Unit 3 – Chapter 7

Unit 3 Section 7.1 – Solve Linear Equations by graphing Section 7.2 – Solve Linear Equations by substitution. Section 7.3 – Solve Linear Equations by Adding or Subtracting Section 7.4 – Solve Linear Systems by multiplying first Section 7.5 – Solve Special Types of Linear Systems Section 7.6 – Solve systems of linear inequalities

Warm-Up – 7.1

Lesson 7.1, For use with pages 426-434 1. Graph the equation –2x + y = 1. ANSWER 2. It takes 3 hours to mow a lawn and 2 hours to trim hedges. You spend 16 hours doing yard work. What are 2 possible numbers of lawns you mowed and hedges you trimmed? ANSWER 2 lawns and 5 hedges, or 4 lawns and 2 hedges

Vocabulary – 7.1 System of Linear Equations 2 or more linear equations with the same variables Solution of a Linear Equation The solution set that MAKES ALL THE EQUATIONS TRUE AT THE SAME TIME!! Usually a single point! Consistent Independent System A linear system that has EXACTLY one solution.

Notes – 7.1 – Solving Linear Systems by Graphing Remember the steps to convert English to Mathlish: Read and highlight key words DEFINE THE VARIABLES (MOST CRITICAL STEP!!) Write Mathlish sentence left to right (careful with subtraction and division!) 3 Step Process To Solve Linear Eqns by graphing Write equations in slope-intercept form Graph them on calculator Find intersection of lines CHECK ANSWERS!

Examples 7.1

Use the graph-and-check method EXAMPLE 2 Use the graph-and-check method Solve the linear system: – x + y = – 7 Equation 1 x + 4y = – 8 Equation 2 SOLUTION STEP 1 Graph both equations.

Use the graph-and-check method EXAMPLE 2 Use the graph-and-check method STEP 2 Estimate the point of intersection. The two lines appear to intersect at (4, – 3). STEP 3 Check whether (4, – 3) is a solution by substituting 4 for x and – 3 for y in each of the original equations. Equation 1 Equation 2 – x + y = – 7 x + 4y = – 8 –(4) + (– 3) – 7 = ? 4 + 4(– 3) – 8 = ? – 7 = – 7 – 8 = – 8

EXAMPLE 2 Use the graph-and-check method ANSWER Because (4, – 3) is a solution of each equation, it is a solution of the linear system.

EXAMPLE 2 GUIDED PRACTICE Use the graph-and-check method for Examples 1 and 2 Solve the linear system by graphing. Check your solution. – 5x + y = 0 1. 5x + y = 10 Put eqns in slope- Intercept form y = 5x 5x + y = -5x + 10 The Intersection point is at (1,5).

EXAMPLE 3 Standardized Test Practice The parks and recreation department in your town offers a season pass for $90. As a season pass holder, you pay $4 per session to use the town’s tennis courts. • • Without the season pass, you pay $13 per session to use the tennis courts. • Write a system of equations to model these situations and find out where they are equal.

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

EXAMPLE 3 Standardized Test Practice EQUATION 2 y = 90 + 4 x ANSWER The correct answer is y = 13x and y = 4x + 90.

GUIDED PRACTICE for Example 3 4. 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. SOLUTION Let the number of sessions be x So, 13 x = 90 + 4 x 13x = 90 + 4x 9x = 90 x = 10 ANSWER 10 sessions

GUIDED PRACTICE for Example 3 5. 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? SOLUTION Let the number of sessions be x So, 13 x = 135 + 4 x 13x = 135 + 4x 9x = 135 x = 15 ANSWER 15 sessions

EXAMPLE 4 Solve a multi-step problem RENTAL BUSINESS A 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.

Solve a multi-step problem EXAMPLE 4 Solve a multi-step problem SOLUTION STEP 1 Write a linear system. Let x be the number of pairs of skates rented, and let y be the number of bicycles rented. x + y =25 Equation for number of rentals 15x + 30y = 450 Equation for money collected from rentals STEP 2 Graph both equations.

