examples and guided practice come from the algebra 1 powerpoint presentations available at

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Examples and Guided Practice come from the Algebra 1 PowerPoint Presentations available at www.classzone.comLesson 7.17 = 7

SOLUTION

EXAMPLE 1

Check the intersection pointUse the graph to solve the system. Then check your solution algebraically.x + 2y = 7Equation 13x 2y = 5Equation 2

The lines appear to intersect at the point (3, 2).CHECKSubstitute 3 for x and 2 for y in each equation.x + 2y = 73 + 2(2)=?7ANSWERBecause the ordered pair (3, 2) is a solution of each equation, it is a solution of the system.

EXAMPLE 1

Check the intersection point3x 2y = 55 = 5

3(3) 2(2)5=?

EXAMPLE 2

Use the graph-and-check methodSolve the linear system:x + y = 7Equation 1x + 4y = 8Equation 2SOLUTIONSTEP 1Graph both equations.

EXAMPLE 2

STEP 2Use the graph-and-check methodEstimate the point of intersection. The two lines appear to intersect at (4, 3).STEP 3Check whether (4, 3) is a solution by substituting 4 for x and 3 for y in each of the original equations.Equation 1x + y = 77 = 7

(4) + (3) 7=?Equation 2x + 4y = 88 = 8

4 + 4(3) 8=?ANSWERBecause (4, 3) is a solution of each equation, it is a solution of the linear system.

EXAMPLE 2

Use the graph-and-check method

EXAMPLE 2

Use the graph-and-check methodSolve the linear system by graphing. Check your solution.

GUIDED PRACTICEfor Examples 1 and 2 5x + y = 01.5x + y = 10ANSWER(1, 5)

EXAMPLE 2


GUIDED PRACTICEfor Examples 1 and 22x + y = 4x + 2y = 32.ANSWER(1, 2)

EXAMPLE 2


GUIDED PRACTICEfor Examples 1 and 23x + y = 3x y = 53.ANSWER(2, 3)

EXAMPLE 3Standardized Test PracticeAs a season pass holder, you pay $4 per session to use the towns tennis courts. Without the season pass, you pay $13 per session to use the tennis courts.The parks and recreation department in your town offers a season pass for $90.11

EXAMPLE 3Standardized Test PracticeWhich system of equations can be used to find the number x of sessions of tennis after which the total cost y with a season pass, including the cost of the pass, is the same as the total cost without a season pass? y = 13x y = 4xA y = 13x y = 90 + 4xCy = 4xy = 90 + 13xBy = 90 + 4xy = 90 + 13xD12SOLUTION

EXAMPLE 3Standardized Test PracticeWrite a system of equations where y is the total cost (in dollars) for x sessions.EQUATION 1 y = 13 x

13

EXAMPLE 3Standardized Test PracticeEQUATION 2 y = 90 + 4 x

ANSWERThe correct answer is C.ACBD14

GUIDED PRACTICEfor Example 34. Solve the linear system in Example 3 to find the number of sessions after which the total cost with a season pass, including the cost of the pass, is the same as the total cost without a season pass.ANSWER10 sessions

GUIDED PRACTICEfor Example 35. WHAT IF? In Example 3, suppose a season pass costs $135. After how many sessions is the total cost with a season pass, including the cost of the pass, the same as the total cost without a season pass?ANSWER15 sessions

EXAMPLE 4

Solve a multi-step problemA business rents in-line skates and bicycles. During one day, the business has a total of 25 rentals and collects $450 for the rentals. Find the number of pairs of skates rented and the number of bicycles rented.RENTAL BUSINESSSOLUTION

EXAMPLE 4

Solve a multi-step problemSTEP 1Write a linear system. Let x be the number of pairs of skates rented, and let y be the number of bicycles rented. x + y = 2515x + 30y = 450Equation for number of rentalsEquation for money collected from rentalsSTEP 2Graph both equations.

EXAMPLE 4

Solve a multi-step problemSTEP 3Estimate the point of intersection. The two lines appear to intersect at (20, 5).

