 a 40. Answers may vary. Sample: 46. a. 2. 8 g of gold You solve a system using the b. about 2. 7% elimination method by adding 47. D or subtracting the eqs. to 48. H eliminate one of the variables. This sum or difference is one eq. with one variable that can be solved. Use addition: Use subtraction: Use multiplication: 3 x + 2 y = 6 5 x + 3 y = 15 4 x + 5 y = 20 –x – 2 y = 4 5 x – 2 y = 10 2 x – y = 10 7 -3
Solving Systems Using Elimination ALGEBRA 1 LESSON 7 -3 51– 53. Coordinates given in alphabetical order. 51. (6, 26) 49. [2] y – x = 13 7 y + x = 11 8 y = 24 y = 3 3 – x = 13 x = – 10; (– 10, 3) [1] no work shown 50. [4] 52. (– 1, 4) 53. (9, – 5) 1 1 A = h(b 1 + b 2) 2 1 = (4)(4 + 2) 2 = 2(6) = 12 The area is 12 square units. [3] graph and formula with one mathematical error [2] graph but error in formula [1] graph only 7 -3 54. 9 2 55. 9 56. 1 15 57. 71 58. – 18 59. – 44 60. 22 61. 83 62. – 6
Solving Systems Using Elimination ALGEBRA 1 LESSON 7 -3 Solve using elimination. 1. 3 x – 4 y = 7 2 x + 4 y = 8 (3, 0. 5) 3. – 6 x + 5 y = 4 (1, 2) 2. 5 m + 3 n = 22 (2, 4) 5 m + 6 n = 34 4. 7 p + 5 q = 2 3 x + 4 y = 11 8 p – 9 q = 17 7 -3 (1, 1)
Applications of Linear Systems ALGEBRA 1 LESSON 7 -4 (For help, go to Lesson 2 -5. ) 1 1. Two trains run on parallel tracks. The first train leaves a city hour 2 before the second train. The first train travels at 55 mi/h. The second train travels at 65 mi/h. Find how long it takes for the second train to pass the first train. 2. Luis and Carl drive to the beach at an average speed of 50 mi/h. They return home on the same road at an average speed of 55 mi/h. The trip home takes 30 min. less. What is the distance from their home to the beach? 7 -4
Applications of Linear Systems ALGEBRA 1 LESSON 7 -4 Solutions 1. 55 t = 65(t – 0. 5) 55 t = 65 t – 32. 5 = 10 t 3. 25 = t It takes 3. 25 hours for the second train to pass the first train. 2. 50 t = 55(t – 0. 5) 50 t = 55 t – 27. 5 = 5 t 5. 5 = t 50 t = 50(5. 5) = 275 It is 275 miles to the beach from their home. 7 -4
Applications of Linear Systems ALGEBRA 1 LESSON 7 -4 A chemist has one solution that is 50% acid. She has another solution that is 25% acid. How many liters of each type of acid solution should she combine to get 10 liters of a 40% acid solution? Define: Let a = volume of the 50% solution. Let b = volume of the 25% solution. Relate: volume of solution amount of acid Write: 0. 5 a + 0. 25 b = 0. 4(10) a + b = 10 Step 1: Choose one of the equations and solve for a variable. a + b = 10 Solve for a. a = 10 – b Subtract b from each side. 7 -4
Applications of Linear Systems ALGEBRA 1 LESSON 7 -4 (continued) Step 2: Find b. 0. 5 a + 0. 25 b = 0. 4(10) 0. 5(10 – b) + 0. 25 b = 0. 4(10) Substitute 10 – b for a. Use parentheses. 5 – 0. 5 b + 0. 25 b = 0. 4(10) Use the Distributive Property. 5 – 0. 25 b = 4 Simplify. – 0. 25 b = – 1 Subtract 5 from each side. b = 4 Divide each side by – 0. 25. Step 3: Find a. Substitute 4 for b in either equation. a + 4 = 10 a = 10 – 4 a = 6 To make 10 L of 40% acid solution, you need 6 L of 50% solution and 4 L of 25% solution. 7 -4