Solve a multi-step problem EXAMPLE 4 Solve a multi-step problem STEP 3 Estimate the point of intersection. The two lines appear to intersect at (20, 5). STEP 4 Check whether (20, 5) is a solution. 20 + 5 25 = ? 15(20) + 30(5) 450 = ? 25 = 25 450 = 450 ANSWER The business rented 20 pairs of skates and 5 bicycles.

Solve a multi-step problem GUIDED PRACTICE for Example 4 In Example 4, suppose the business has a total of 20 rentals and collects $ 420. Find the number of bicycles rented. 6. SOLUTION STEP 1 Write a linear system. Let x be the number of pairs of skates rented, and let y be the number of bicycles rented. x + y =20 Equation for number of rentals. 15x + 30y = 420 Equation for money collected from rentals.

EXAMPLE 4 Solve a multi-step problem GUIDED PRACTICE for Example 4 STEP 2 Graph both equations. STEP 3 Estimate the point of intersection. The two lines appear to intersect at (12, 8). STEP 4 Check whether (12, 8) is a solution. x + y 20 = 15x + 30y 420 = 12 + 8 = 20 15(12) + 30(8) 420 = 20 = 20 420 = 420

EXAMPLE 4 EXAMPLE 4 Solve a multi-step problem GUIDED PRACTICE for Example 4 ANSWER The business rented number of bicycle is 8.

Warm-Up – 7.2

Lesson 7.2, For use with pages 435-441 Solve the equation. 1. 6a – 3 + 2a = 13 ANSWER a = 2 2. 4(n + 2) – n = 11 ANSWER n = 1

Lesson 7.2, For use with pages 435-441 Solve the equation. 3. You burned 8 calories per minute on a treadmill and 10 calories per minute on an elliptical trainer for a total of 560 calories in 60 minutes. How many minutes did you spend on each machine? HINT Define the variables Create two equations (one for total minutes And one for total calories) ANSWER EQN 1 – 8x + 10y = 560 EQN 2 – x + y = 60 treadmill: 20 min, elliptical trainer: 40 min

Do the Dance!!!! Lesson 7.2, For use with pages 435-441 Solve the system of equations BUT YOU CAN’T USE THE CALCULATOR OR GRAPH IT!!! WORK WITH YOUR GROUP!!. 2x + 3y = 40 y = x + 5 HINT! Do the Dance!!!! ANSWER X = 5 Y = 10 26

Vocabulary – 7.2 None!!

Notes – 7.2 – Solving Systems w/Substitution If it’s EASY to get one of the variables in an equation by itself, substitution may be the easiest way to solve the system of equations. To Solve Linear Systems with Substitution Get ONE of the variables in ONE of the equations by itself. Substitute that variable into the OTHER equation. Solve the equation from #2. Plug the answer back into the original equation.

Examples 7.2

Use the substitution method EXAMPLE 2 Use the substitution method Solve the linear system: x – 2y = – 6 Equation 1 4x + 6y = 4 Equation 2 SOLUTION STEP 1 Solve Equation 1 for x. x – 2y = – 6 Write original Equation 1. x =2y – 6 Revised Equation 1

Use the substitution method EXAMPLE 2 Use the substitution method STEP 2 Substitute 2y – 6 for x in Equation 2 and solve for y. 4x + 6y = 4 Write Equation 2. 4(2y – 6) + 6y = 4 Substitute 2y – 6 for x. 8y – 24 + 6y = 4 Distributive property 14y – 24 = 4 Simplify. 14y = 28 Add 24 to each side. y = 2 Divide each side by 14.

Use the substitution method EXAMPLE 2 Use the substitution method STEP 3 Substitute 2 for y in the revised Equation 1 to find the value of x. x = 2y – 6 Revised Equation 1 x = 2(2) – 6 Substitute 2 for y. x = – 2 Simplify. ANSWER The solution is (– 2, 2).

Use the substitution method EXAMPLE 2 GUIDED PRACTICE Use the substitution method CHECK Substitute –2 for x and 2 for y in each of the original equations. Equation 1 Equation 2 4x + 6y = 4 x – 2y = – 6 –2 –2 (2)= – 6 ? 4( –2 )+ 6 (2 ) = 4 ? – 6 = – 6 4 = 4

Use the substitution method for Examples 1 and 2 GUIDED PRACTICE Use the substitution method for Examples 1 and 2 Solve the linear system using the substitution method. y = 2x + 5 1. Equation 1 3x + y = 10 Equation 2 SOLUTION STEP 1 Solve for y. Equation 1 is already solved for y.