STEP 4Check whether (20, 5) is a solution.20 + 5 25=?15(20) + 30(5) 450=?450 = 450

25 = 25

ANSWERThe business rented 20 pairs of skates and 5 bicycles.

EXAMPLE 4

Solve a multi-step problem

GUIDED PRACTICEfor Example 4In Example 4, suppose the business has a total of 20 rentals and collects $420. Find the number of bicycles rented. 6.ANSWER8 bicyclesLesson 7.2

EXAMPLE 1Use the substitution methodSolve the linear system:y = 3x + 2Equation 2Equation 1x + 2y = 11Solve for y. Equation 1 is already solved for y.SOLUTIONSTEP 1

EXAMPLE 1Use the substitution method

7x + 4 = 11Simplify.7x = 7Subtract 4 from each side.x = 1Divide each side by 7.Substitute 3x + 2 for y.x + 2(3x + 2) = 11Write Equation 2.x + 2y = 11Substitute 3x + 2 for y in Equation 2 and solve for x.STEP 2

EXAMPLE 1Use the substitution methodANSWERThe solution is (1, 5).Substitute 1 for x in the original Equation 1 to find the value of y.y = 3x + 2 = 3(1) + 2 = 3 + 2 = 5STEP 3

GUIDED PRACTICECHECKy = 3x + 2 5 = 3(1) + 2? 5 = 5

Substitute 1 for x and 5 for y in each of the original equations.x + 2y = 11 1 + 2 (5) = 11? 11 = 11EXAMPLE 1Use the substitution method


Solve the linear system:x 2y = 6Equation 14x + 6y = 4Equation 2SOLUTIONSolve Equation 1 for x.x 2y = 6Write original Equation 1.x = 2y 6Revised Equation 1STEP 1


Substitute 2y 6 for x in Equation 2 and solve for y.4x + 6y = 4Write Equation 2.4(2y 6) + 6y = 4Substitute 2y 6 for x.Distributive property8y 24 + 6y = 414y 24 = 4Simplify.14y = 28Add 24 to each side.y = 2Divide each side by 14.STEP 2

EXAMPLE 2Use the substitution methodSubstitute 2 for y in the revised Equation 1 to find the value of x.x = 2y 6Revised Equation 1x = 2(2) 6Substitute 2 for y.x = 2Simplify.ANSWERThe solution is (2, 2).STEP 3 4(2) + 6 (2) = 4 ?

GUIDED PRACTICECHECK2 2(2) = 6?6 = 6

Substitute 2 for x and 2 for y in each of the original equations.4x + 6y = 4 4 = 4Equation 1Equation 2x 2y = 6EXAMPLE 2Use the substitution method


Solve the linear system using the substitution method.3x + y = 10

y = 2x + 51.GUIDED PRACTICEfor Examples 1 and 2ANSWER(1, 7)


x + 2y = 6

GUIDED PRACTICEfor Examples 1 and 2x y = 32.ANSWER(0, 3)Solve the linear system using the substitution method.


2x + 4y = 0

GUIDED PRACTICEfor Examples 1 and 23x + y = 73.Solve the linear system using the substitution method.ANSWER(2, 1)

EXAMPLE 3Solve a multi-step problem

Many businesses pay website hosting companies to store and maintain the computer files that make up their websites. Internet service providers also offer website hosting. The costs for website hosting offered by a website hosting company and an Internet service provider are shown in the table. Find the number of months after which the total cost for website hosting will be the same for both companies. WEBSITES33

Solve a multi-step problem EXAMPLE 3SOLUTIONWrite a system of equations. Let y be the totalcost after x months.Equation 1: Internet service provider

y = 10 + 21.95 x

STEP 134

Solve a multi-step problem EXAMPLE 3Equation 2: Website hosting company

y = 22.45 x

The system of equations is:y = 22.45xEquation 1y = 10 + 21.95xEquation 235

Solve a multi-step problem EXAMPLE 3Substitute 22.45x for y in Equation 1 and solvefor x.y = 10 + 21.95x22.45x = 10 + 21.95x0.5x = 10x = 20The total cost will be the same for both companies after 20 months.ANSWERSTEP 2Write Equation 1.Substitute 22.45x for y.Subtract 21.95x from each side.Divide each side by 0.5.36

