Solving Systems by Graphing ALGEBRA 1 LESSON 7

Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 (For help, go to Lessons 3 -3 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 Check Skills You’ll Need 7 -1

Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 Solutions 1. 2 n + 3 = 5 n – 2 2. 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 12 = n 31 = z 3 3. 8 – 4 z = 2 z – 13 2 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 y = – 6 x 15 – 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). y = 3 x – 1 5 3(2) – 1 5 6– 1 5=5 Quick Check 7 -1

Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 Suppose you plan to start taking an aerobics class. Non-members pay $4 per class while members pay $10 a month plus an additional $2 per class. Which system of equations models the cost as a function of classes? A C(a) = 4 a C(a) = 10 + 2 a B C(a) = a + 4 C(a) = 10 + 2 a C C(a) = 4 a C(a) = 10 a + 2 D C(a) = a + 4 C(a) = 10 a + 2 Define: Let a = number of classes. Let C(a) = total cost of the classes. 7 -1

Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 (continued) Relate: cost is Write: member C(a) = 10 + 2 a = 0 + 4 a non-member C(a) membership fee plus cost of classes attended The system is C(a) = 4 a C(a) = 10 + 2 a Quick Check 7 -1

Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 The system below models the cost of taking an aerobics class as a function of the number of classes. Find the solution of the system by graphing. What does the solution mean in terms of the situation? Part 1: Find the solutions by graphing. C(a) = 2 a + 10 The slope is 2. The intercept on the vertical axis is 10. C(a) = 4 a The slope is 4. The intercept on the vertical axis is 0. Graph the equations. C(a) = 2 a + 10 C(a) = 4 a The lines intersect at (5, 20). 7 -1

Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 (continued) Part 2: Interpret the solution. After 5 classes, both classes will cost the same, $20. Quick Check 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. Quick Check 7 -1

Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 Solve by graphing. 3 x + 4 y = 12 y=– 3 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. y = – 3 x + 3 The slope is – 3. 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 y = – 3 x + 3. 4 7 -1 Quick Check

Solving Systems by Graphing ALGEBRA 1 LESSON 7 -1 Solve by graphing. 1. y = –x – 2 y=2 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 3 -3 and 7 -1. ) Solve each equation. 1. m – 6 = 4 m + 8 2. 4 n = 9 – 2 n 1 3. 3 t + 5 = 10 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 Check Skills You’ll Need 7 -2

Solving Systems Using Substitution ALGEBRA 1 LESSON 7 -2 Solutions 1. m – 6 = 4 m + 8 2. m – 6 = 4 m – m + 8 – 6 = 3 m + 8 4 n + 2 n = 9 – 2 n + 2 n 6 n = 9 n = 11 – 14 = 3 m – 4 2 = m 2 3 3. 4 n = 9 – 2 n 1 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 1 (5, 1) in y = x – 6 – 5 + 4 1 =/ – 1 1 5– 6 1 =/ – 1 no 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 = – 3 x + 4 y = 2 x + 2 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. Quick Check 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). Quick Check 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 Persons v 7 v + + c 5 c = 5 = 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 = 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. Quick Check 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 3. y = – 2 x + 2 3 x – 17 = 2 y Check Skills You’ll Need 7 -3

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 Quick Check 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 Solve by elimination. 7 -3 + s = 3067

Solving Systems Using Elimination ALGEBRA 1 LESSON 7 -3 Quick Check (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). Quick Check 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 p + n = 240 7 -3 + 8 n = 1614

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. Quick Check 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). Quick Check 7 -3

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 3 -6. ) 1. Two trains run on parallel tracks. The first train leaves a city 1 2 hour before the second train. The first train travels at 55 mi/h. The second train travels at 65 mi/h. How long does it take for the second train to pass the first train? 2. Carl drives to the beach at an average speed of 50 mi/h. He returns 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 his home to the beach? Check Skills You’ll Need 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 t – 0. 5 = 3. 25 – 0. 5 = 2. 75 It takes 2. 75 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 from his home to the beach. 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 a + b = 10 + 0. 25 b = 0. 4(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 Quick Check (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 7 -4 = 5. 5 p

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 = 5. 5 p = p= 0. 5 p + 1750 350 Start with one equation. Substitute 5. 5 p for d. Solve for p. To break even, you must type 350 pages. Quick Check 7 -4