Use the substitution method for Examples 1 and 2 GUIDED PRACTICE Use the substitution method for Examples 1 and 2 STEP 2 Substitute 2x + 5 for y in Equation 2 and solve for x. 3x + y = 10 Write Equation 2. 3x + (2x + 5) = 10 Substitute 2x+5 for x. 5x + 5 = 10 Simplify. 5x = 5 x = 2

EXAMPLE 2 GUIDED PRACTICE for Examples 1 and 2 STEP 3 Substitute 1 for x in the revised Equation 1 to find the value of y. y = 2x + 5 = 2(1) + 5 = 7 ANSWER The solution is ( 1, 7).

Use the substitution method for Examples 1 and 2 for Example 1 GUIDED PRACTICE GUIDED PRACTICE Use the substitution method for Examples 1 and 2 for Example 1 CHECK Substitute 1 for x and 7 for y in each of the original equations. Equation 1 Equation 2 y = 2x + 5 3x + y = 10 7 = 2(1) + 5 ? 3 ( 1 ) + 7 = 11 ? 7 = 7 10 = 10

Use the substitution method for Examples 1 and 2 GUIDED PRACTICE Use the substitution method for Examples 1 and 2 x – y = 3 2. Equation 1 x + 2y = – 6 Equation 2 SOLUTION STEP 1 Solve Equation 1 for x. x – y = 3 Write original Equation 1. x = y + 3 Revised Equation 1

Use the substitution method for Examples 1 and 2 GUIDED PRACTICE Use the substitution method for Examples 1 and 2 STEP 2 Substitute y + 3 for x in Equation 2 and solve for y. x + 2y = – 6 Write Equation 2. ( y + 3) + 2y = – 6 Substitute y + 3 for x. 3y + 3 = – 6 Simplify. 3y = – 9 Add 3 to each side. y = – 3 Divide each side by 3.

Use the substitution method for Examples 1and 2 GUIDED PRACTICE Use the substitution method for Examples 1and 2 STEP 3 Substitute – 3 for y in the revised Equation 1 to find the value of x. x = y + 3 Revised Equation 1 x = – 3 + 3 Substitute – 3 for y. x = 0 Simplify. ANSWER The solution is ( 0, – 3 ).

Use the substitution method GUIDED PRACTICE for Examples 1 and 2 CHECK Substitute 0 for x and – 3 for y in each of the original equations. Equation 1 Equation 2 x + 2y = – 6 x – y = 3 0 – (–3) = 3 ? 0 + 2 (–3 ) = – 6 ? 3 = 3 – 6 = – 6

EXAMPLE 3 Solve a multi-step problem WEBSITES 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.

EXAMPLE 3 Solve a multi-step problem SOLUTION STEP 1 Write a system of equations. Let y be the total cost after x months. Equation 1: Internet service provider y = 10 + 21.95 x

Solve a multi-step problem EXAMPLE 3 Solve a multi-step problem Equation 2: Website hosting company y = 22.45 x The system of equations is: y = 10 + 21.95x Equation 1 y = 22.45x Equation 2

Solve a multi-step problem EXAMPLE 3 Solve a multi-step problem STEP 2 Substitute 22.45x for y in Equation 1 and solve for x. y = 10 + 21.95x Write Equation 1. 22.45x = 10 + 21.95x Substitute 22.45x for y. 0.5x = 10 Subtract 21.95x from each side. x = 20 Divide each side by 0.5. The total cost will be the same for both companies after 20 months. ANSWER

GUIDED PRACTICE for Example 3 4. In Example 3, what is the total cost for website hosting for each company after 20 months? SOLUTION Let y be the total cost. y = 22.45 20 = 449

GUIDED PRACTICE for Example 3 The total cost for website hosting for each company after 20 months is $ 449. ANSWER

EXAMPLE 4 Solve a mixture problem ANTIFREEZE 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? SOLUTION STEP 1 Write 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.