GUIDED PRACTICEfor Example 34. In Example 3, what is the total cost for website hosting for each company after 20 months? $449ANSWER37

GUIDED PRACTICEfor Example 35. WHAT IF? In Example 3, suppose the Internet service provider offers $5 off the set-up fee. After how many months will the total cost for website hosting be the same for both companies? 10 moANSWER38

For extremely cold temperatures, an automobile manufacturer recommends that a 70% antifreeze and 30% water mix be used in the cooling system of a car. How many quarts of pure (100%) antifreeze and a 50% antifreeze and 50% water mix should be combined to make 11 quarts of a 70% antifreeze and 30% water mix?ANTIFREEZESolve a mixture problemEXAMPLE 439

SOLUTIONWrite an equation for the total number of quarts and an equation for the number of quarts of antifreeze. Let x be the number of quarts of 100% antifreeze, and let y be the number of quarts of a 50% antifreeze and 50% water mix.STEP 1Solve a mixture problemEXAMPLE 440

Equation 1: Total number of quartsx + y = 11 Equation 2: Number of quarts of antifreeze x quarts of100% antifreeze y quarts of50%50% mix 11 quarts of70%30% mix

1 x + 0.5 y = 0.7(11)

x + 0.5y = 7.7Solve a mixture problemEXAMPLE 441

The system of equations is:x + 0.5y = 7.7Solve Equation 1 for x.x + y = 11x = 11 ySubstitute 11 y for x in Equation 2 and solvefor y.x + 0.5y = 7.7STEP 2STEP 3Equation 1x + y =11Equation 2Write Equation 1Revised Equation 1Write Equation 2.Solve a mixture problemEXAMPLE 442

Solve a mixture problem EXAMPLE 4(11 y) + 0.5y = 7.7y = 6.6Substitute 6.6 for y in the revised Equation 1 tofind the value of x.STEP 4x = 11 y = 11 6.6 = 4.4ANSWERMix 4.4 quarts of 100% antifreeze and 6.6 quarts of a 50%antifreeze and 50% water mix to get 11 quarts of a 70%antifreeze and 30% water mix.Substitute 11 y for x.Solve for y.43

GUIDED PRACTICEfor Example 4WHAT IF? How many quarts of 100% antifreeze and a 50% antifreeze and 50% water mix should be combined to make 16 quarts of a 70% antifreeze and 30% water mix? 6. ANSWER6.4 quarts of 100% antifreeze and 9.6 quarts of a 50%antifreeze and 50% water mix44Lesson 7.3

Use addition to eliminate a variable EXAMPLE 1Solve the linear system:2x + 3y = 11 2x + 5y = 13Equation 1Equation 2SOLUTIONAdd the equations toeliminate one variable.2x + 3y = 112x + 5y = 13Solve for y.8y = 24y = 3STEP 1STEP 2

Use addition to eliminate a variable EXAMPLE 12x + 3y = 11Write Equation 12x + 3(3) = 11Substitute 3 for y.x = 1Solve for x.ANSWERThe solution is (1, 3).Substitute 3 for y in either equation and solve for x.STEP 3

Use addition to eliminate a variable EXAMPLE 12x + 3y = 1111 = 11Substitute 1 for x and 3 for y in each of the original equations.CHECK2(1) + 3(3) = 11?2x + 5y = 1313 = 132(1) + 5(3) = 13?