Applications of Linear Systems ALGEBRA 1 LESSON 7 -4 Suppose you have a typing service. You buy a personal computer for $1750 on which to do your typing. You charge $5. 50 per page for typing. Expenses are $. 50 per page for ink, paper, electricity, and other expenses. How many pages must you type to break even? Define: Let p = the number of pages. Let d = the amount of dollars of expenses or income. Relate: Expenses are per-page Income is price expenses plus times pages typed. computer purchase. Write: d = 0. 5 p + 1750 d = 5. 5 p 7 -4
Applications of Linear Systems ALGEBRA 1 LESSON 7 -4 (continued) Choose a method to solve this system. Use substitution since it is easy to substitute for d with these equations. d = 0. 5 p + 1750 5. 5 p = 0. 5 p + 1750 5 p = 1750 p = 350 Start with one equation. Substitute 5. 5 p for d. Solve for p. To break even, you must type 350 pages. 7 -4
Applications of Linear Systems ALGEBRA 1 LESSON 7 -4 Suppose it takes you 6. 8 hours to fly about 2800 miles from Miami, Florida to Seattle, Washington. At the same time, your friend flies from Seattle to Miami. His plane travels with the same average airspeed, but this flight only takes 5. 6 hours. Find the average airspeed of the planes. Find the average wind speed. Define: Let A = the airspeed. Let W = the wind speed. Relate: with tail wind with head wind (rate)(time) = distance (A + W) (time) = distance Write: (rate)(time) = distance (A + W) (time) = distance (A + W)5. 6 = 2800 (A + W)6. 8 = 2800 7 -4
Applications of Linear Systems ALGEBRA 1 LESSON 7 -4 (continued) Solve by elimination. First divide to get the variables on the left side of each equation with coefficients of 1 or – 1. (A + W)5. 6 = 2800 A + W = 500 Divide each side by 5. 6. (A – W)6. 8 = 2800 A – W 412 Divide each side by 6. 8. Step 1: Eliminate W. A + W = 500 A – W = 412 Add the equations to eliminate W. 2 A + 0 = 912 Step 2: Solve for A. A = 456 Divide each side by 2. Step 3: Solve for W using either of the original equations. A + W = 500 Use the first equation. 456 + W = 500 Substitute 456 for A. W = 44 Solve for W. The average airspeed of the planes is 456 mi/h. The average wind speed is 44 mi/h. 7 -4
Applications of Linear Systems ALGEBRA 1 LESSON 7 -4 pages 365– 368 Exercises 1. a. 4 a + 5 b = 6. 71 b. 5 a + 3 b = 7. 12 c. pen: $1. 19, pencil: $. 39 2. D 3. a. b. a + b = 24; 0. 04 a + 0. 08 b = 1. 2 c. 18 kg A, 6 kg B 4. a. at 8 wk b. $160; $62 5. 600 games 7. a. b. c. d. 8. a. (A + W )4. 8 = 2100 (A – W )5. 6 = 2100 b. 406. 25 mi/h c. 31. 25 mi/h 9– 14. Answers may vary. Samples are given. 9. Substitution; one eq. is solved for t. 10. Substitution; both eqs. are solved for y. 11. Elimination; subtract to eliminate m. 12. Substitution; both eqs. are solved for y. 6. 40 shirts s + c = 2. 75 s – c = 1. 5 2. 125 mi/h 0. 625 mi/h 7 -4