EXAMPLE 4 Solve a mixture problem Equation 1: Total number of quarts x + y = 11 Equation 2: Number of quarts of antifreeze x quarts of 100% antifreeze y quarts of 50% –50% mix 11 quarts of 70% – 30% mix 1 x + 0.5 y = 0.7(11) x + 0.5y = 7.7

Solve a mixture problem EXAMPLE 4 Solve a mixture problem The system of equations is: x + y =11 Equation 1 x + 0.5y = 7.7 Equation 2 STEP 2 Solve Equation 1 for x. Write Equation 1. x + y = 11 x = 11 – y Revised Equation 1 STEP 3 Substitute 11 – y for x in Equation 2 and solve for y. x + 0.5y = 7.7 Write Equation 2.

Solve a mixture problem EXAMPLE 4 Solve a mixture problem (11 – y) = 0.5y = 7.7 Substitute 11 – y for x. Solve for y. y = 6.6 STEP 4 Substitute 6.6 for y in the revised Equation 1 to find the value of x. x = 11 – y = 11 – 6.6 = 4.4 ANSWER Mix 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.

GUIDED PRACTICE for Example 4 WHAT 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. SOLUTION STEP 1 Write 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.

GUIDED PRACTICE for Example 4 Equation 1: Total number of quarts x + y = 16 Equation 2: Number of quarts of antifreeze x quarts of 100% antifreeze y quarts of 50% –50% mix 11 quarts of 70% – 30% mix 1 x + 0.5 y = 0.7(16) x + 0.5y = 11.2

The system of equations is: x + y =16 GUIDED PRACTICE for Example 4 The system of equations is: x + y =16 Equation 1 x + 0.5y = 11.2 Equation 2. STEP 2 Solve Equation 1 for x. Write Equation 1. x + y = 16 x = 16 – y Revised Equation 1

Substitute 16 – y for x in Equation 1 and solve for x. GUIDED PRACTICE for Example 4 STEP 3 Substitute 16 – y for x in Equation 1 and solve for x. x + 0.5y = 7.7 Write Equation 2. (16 – y) + 0.5y = 7.7 Substitute 16 – y for x. Solve for y. y = 9.6

GUIDED PRACTICE for Example 4 STEP 4 Substitute 9.6 for y in the revised Equation 1 to find the value of x. x = 16 – y = 16 – 9.6 = 6.4 ANSWER Mix 6.4 quarts of 100% antifreeze and 9.6 quarts of a 50% Antifreeze.

Warm-Up – 7.3

Lesson 7.2, For use with pages 435-441 Solve the linear systems by GRAPHING!!! x + y = -2 -x + y = 6 ANSWER x = -4 y = 2 x – y = 0 5x + 2y = -7 ANSWER x = -1 y = -1

Lesson 7.2, For use with pages 435-441 Solve the linear systems by SUBSTITUTION!!!!!! y = x – 4 -2x + y = 18 ANSWER x = -22 y = -26 5x – 4y = 27 -2x + y = 3 ANSWER x= -13 y = -23

Lesson 7.3, For use with pages 443-450 1. Solve the linear system using substitution. 2x + y = 12 3x – 2y = 11 ANSWER (5, 2) One auto repair shop charges $30 for a diagnosis and $25 per hour for labor. Another auto repair shop charges $35 per hour for labor. For how many hours are the total charges for both of the shops the same? HINT: FIND EQUATIONS FOR TOTAL AND SUBSTITUTE! ANSWER 3 h

Lesson 7.3, For use with pages 443-450 1. Add the two equations together (combine like terms) and solve for x and y. 2x + 3y = 11 -2x + 5y = 13 ANSWER (1,3)

Vocabulary – 7.3 - REVIEW System of Linear Equations 2 or more linear equations with the same variables Solution of a Linear Equation The solution set that MAKES ALL THE EQUATIONS TRUE AT THE SAME TIME!! Consistent Independent System A linear system that has EXACTLY one solution.

Notes – 7.3 – Solving systems with elimination If we have two equations and two variables, how many solutions should we USUALLY have?? What’s the goal of solving every algebra eqn. you will ever see? You can eliminate variables from some systems by adding or subtracting equations to eliminate variables. RULES/HINTS TO MAKE PROCESS EASIER! Remember the goal: You are trying to eliminate one variable! Line up like terms under each other. NEVER subtract. Add the negative instead!