Use subtraction to eliminate a variable EXAMPLE 2Solve the linear system:4x + 3y = 2Equation 15x + 3y = 2Equation 2SOLUTIONSubtract the equations toeliminate one variable.4x + 3y = 25x + 3y = 2Solve for x. x = 4STEP 1STEP 2 x = 4 49

Use subtraction to eliminate a variable EXAMPLE 24x + 3y = 2Write Equation 1. 4(4) + 3y = 2Substitute 4 for x.y = 6Solve for y.ANSWERThe solution is (4, 6).Substitute 4 for x in either equation and solvefor y.STEP 350

Arrange like terms EXAMPLE 3Solve the linear system:8x 4y = 4Equation 14y = 3x + 14Equation 2SOLUTIONSTEP 1Rewrite Equation 2 so that the like terms are arranged in columns.8x 4y = 44y = 3x + 14

8x 4y = 43x + 4y = 14STEP 2Add the equations. 5x = 10STEP 3Solve for x. x = 251

Arrange like terms EXAMPLE 34y = 3x + 14Write Equation 2.4y = 3(2) + 14Substitute 2 for x.y = 5Solve for y.ANSWERThe solution is (2, 5).STEP 4Substitute 2 for x in either equation and solve for y.52

GUIDED PRACTICEfor Example 1,2 and 3Solve the linear system:4x 3y = 5 2x + 3y = 7`1. ANSWER(1, 3)53

GUIDED PRACTICEfor Example 1,2 and 3Solve the linear system:5x + 2y = 45x 6y = 82. ANSWER(2, 3)54

GUIDED PRACTICEfor Example 1,2 and 3Solve the linear system:6x 4y = 143x + 4y = 13. ANSWER(5, 4)55

GUIDED PRACTICEfor Example 1,2 and 3Solve the linear system:7x 2y = 57x 3y = 44. ANSWER(1, 1)56

GUIDED PRACTICEfor Example 1,2 and 3Solve the linear system:3x + 4y = 65. =3x + 62yANSWER(2, 0)57

GUIDED PRACTICEfor Example 1,2 and 3Solve the linear system:2x + 5y = 126. =4x + 65yANSWER(1, 2)58KAYAKING

EXAMPLE 4Write and solve a linear systemDuring a kayaking trip, a kayaker travels 12 miles upstream (against the current) and 12 miles downstream (with the current), as shown. The speed of the current remained constant during the trip. Find the average speed of the kayak in still water and the speed of the current.

EXAMPLE 4Write and solve a linear systemSTEP 1Write a system of equations. First find the speed of the kayak going upstream and the speed of the kayak going downstream.Upstream:d = rt12 = r 3

4 = rDownstream:d = rt12 = r 2

6 = r

EXAMPLE 4Write and solve a linear systemUse the speeds to write a linear system. Let x be the average speed of the kayak in still water, and let y be the speed of the current.

xy4=Equation 1:Going upstream

EXAMPLE 4Write and solve a linear systemEquation 2:Going downstream

xy6=+

EXAMPLE 4Write and solve a linear systemSTEP 2Solve the system of equations.x y = 42x = 10x = 5Write Equation 1.Write Equation 2.Add equations.Solve for x.Substitute 5 for x in Equation 2 and solve for y.x + y = 6

EXAMPLE 4Write and solve a linear system5 + y = 6y = 1Substitute 5 for x in Equation 2.Subtract 5 from each side.

GUIDED PRACTICEfor Example 47. WHAT IF? In Example 4, suppose it takes the kayaker 5 hours to travel 10 miles upstream and 2 hours to travel 10 miles downstream. The speed of the current remains constant during the trip. Find the average speed of the kayak in still water and the speed of the current.ANSWERaverage speed of the kayak: 3.5 mi/h, speed of the current 1.5 mi/hLesson 7.4SOLUTION

EXAMPLE 1Multiply one equation, then add

Solve the linear system:6x + 5y = 19Equation 12x + 3y = 5Equation 2STEP 1Multiply Equation 2 by 3 so that the coefficients of x are opposites.6x + 5y = 192x + 3y = 56x + 5y = 19STEP 2Add the equations.4y = 4