Applications of Linear Systems ALGEBRA 1 LESSON 7 -4 18. 19 small mowers, 11 large mowers 19. a. 42 mi/h b. 12 mi/h 13. Elimination; mult. first eq. by 3 and add to elim. y. 14. Substitution; one eq. is solved for u. 15. a. t = 99 – 3. 5 m; t = 0 + 2. 5 m; t = 41. 25°, m = 16. 5 min b. After 16. 5 min, the temp. of either piece will be 41. 25°C. 16. 5 cm; 12 cm 19 20. a. g = b 17 b. g + b= 1908 19 g = b 17 901 boys, 1007 girls 21. a. 16 days 17. Answers may vary. Sample: b. Answers may vary. Sample: You have 10 coins, all dimes and If you plan to ski for many years, quarters. The value of the coins is you should buy the equipment, $1. 75. How many dimes do you have? since you will break even at How many quarters do you have? 16 days. q + d = 10 22. x = 2 0. 25 q + 0. 10 d = 1. 75 y = 4 You have 5 dimes and 5 quarters. 7 -4
Applications of Linear Systems ALGEBRA 1 LESSON 7 -4 24. a. 2. 50 s + 4. 00 = 10, 000 29. (4, 1) 5 2 = s 30. (– 6, 7) 3 31. (3, ) 800 small, 2000 large b. 560 h c. $17. 86/h 33. 1 25. C 34. 1 26. H 3 35. – 2 27. B 28. [2] 2 32. 3 2 3 V + C = 22 2 V + 4 C = 28 12 V + 4 C 10 V V = 88 = 28 = 60 6 people/van [1]correct answer, no work shown 7 -4 36. – 4 11 37. undefined
Applications of Linear Systems ALGEBRA 1 LESSON 7 -4 38. 6 < y < 10 39. – 8 < n 3 < – 40. 1 < k < 6 41. 1 p 4 < < – – < 42. – 5 < c 5 – 43. 5 > w > 1 7 -4
Applications of Linear Systems ALGEBRA 1 LESSON 7 -4 1. One antifreeze solution is 10% alcohol. Another antifreeze solution is 18% alcohol. How many liters of each antifreeze solution should be combined to create 20 liters of antifreeze solution that is 15% alcohol? 7. 5 L of 10% solution; 12. 5 L of 18% solution 2. A local band is planning to make a compact disk. It will cost $12, 500 to record and produce a master copy, and an additional $2. 50 to make each sale copy. If they plan to sell the final product for $7. 50, how many disks must they sell to break even? 2500 disks 3. Suppose it takes you and a friend 3. 2 hours to canoe 12 miles downstream (with the current). During the return trip, it takes you and your friend 4. 8 hours to paddle upstream (against the current) to the original starting point. Find the average paddling speed in still water of you and your friend and the average speed of the current of the river. Round answers to the nearest tenth. still water: 3. 1 mi/h; current: 0. 6 mi/h 7 -4
Linear Inequalities ALGEBRA 1 LESSON 7 -5 (For help, go to Lesson 3 -1 and 6 -3. ) Describe each statement as always, sometimes, or never true. 1. – 3 > – 2 < 2. 8 8 – 3. 4 n n > – Write each equation is slope-intercept form. 4. 2 x – 3 y = 9 5. y + 3 x = 6 7 -5 6. 4 y – 3 x = 1
Linear Inequalities ALGEBRA 1 LESSON 7 -5 Solutions 1. – 3 < – 2 is true, so – 3 > – 2 is never true. < 2. 8 = 8 is true, so 8 8 is always true. – 3. 4(1) 1 is true and 4(– 1) – 1 is false, so 4 n n is sometimes > > > – – – true. 4. 2 x – 3 y = 9 – 3 y = – 2 x + 9 5. y + 3 x = 6 y = – 3 x + 6 – 2 x + 9 – 3 2 y = x – 3 3 y = 6. 4 y – 3 x = 1 4 y = 3 x + 1 y = 3 x 1 4 3 y = x + 1 4 4 7 -5