Examples 7.3

Use addition to eliminate a variable EXAMPLE 1 Use addition to eliminate a variable Solve the linear system: 2x + 3y = 11 Equation 1 – 2x + 5y = 13 Equation 2 SOLUTION STEP 1 Add the equations to eliminate one variable. 2x + 3y = 11 – 2x +5y = 13 STEP 2 Solve for y. 8y = 24 y = 3 STEP 3 Substitute 3 for y in either equation and Solve for x.

Use addition to eliminate a variable EXAMPLE 1 Use addition to eliminate a variable 2x + 3y = 11 Write Equation 1 2x + 3(3) = 11 Substitute 3 for y. x = 1 Solve for x. ANSWER The solution is (1, 3).

Use subtraction to eliminate a variable EXAMPLE 2 Use subtraction to eliminate a variable Solve the linear system: 4x + 3y = 2 Equation 1 5x + 3y = – 2 Equation 2 SOLUTION STEP 1 Subtract the equations to eliminate one variable. 4x + 3y = 2 5x + 3y = – 2 STEP 2 Solve for x. – x = 4 x = 4 STEP 3 Substitute 4 for x in either equation and solve for y.

Use subtraction to eliminate a variable EXAMPLE 2 Use subtraction to eliminate a variable 4x + 3y = 2 Write Equation 1. 4(– 4) + 3y = 2 Substitute – 4 for x. y = 2 Solve for y. ANSWER The solution is (– 4, 6).

Solve the linear system: 8x – 4y = –4 4y = 3x + 14 EXAMPLE 3 Arrange like terms Solve the linear system: 8x – 4y = –4 Equation 1 4y = 3x + 14 Equation 2 SOLUTION STEP 1 Rewrite Equation 2 so that the like terms are arranged in columns. 8x – 4y = –4 8x – 4y = –4 4y = 3x + 14 3x + 4y = 14 STEP 2 Add the equations. 5x = 10 STEP 3 Solve for x. x = 2 STEP 4 Substitute 2 for x in either equation and solve for y.

EXAMPLE 3 Arrange like terms 4y = 3x + 14 4y = 3(2) + 14 y = 5 ANSWER Write Equation 2. 4y = 3(2) + 14 Substitute 2 for x. y = 5 Solve for y. ANSWER The solution is (2, 5).

Solve the linear system: GUIDED PRACTICE for Example 1,2 and 3 Solve the linear system: ` 1. 4x – 3y = 5 Equation 1 – 2x + 3y = – 7 Equation 2 SOLUTION STEP 1 Add the equations to eliminate one variable. 4x – 3y = 5 – 2x +3y = – 7 STEP 2 Solve for x. 2x = – 2 x = – 1 STEP 3 Substitute – 1 for y in either equation and Solve for x.

GUIDED PRACTICE for Example 1,2 and 3 4x – 3y = 5 2(– 1) – 3y = 5 Write Equation 1. 2(– 1) – 3y = 5 Substitute – 1 for x. y = – 3 Solve for x. ANSWER The solution is (– 1, – 3).

GUIDED PRACTICE for Example 1,2 and 3 CHECK Substitute 1 for x and 3 for y in each of the. original equation 4x – 3y = 5 – 2x + 3y = – 7 4(– 1) – 3(– 3) = 5 ? – 2(– 1) + 5(– 3) – 7 ? = 5 = 5 – 7 = – 7

Solve the linear system: 4. 7x – 2y = 5 7x – 3y = 4 GUIDED PRACTICE for Example 1,2 and 3 Solve the linear system: 4. 7x – 2y = 5 Equation 1 7x – 3y = 4 Equation 2 SOLUTION STEP 1 Subtract the equations to eliminate one variable. 7x – 2y = 5 7x – 3y = 4 STEP 2 y = 1 Solve for y. STEP 3 Substitute 1 for y in either and solve for x.

GUIDED PRACTICE for Example 1,2 and 3 7x – 2y = 5 7x – 2(1) = 5 x = 1 Write Equation 1. 7x – 2(1) = 5 Substitute 1 for y. x = 1 Solve for x. ANSWER The solution is (1, 1).