6x 9y = 15

EXAMPLE 1Multiply one equation, then addSTEP 3STEP 42x = 8Write Equation 2.2x + 3(1) = 5Substitute 1 for y.2x + 3y = 5x = 4Multiply.Subtract 3 from each side.Solve for y.Substitute 1 for y in either of the original equations and solve for x.2x + (3) = 5Divide each side by 2.y = 1

EXAMPLE 1Multiply one equation, then add

ANSWERThe solution is (4, 1).CHECKEquation 22x + 3y = 5Substitute 4 for x and 1 for y in each of the original equations.Equation 16x + 5y = 196(4) + 5(1) = 19?2(4) + 3(1) = 5?19 = 19

5 = 5

EXAMPLE 2Multiply both equations, then subtractSolve the linear system:4x + 5y = 35Equation 12y = 3x 9Equation 2SOLUTIONSTEP 14x + 5y = 35Write Equation 1.3x + 2y = 9Rewrite Equation 2.Arrange the equations so that like terms are in columns.70

EXAMPLE 2Multiply both equations, then subtractSTEP 24x + 5y = 353x + 2y = 923x = 115STEP 3STEP 48x + 10y = 70

15x +10y = 45

Multiply Equation 1 by 2 and Equation 2 by 5 so that the coefficient of y in each equation is the least common multiple of 5 and 2, or 10.Subtract: the equations.x = 5Solve: for x.71

EXAMPLE 2Multiply both equations, then subtractSTEP 54x + 5y = 354(5) + 5y = 35y = 3Write Equation 1.Substitute 5 for x.Solve for y.ANSWERThe solution is (5, 3).Substitute 5 for x in either of the original equations and solve for y.72

EXAMPLE 2Multiply both equations, then subtractCHECK4x + 5y = 35ANSWERThe solution is (5, 3).Substitute 5 for x and 3 for y in each of the original equations.4(5) + 5(3) = 35?Equation 1Equation 22y = 3x 92(3) = 3(5) 9?35 = 35

6 = 6

73

GUIDED PRACTICEfor Examples 1 and 2Solve the linear system using elimination.2x + 3y = 56x 2y = 11.ANSWERThe solution is (0.5, 2).74

GUIDED PRACTICEfor Examples 1 and 23x + 10y = 32x + 5y = 32.ANSWERThe solution is (9, 3).Solve the linear system using elimination.75

GUIDED PRACTICEfor Examples 1 and 29y = 5x + 53x 7y = 53.Solve the linear system using elimination.ANSWERThe solution is (10, 5).76

Standardized Test PracticeEXAMPLE 3Darlene is making a quilt that has alternating stripes of regular quilting fabric and sateen fabric. She spends $76 on a total of 16 yards of the two fabrics at a fabric store. Which system of equations can be used to find the amount x (in yards) of regular quilting fabric and the amount y (in yards) of sateen fabric she purchased?x + y = 16Ax + y = 16 Bx + y = 16 Dx + y = 76 Cx + y = 764x + 6y = 766x + 4y = 764x + 6y = 16

Standardized Test PracticeEXAMPLE 3SOLUTIONWrite a system of equations where x is the number of yards of regular quilting fabric purchased and y is the number of yards of sateen fabric purchased.Equation 1: Amount of fabricx+y=16

Standardized Test PracticeEXAMPLE 3Equation 2: Cost of fabricThe system of equations is:x + y = 164x + 6y = 76Equation 1Equation 2ANSWERADCBThe correct answer is B.4766+=yx

GUIDED PRACTICEfor Example 3SOCCER A sports equipment store is having a sale on soccer balls. A soccer coach purchases 10 soccer balls and 2 soccer ball bags for $155. Another soccer coach purchases 12 soccer balls and 3 soccer ball bags for $189. Find the cost of a soccer ball and the cost of a soccer ball bag.4. ANSWERsoccer ball $14.50, soccer ball bag: $5Lesson 7.6SOLUTION

EXAMPLE 1Graph a system of two linear inequalitiesGraph the system of inequalities.y > x 2 y 3x + 6 Inequality 1Inequality 2Graph both inequalities in the same coordinate plane. The graph of the system is the intersection of the two half-planes, which is shown as the darker shade of blue.