Linear Inequalities ALGEBRA 1 LESSON 7 -5 Graph y > – 2 x + 1. First, graph the boundary line y = – 2 x + 1. The coordinates of the points on the boundary line do not make the inequality true. So, use a dashed line. Shade above the boundary line. Check The point (0, 2) is in the region of the graph of the inequality. See if (0, 2) satisfies the inequality. y > – 2 x + 1 Substitute (0, 2) for (x, y). 2 > – 2(0) + 1 2 > 1 7 -5
Linear Inequalities ALGEBRA 1 LESSON 7 -5 > Graph 4 x – 3 y 6. – Solve 4 x – 3 y 6 for y. > – 4 x – 3 y 6 > – > – 3 y – 4 x + 6 – < 4 y x – 2 – 3 Subtract 4 x from each side. Divide each side by – 3. Reverse the inequality symbol. 4 3 Graph y = x – 2. The coordinates of the points on the boundary line make the inequality true. So, use a solid line. < 4 Since y x – 2, shade below the boundary – 3 line. 7 -5
Linear Inequalities ALGEBRA 1 LESSON 7 -5 Suppose your budget allows you to spend no more than $24 for decorations for a party. Streamers cost $2 a roll and tablecloths cost $6 each. Use intercepts to graph the inequality that represents the situation. Find three possible combinations of streamers and tablecloths you can buy. Relate: cost of streamers plus cost of tablecloths is less than or equal to total budget Define: Let s = the number of rolls of streamers. Let t = the number of rolls of streamers. Write: 2 s + 6 t 7 -5 < – 24
Linear Inequalities ALGEBRA 1 LESSON 7 -5 (continued) < Graph 2 s + 6 t 24 by graphing the – intercepts (12, 0) and (0, 4). The coordinates of the points on the boundary line make the inequality true. So, use a solid line. Graph only in Quadrant I, since you cannot buy a negative amount of decorations. Test the point (0, 0). < 2 s + 6 t 24 – < 2(0) + 6(0) 24 – < 0 24 – Substitute (0, 0) for (s, t). Since the inequality is true, (0, 0) is a solution. 7 -5
Linear Inequalities ALGEBRA 1 LESSON 7 -5 (continued) Shade the region containing (0, 0). The graph below shows all the possible solutions of the problem. Since the boundary line is included in the graph, the intercepts are also solutions to the inequality. The solution (9, 1) means that if you buy 9 rolls of streamers, you can buy 1 tablecloth. Three solutions are (9, 1), (6, 2), and (3, 3). 7 -5
Linear Inequalities ALGEBRA 1 LESSON 7 -5 pages 373– 376 Exercises 1. no 14. 17. 15. 18. 11. 2. yes 3. yes 4. yes 12. 5. no 6. yes 7. A 8. B 13. 16. 9. B 10. A 7 -5 2 7 19. y x – ; < – 3 3
Linear Inequalities ALGEBRA 1 LESSON 7 -5 20. y x – 2; >5 – 3 23. a. 3 x + 5 y 48 < – b. 8 21. y x – ; < 2 – 3 3 c. Answers may vary. Sample: 8 blue and 4 gold, 2 blue and 8 gold, 12 blue and 2 gold d. No; you cannot buy – 2 rolls of paper. 22. y – x – 2; < 3 – 2 7 -5
Linear Inequalities ALGEBRA 1 LESSON 7 -5 24. a. 3 n + 10 c > 250 b. 25. 26. c. Answers may vary. Sample: 30 canvas and 10 nylon, 26 canvas and 20 nylon, 27. 35 canvas and 10 nylon d. Domain and range values must be positive integers, since you cannot buy portions of packs or a negative number of packs. 7 -5 28. 29. 30.