Solve the linear system: 5. 3x + 4y = – 6 = 3x + 6 2y GUIDED PRACTICE for Example 1,2 and 3 Solve the linear system: 5. 3x + 4y = – 6 Equation 1 = 3x + 6 2y Equation 2 SOLUTION STEP 1 Rewrite Equation 2 so that the like terms are arranged in columns. 3x + 4y = – 6 3x – 2y = – 6 3x + 4y = – 6 = 3x + 6 2y STEP 2 Subtract the equations. 6y = 0 STEP 3 Solve for y. y = 0 STEP 4 Substitute 0 for y in either equation and solve for x.

GUIDED PRACTICE for Example 1,2 and 3 2y = 3x + 6 2(0) = 3x + 6 Write Equation 2. = 3x + 6 2(0) Substitute 0 for y. x = – 2 Solve for x . ANSWER The solution is (– 2, 0).

Solve the linear system: 6. 2x + 5y = 12 = 4x + 6 5y GUIDED PRACTICE for Example 1,2 and 3 Solve the linear system: 6. 2x + 5y = 12 Equation 1 = 4x + 6 5y Equation 2 SOLUTION STEP 1 Rewrite Equation 2 so that the like terms are arranged in columns. 2x + 5y = 12 – 4x + 5y = 6 2x + 5y = 12 = 4x + 6 5y STEP 2 Subtract the equations. 6x = 6 STEP 3 Solve for x. x = 1 STEP 4 Substitute 1 for x in either equation and solve for y.

GUIDED PRACTICE for Example 1,2 and 3 2x + 5y = 12 2(1) + 5y = 6 x = 2 Write Equation 2. + 5y = 6 2(1) Substitute 1 for x. x = 2 Solve for y . ANSWER The solution is (1, 2).

EXAMPLE 4 Write and solve a linear system KAYAKING During 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 4 Write and solve a linear system STEP 1 Write a system of equations. First find the speed of the kayak going upstream and the speed of the kayak going downstream. Upstream: d = rt Downstream: d = rt 12 = r 3 12 = r 2 4 = r 6 = r

EXAMPLE 4 Write and solve a linear system Use 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. Equation 1: Going upstream x y 4 = –

EXAMPLE 4 Write and solve a linear system Equation 2: Going downstream x y 6 = +

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

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

GUIDED PRACTICE for Example 4 7. 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. SOLUTION STEP 1 Write a system of equations. First find the speed of the kayak going downstream.

GUIDED PRACTICE for Example 4 Upstream: d = rt Downstream: d = rt 10 = r 5 10 = r 2 2 = r 5 = r Use the speeds to write a linear system. Let x be the average speed of the kayakar in still water, and let y be the speed of the current.

GUIDED PRACTICE for Example 4 Going upstream Equation 1: x y 2 = – Equation 2: Going downstream x y 5 = +

Solve the system of equations. GUIDED PRACTICE for Example 4 STEP 2 Solve the system of equations. x – y= 2 Equation 1. x + y = 5 Equation 2. 2x = 7 Add equations. x = 3.5 Solve for x. Substitute 3.5 for x in Equation 2 and solve for y.

GUIDED PRACTICE for Example 4 3.5 + y = 6 y = 1.5 ANSWER Substitute 3.5 for x in Equation 2. y = 1.5 Subtract 3.5 from each side. ANSWER The average speed of the kayakar in still water is 3.5 miles per hour, and the speed of the current is 1.5 mile per hour.

Warm-Up – 7.4

Lesson 7.4, For use with pages 451-457 Solve the linear system. 1. 4x – 3y = 15 2x – 3y = 9 ANSWER (3, – 1) 2. –2x + y = – 8 2x – 2y = 8 ANSWER (4, 0)

Lesson 7.4, For use with pages 451-457 Solve the linear system. 3. You can row a canoe 10 miles upstream in 2.5 hours and 10 miles downstream in 2 hours. What is the average speed of the canoe in still water? ANSWER 4.5 mi/h Multiply the second equation by -2 and rewrite it. Then use it to solve the system. 8x – 6y = 30 2x – 3y = 9 ANSWER (3, – 1)