Applications of Linear Systems ALGEBRA 1 LESSON 7 -4 Suppose it takes you 7. 0 hours to fly 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. Relate: Let W = the wind speed. with tail wind with head wind (rate)(time) = distance (A + W) (time) = distance Write: (A + W)5. 6 = 2800 (A W) (time) = distance (A + W)7. 0 = 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)7. 0 = 2800 A – W = 400 Divide each side by 7. 0. Step 1: Eliminate W. A + W = 500 A – W = 400 Add the equations to eliminate W. 2 A + 0 = 900 Step 2: Solve for A. A = 450 Divide each side by 2. Step 3: Solve for W using either of the original equations. A + W = 500 Use the first equation. 450 + W = 500 Substitute 450 for A. W = 50 Solve for W. The average airspeed of the planes is 450 mi/h. The Quick Check average wind speed is 50 mi/h. 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 Lessons 4 -1 and 6 -2. ) Describe each statement as always, sometimes, or never true. 1. – 3 > – 2 2. 8 < – 8 3. 4 n > – n Write each equation in slope-intercept form. 4. 2 x – 3 y = 9 5. y + 3 x = 6 6. 4 y – 3 x = 1 Check Skills You’ll Need 7 -5

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 y = 2 x – 3 3 y = 6. 4 y – 3 x = 1 4 y = 3 x + 1 y = 3 x 1 4 y = 3 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 Quick Check 7 -5

Linear Inequalities ALGEBRA 1 LESSON 7 -5 Quick Check Graph 4 x – 3 y > – 9. Solve 4 x – 3 y > – 9 for y. 4 x – 3 y > – 9 – 3 y > – – 4 x + 9 4 y< – x – 3 3 Graph y = Subtract 4 x from each side. Divide each side by – 3. Reverse the inequality symbol. 4 x – 3. 3 The coordinates of the points on the boundary line make the inequality true. So, use a solid line. 4 Since y < – 3 x – 3, shade below the boundary 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 Define: Let s = the number of rolls of streamers. Let t = the number of tablecloths. Write: 2 s + 6 t 7 -5 < – total budget 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). Quick Check 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 2. y = – 1 x + 4 y = –x + 2 y = – 1 x + 3 2 2 3. x + y = 4 2 x – y = 8 Graph each inequality. 4. y > 5 2 5. y < – 3 x – 1 6. 4 x – 8 y > – 4 Check Skills You’ll Need 7 -6

Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 Solutions 1. y = 3 x – 6 2. y = – 1 x + 4 y = –x + 2 y = – x + 3 The solution is (2, 0) There is no solution. 2 1 2 3. x + y = 4 4. 2 x – y = 8 Solve equations for y: y = –x + 4 y = 2 x – 8 The solution is (4, 0). 7 -6 y > 5

Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 Solutions (continued) 2 5. y < – 3 x – 1 6. 4 x – 8 y > – 4 – 8 y > – – 4 x + 4 4 x 4 y < – 8 1 1 y < – 2 x – 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 1 < –(– 1) + 3 – 2 x + 4 y > – 0 Substitute (– 1, 1) for (x, y). – 2(– 1) + 4(1) > – 0 2 + 4 > – 0 1 < 4 The coordinates of the points in the [lavender] region where the graphs of the two inequalities overlap are solutions of the system. Quick Check 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 boundary: y = – 1 x + 2 boundary: y = 4 The region lies below the boundary line, so the inequality is y < – 1 x + 2. y < 4. 2 2 System for the [lavender] region: y < – 1 x + 2 2 y < 4 7 -6 Quick Check

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. w Relate: The length = width of the dog run. is no more than Write: Solve by graphing. < – 60 ft. The perimeter 60 2 < – 60 2 + 2 w < – 136 7 -6 + 2 w is no more than < – 136 ft. 136

Systems of Linear Inequalities ALGEBRA 1 LESSON 7 -6 Quick Check (continued) 2 < – 60 m = 0: b = 60 + 2 w < – 136 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 Write: b + s is less 20. than or equal to < – 20 7 -6 The amount earned 5 b + 6 s is at least 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. Quick Check 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 7 -6 2 3. y > – 3 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
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