EXAMPLE 1Graph a system of two linear inequalitiesCHECKChoose a point in the dark blue region, such as (0, 1). To check this solution, substitute 0 for x and 1 for y into each inequality.1 > 0 2 ?1 > 2

1 6

1 0 + 6 ?

EXAMPLE 2Graph a system of three linear inequalitiesGraph the system of inequalities.y > 1 x > 2Inequality 1Inequality 2x + 2y 4Inequality 3SOLUTIONGraph all three inequalities in the same coordinate plane. The graph of the system is the triangular region shown.

GUIDED PRACTICEfor Examples 1 and 2ANSWERGraph the system of linear inequalities.1.y < x 4 y x + 3

GUIDED PRACTICEANSWERGraph the system of linear inequalities.2.y x + 2 y < 4x < 3for Examples 1 and 2

GUIDED PRACTICEANSWERGraph the system of linear inequalities.3.y > x y x 4 y < 5for Examples 1 and 2

EXAMPLE 3Write a system of linear inequalitiesWrite a system of inequalities for the shaded region.

SOLUTIONINEQUALITY 1: One boundary line for the shaded region is y = 3. Because the shaded region is above the solid line, the inequality is y 3.INEQUALITY 2: Another boundary line for the shaded region has a slope of 2 and a y-intercept of 1. So, its equation is y = 2x + 1. Because the shaded region is above the dashed line, the inequality is y > 2x + 1.

EXAMPLE 3Write a system of linear inequalitiesANSWERThe system of inequalities for the shaded region is:y 3 y > 2x + 1 Inequality 1Inequality 2

EXAMPLE 4Write and solve a system of linear inequalitiesBASEBALLThe National Collegiate Athletic Association (NCAA) regulates the lengths of aluminum baseball bats used by college baseball teams. The NCAA states that the length (in inches) of the bat minus the weight (in ounces) of the bat cannot exceed 3. Bats can be purchased at lengths from 26 to 34 inches.a. Write and graph a system of linear inequalities that describes the information given above.b. A sporting goods store sells an aluminum bat that is 31 inches long and weighs 25 ounces. Use the graph to determine if this bat can be used by a player on an NCAA team.

EXAMPLE 4Write and solve a system of linear inequalitiesSOLUTIONa. Let x be the length (in inches) of the bat, and let y be the weight (in ounces) of the bat. From the given information, you can write the following inequalities:The difference of the bats length and weight can be at most 3.x y 3x 26The length of the bat must be at least 26 inches.x 34y 0The length of the bat can be at most 34 inches.The weight of the bat cannot be a negative number.Graph each inequality in the system. Then identify the region that is common to all of the graphs of the inequalities. This region is shaded in the graph shown.

EXAMPLE 4Write and solve a system of linear inequalitiesb. Graph the point that represents a bat that is 31 inches long and weighs 25 ounces.

ANSWERBecause the point falls outside the solution region, the bat cannot be used by a player on an NCAA team.

GUIDED PRACTICEWrite a system of inequalities that defines the shaded region.for Examples 3 and 44.

ANSWERx 3, y > x 132

GUIDED PRACTICEWrite a system of inequalities that defines the shaded region.for Examples 3 and 45.

ANSWERy 4, x < 2

GUIDED PRACTICE6. WHAT IF? In Example 4, suppose a Senior League (ages 1014) player wants to buy the bat described in part (b). In Senior League, the length (in inches) of the bat minus the weight (in ounces) of the bat cannot exceed 8. Write and graph a system of inequalities to determine whether the described bat can be used by the Senior League player.for Examples 3 and 4ANSWER

x y 8, x 26, x 34, y 0The bat can be used.

examples and guided practice come from the algebra 1 powerpoint presentations available at

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