Vocabulary – 7.4 Least Common Multiple Smallest POSITIVE number that is a multiple of two or more factors

Notes – 7.4 – Solving systems by multiplying first. In order to add equations and eliminate a variable, two of the coefficients must be opposite signs. Learned 3 ways to solve systems of linear eqns: Graphing Easiest when I can get y by itself and have a calculator! Substitution Easiest when I can get one variable by itself. Elimination Easiest when I can get opposite coefficients. There is a 4th way - multiply and then eliminate. To get coefficients with opposite signs, you can multiply one or more equations by constants. May need to identify LCM of two coefficients.

Examples 7.4

Multiply one equation, then add EXAMPLE 1 Multiply one equation, then add Solve the linear system: 6x +5y = 19 Equation 1 2x +3y = 5 Equation 2 SOLUTION STEP 1 Multiply: Equation 2 by –3 so that the coefficients of x are opposites. 6x + 5y = 19 6x + 5y = 19 2x + 3y = 5 –6x – 9y = –15 STEP 2 Add: the equations. –4y = 4

EXAMPLE 1 Multiply one equation, then add STEP 3 Solve: for y. y = –1 STEP 4 Substitute: –1 for y in either of the original equations and solve for x. ANSWER The solution is (4, –1).

Multiply both equations, then subtract EXAMPLE 2 Multiply both equations, then subtract Solve the linear system: 4x + 5y = 35 Equation 1 2y = 3x – 9 Equation 2 SOLUTION STEP 1 Arrange: the equations so that like terms are in columns. 4x + 5y = 35 Write Equation 1. –3x + 2y = –9 Rewrite Equation 2.

EXAMPLE 2 Multiply both equations, then subtract STEP 2 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. 4x + 5y = 35 8x + 10y = 70 –3x + 2y = –9 –15x +10y = –45 STEP 3 Subtract: the equations. 23x = 115 STEP 4 Solve: for x. x = 5 ANSWER The solution is (5, 3).

Solve the linear system using elimination: GUIDED PRACTICE for Examples 1 and 2 Solve the linear system using elimination: 6x – 2y = 1 1. Equation 1 –2x + 3y = –5 Equation 2 SOLUTION ANSWER The solution is (–0.5, –2). y = –2

GUIDED PRACTICE for Examples 1 and 2 2x + 5y = 3 2. 3x + 10y = –3 Equation 1 3x + 10y = –3 Equation 2 SOLUTION ANSWER The solution is (9, –3).

GUIDED PRACTICE for Examples 1 and 2 3. 3x – 7y = 5 9y = 5x +5 Equation 1 9y = 5x +5 Equation 2 SOLUTION ANSWER The solution is (–10, –5).

EXAMPLE 3 Standardized Test Practice Darlene 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. Write a 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?

EXAMPLE 3 Standardized Test Practice SOLUTION Write 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 fabric x + y = 16

Standardized Test Practice EXAMPLE 3 Standardized Test Practice Equation 2: Cost of fabric 4 76 6 + = y x The system of equations is: x + y = 16 Equation 1 4x +6y = 76 Equation 2 ANSWER The correct answer is x = 10 and y = 6.

GUIDED PRACTICE for Example 3 SOCCER 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. SOLUTION Write a system of equations where x is the cost of soccer ball and y is the cost of soccer ball bag.

GUIDED PRACTICE for Example 3 Equation 1: Cost of soccer ball Total Cost Cost of soccer bag + = 10x + 2y = 155

The system of equations is: GUIDED PRACTICE for Example 3 Equation 2: Cost of soccer ball Total Cost Cost of soccer bag + = 12x + 3y = 189 The system of equations is: 10x +2y = 155 Equation 1 12x +3y = 189 Equation 2

GUIDED PRACTICE for Example 3 STEP 1 Multiply equation 1 by – 3 and equation 2 by 2 so that the coefficient of y in each equation is the least common multiple of – 3 and 2 . STEP 2 10x +2y = 155 – 30x – 6y = –465 12x +3y = 189 24x + 6y = 378 STEP 3 Add the equation 6x = 87 STEP 4 Solve for x x = 14.50

Cost of soccer ball is $ 14.50 and soccer ball bag is $5. GUIDED PRACTICE for Example 3 STEP 5 Substitute 14.50 for x in either of the original equations and solve for y. 10x + 2y = 155 Write Equation 1. 10(14.50) + 2y = 155 Substitute 14.50 for x. y = 5 solve for y ANSWER Cost of soccer ball is $ 14.50 and soccer ball bag is $5.