Linear Inequalities ALGEBRA 1 LESSON 7 -5 31. 34. y > 2 x – 1 35. x – 3 < – 36. y x – 2 < 1 – 3 32. 37. a. 12 x + 8 y 180 < – b. 33. For an inequality written in the form y < or y , shade below the < – boundary line. For an inequality > written in the form y > or y , – shade above the boundary line. c. Yes; you can buy 8 CDs and 9 tapes. d. 43 7 -5
Linear Inequalities ALGEBRA 1 LESSON 7 -5 38. x > 0; 39. y < 0; 41. x < y; < 45. a. 2 w + 2 50; – 42. It should be shaded above the line, and the line should be dashed. b. Answers may vary. 43. y < x + 2 Sample: 10 ft by 10 ft, 44. a. 5 ft by 5 ft 40. y 0; > – c. No; (12, 15) is not in the shaded region and is not a sol. of the inequality. b. c. 46. y < x – 3 y > x 47. y 2 x – 2 > – 5 12 7 -5
Linear Inequalities ALGEBRA 1 LESSON 7 -5 48. a. Answers may vary. Sample: 2 x + y > 3 49. a. yes b. yes c. Answers may vary. Sample: (2, 3) d. b. Answers may vary. Sample: 3 x + y < 1 c. If an inequality is in standard form, where A, B, and C are all positive, you shade above the line for > or and below the > – < line for < or . – d. yes 50. A 51. H 52. D 53. F 7 -5
![Linear Inequalities ALGEBRA 1 LESSON 7 -5 54. [2] First, graph the solid line](https://present5.com/presentation/3e20e6a43492863328b82d3ab2bf5a90/image-80.jpg "Linear Inequalities ALGEBRA 1 LESSON 7 -5 54. [2] First, graph the solid line")
Linear Inequalities ALGEBRA 1 LESSON 7 -5 54. [2] First, graph the solid line y = 3 x – 4. Then shade below the line. 59. 11 60. – 6 61. 8, 18 62. – 6. 7, – 2. 1 63. 12 64. 28 65. 11 [1] correct graph given, no explanation 66. 16 2 55. about 775 games 3 67. – 1 3 56. 0. 75 mi/h, 3. 25 mi/h 68. 4 57. 5 69. 13 58. 7 70. 6. 5 7 -5
Linear Inequalities ALGEBRA 1 LESSON 7 -5 1. Determine whether (4, 1) is a solution of 3 x + 2 y 10. > – yes Graph each inequality. 2. x > – 2 3. 5 x – 2 y > 10 7 -5 4. 2 x + 6 y 0 < –
Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 (For help, go to Lessons 7 -1 and 7 -5. ) Solve each system by graphing. 1. y = 3 x – 6 1 2. y = – x + 4 3. x + y = 4 y = –x + 2 1 y = – x + 3 2 x – y = 8 2 2 Graph each inequality. 4. y > 5 2 5. y x – 1 < 3 – 7 -6 6. 4 x – 8 y > 4 –
Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 Solutions 1 2. y = – x + 4 1. y = 3 x – 6 2 1 2 y = –x + 2 y = – x + 3 The solution is (2, 0) There is no solution. 3. x + y = 4 4. y > 5 2 x – y = 8 Solve equations for y: y = –x + 4 y = 2 x – 8 The solution is (4, 0). 7 -6
Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 Solutions (continued) < 2 5. y x – 1 – 3 6. 4 x – 8 y > 4 – – 8 y > – 4 x + 4 – y 4 x 4 < – 8 1 y x – 1 < 2 – 2 7 -6
Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 Solve by graphing. – 2 x + 4 y 0 y < –x + 3 > – > Graph y < –x + 3 and – 2 x + 4 y 0. – Check The point (– 1, 1) is in the region graphed by both inequalities. See if (– 1, 1) satisfies both inequalities. 7 -6
Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 (continued) y < –x + 3 – 2 x + 4 y 0 > – 1 < –(– 1) + 3 Substitute (– 1, 1) for (x, y). > – 2(– 1) + 4(1) 0 – 1 < 4 > 2 + 4 0 – The coordinates of the points in the [lavender] region where the graphs of the two inequalities overlap are solutions of the system. 7 -6
Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 Write a system of inequalities for each shaded region below. [red] region [blue] region 1 boundary: y = – x + 2 boundary: y = 4 The region lies below the boundary line, so the inequality is 1 y < – x + 2. y < 4. 2 2 System for the [lavender] region: 1 y < – x + 2 2 y < 4 7 -6
Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 You need to make a fence for a dog run. The length of the run can be no more than 60 ft, and you have 136 feet of fencing that you can use. What are the possible dimensions of the dog run? Define: Let = length of the dog run. Let w = width of the dog run. Relate: The length is no more than 60 ft. The perimeter Write: < – 60 2 + 2 w Solve by graphing. 60 < – 2 + 2 w 136 < – 7 -6 is no more than < – 136 ft. 136
Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 (continued) 2 + 2 w 136 < – < 60 – m = 0: b = 60 Graph the intercepts (68, 0) and (0, 68). Shade below = 60. Test (0, 0). 2(0) + 2(0) 136 < – 0 136 < – So shade below 2 + 2 w = 136 The solutions are the coordinates of the points that lie in the region shaded lavender and on the lines = 60 and 2 + 2 w = 136. 7 -6
Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 Suppose you have two jobs, babysitting that pays $5 per hour and sacking groceries that pays $6 per hour. You can work no more than 20 hours each week, but you need to earn at least $90 per week. How many hours can you work at each job? Define: Let b = hours of babysitting. Let s = hours of sacking groceries. Relate: The number of hours worked is less 20. than or equal to The amount is at earned least Write: < – 5 b + 6 s b + s 20 7 -6 > – 90. 90
Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 (continued) Solve by graphing. < b + s 20 – > 5 b + 6 s 90 – The solutions are all the coordinates of the points that are nonnegative integers that lie in the region shaded lavender and on the lines b + s = 20 and 5 b + 6 s = 90. 7 -6
Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 pages 380– 384 Exercises 9. 6. 1. no 2. yes 3. no 7. 4. 10. 8. 11. 5. 7 -6
Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 12. 16. x > 5 and y –x + 3 > – 13. 5 4 19. y – x – 4 and y x – 3 >1 < 2 – 5 – 3 < 22. a. 5. 99 x + 9. 99 y 50, – > > x 0, y 1 – – < 1 17. y > 1 – x – 2 and y x + 2 – 2 b. > 1 > 3 18. y – x + 1 and y – x + 3 – – 20. a. 1. 5 f + 2. 5 c < 9. 50, f + c > 4 b. 14. 15. c. 2 books and 6 CDs; no, (2, 6) is not in the shaded region. d. Answers may vary. Sample: 3 books and 3 CDs for $47. 94 21. 7 -6
Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 23. x + y 30, 1. 25 x + 3 y 60 < > – – 25. x 3, x – 3, y – 3 < < > > – – < 26. y > 2, x < 5, y x – – 2 2 27. y x – 2, y < x + 2 > – 3 3 28. y –x – 3, y –x + 3, y x – 3 > > < < – – 24. a. x + y 12, 6 x + 4 y 60 > < – – 29. Answers may vary. Sample: x – 2, x 4, y 1, y – 2 > < < > – – 30. a. – 1 b. 8 31. a. triangle b. Answers may vary. Sample: b. (2, 2), (– 4, – 1), (– 4, 2) (8, 3), (9, 1), (10, 0) c. 9 units 2 7 -6
Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 32. a. b. c. 33. a. b. c. 34. a. b. c. 35. a. square (1, – 1), (5, – 1), (1, 3), (5, 3) 16 units 2 trapezoid (0, – 4), (0, 2), (2, – 4), (2, 0) 10 units 2 triangle (2, – 3), (2, 2), (7, – 3) 12. 5 units 2 x 1, 10. 99 x + 4. 99 y 45 > < – – 36. a. b. No; they are parallel. c. no d. no 37. a. b. No; they are parallel. c. It is a strip between the lines. b. (3, 0), (3, 1), (3, 2), (4, 0) 7 -6
Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 38– 42. Answers may vary. Samples are given. 38. x 1 and y 2 < < – – 39. x < 0 and y > 0 44. 45. Answers may vary. 1 y > x, y < 2 x 2 46. Answers may vary. y > x + 1, y > –x + 1 47. a. h 2 + 0. 400 a > – b. 40. y > 5 and y < 3 41. x < 2 and y < 5 42. x > 0 and y < 0 43. a. s + d > 10, s + d < 20, > d 3, 10 d + 0. 15 s < 60 – b. Answers may vary. Sample: (8, 4. 5); 12. 5 g; gold: $45. 00, silver: $1. 20 7 -6 c. Answers may vary. Sample: 5 hits, 6 at-bats
Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 48. a. 180 x + 240 y 2700 > – < x + y 17 – x > y y 4 > – 52. [4] a. Let x = number of toppings, and y = cost of pizza. Maria’s: y = 0. 50 x + 8 Tony’s: y = 0. 75 x + 7 b. Answers may vary. Sample: ten 14 -in. drums and six 18 -in. drums 49. D 50. G 51. [2] all points on the line 3 x + 4 y = 12 [1] incorrect description given b. (4, 10) With 4 toppings, the cost is $10 at either Tony’s or Maria’s. c. Answers may vary. Sample: Since I prefer more than 4 toppings, I will go to Maria’s, because the pizza will be less expensive. [3] (a) and (b) only done correctly [2] (a) done correctly, but student makes a computational error in (b) [1] error in (a), but system solved correctly 7 -6
Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 53. 59. 5 56. 2 60. – 3 61. – 8 62. – 1 54. 57. 63. 64. 65. 58. 66. 55. 67. 4 1 – 5 – 8 3 10 9 6 13 15 4 68. ƒ(x) = 7 x 69. ƒ(x) = x + 6 70. ƒ(x) = x 2 7 -6
Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 Solve each system by graphing. > 1. x 0 – y < 3 2. 2 x + 3 y > 12 2 x + 2 y < 12 4. Write a system of inequalities for the following graph. 7 -6 2 3. y = x – 3 3 > 2 x – 3 y – 9 –
Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 Solve each system by graphing. 1. x 0 > – y < 3 2. 2 x + 3 y > 12 2 x + 2 y < 12 7 -6 2 3. y x – 3 > 2 x – 3 y – 9 –
Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 4. Write a system of inequalities for the following graph. y < x + 3 1 2 y > – x – 2 7 -6
Systems of Equations and Inequalities ALGEBRA 1 CHAPTER 7 1. (2, – 1) 9. (8, 25) 19. q + n = 15, 10. (3, – 7) 11. (2, – 9) 2. (0, 4) 3. one 4. 5. 6. 7. 8. one none one inf. many 0. 25 q + 0. 05 n = 2. 75; 10 quarters, 5 nickels 12. (3. 5, 1) 20. Answers may vary. Sample: You solve a 13. (8, – 24) system of linear equations 2 14. (11, – ) by finding a single point that 5 satisfies all the equations in 15. (6, 4) the system. You solve a 16. (– 5, – 2) system of linear inequalities by graphing a region that 17. b + m = 35, contains points that satisfy b + 2 m = 45; $25; $10 all the inequalities in the n 5 18. = , n + p = 24; system. p 3 15 novelists, 9 poets 7 -A 21. C
Systems of Equations and Inequalities ALGEBRA 1 CHAPTER 7 < < 28. a. w 30, 2 + 2 w 180 – – b. 22. 25. 23. 26. Answers may vary. Sample: y = x + 1, y = 3 x – 5; (3, 4) 27. a. 0. 10 d + 0. 25 q < 5. 00; 24. b. 49 items c. 19 items 7 -A
Systems of Equations and Inequalities ALGEBRA 1 CHAPTER 7 29. a. b. x + y = 200, 0. 3 x + 0. 5 y = 84; 120 liters of 50% insecticide and 80 liters of 30% insecticide 7 -A
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