Warm-Up – 7.5

Lesson 7.5, For use with pages 459-465 1. Solve the linear system. 2x + 3y = – 9 x – 2y = 6 ANSWER (0, – 3) 2. You buy 8 pencils for $8 at the bookstore. Standard pencils cost $.85 and specialty pencils cost $1.25. How many specialty pencils did you buy? ANSWER 3 specialty pencils

Lesson 7.5, For use with pages 459-465 1. Solve the linear system by GRAPHING. Describe the lines on your whiteboard. x + y = -2 y = -x+5 1. Solve the linear system by SUBSTITUTION. What do you get? x + y = -2 y = -x+5 1. Solve the linear system by ELIMINATION. What do you get? x + y = -2 y = -x+5 ANSWER No solution.

Vocabulary – 7.5 Consistent Independent System System of equations with ONE solution Inconsistent System System of equations with NO solution. Consistent Dependent System System of equations with INFINITE solutions.

Notes – 7.5–Special Types of Systems Systems can have one soln, no soln, or infinite. Easiest ways to check for solutions: Graph them (put them in slope-intercept form) Intersect = 1 solution Parallel = No solution Same line = Infinite solutions Check if equations are multiples of each other. Yes = infinite solutions No = Check some more! Eliminate the variables (using Add. or Mult.) Always False statement = No solutions Always True statement = Infinite solutions

Notes – 7.5–Special Types of Systems

Examples 7.4

A linear system with no solution EXAMPLE 1 A linear system with no solution Solve the linear system by graphing and by elimination! 3x + 2y = 10 Equation 1 3x + 2y = 2 Equation 2 SOLUTION METHOD 1 Graphing Graph the linear system. Answer: NO SOLUTION.

A linear system with no solution EXAMPLE 1 A linear system with no solution METHOD 2 Elimination Subtract the equation. 3x + 2y = 10 3x + 2y = 2 0 = 8 This is a false statement. ANSWER The variables are eliminated and you are left with a false statement regardless of the values of x and y. This tells you that the system has no solution.

A linear system with infinitely many solutions EXAMPLE 2 A linear system with infinitely many solutions Solve the system by graphing and substitution. x – 2y = – 4 Equation 1 y = x + 2 1 2 Equation 2 SOLUTION Graphing METHOD 1 Graph the linear system.

A linear system with infinitely many solutions EXAMPLE 2 A linear system with infinitely many solutions METHOD 2 Substitution Substitute x + 2 for y in Equation 1 and solve for x. 1 2 x – 2y = – 4 Write Equation 1 x – 2 x + 2 = 1 2 – 4 2 Substitute x + 2 for y. 1 – 4 = – 4 Simplify. The variables are eliminated and you are left with a true statement regardless of the values of x and y. This tells you that the system has infinite solutions. ANSWER

Subtract the equations. GUIDED PRACTICE for Examples 1 and 2 Tell whether the linear system has no solution or infinitely many solutions. Explain. 1. 5x + 3y = 6 Equation 1 – 5x – 3y = 3 Equation 2 METHOD 2 Elimination Subtract the equations. 5x + 3y = 6 – 5x – 3y = 3 0 = 9 This is a false statement.

GUIDED PRACTICE for Examples 1 and 2 ANSWER The variables are eliminated and you are left with a false statement regardless of the values of x and y. This tells you that the system has no solution.

Substitute 2x – 4 for y in Equation 2 and solve for x. GUIDED PRACTICE for Examples 1 and 2 2. y = 2x – 4 Equation 1 – 6x + 3y = – 12 Equation 2 METHOD 2 Elimination Substitute 2x – 4 for y in Equation 2 and solve for x. – 6x + 3y = – 12 Write Equation 2 – 6x + 3(2x – 4) = – 12 Substitute (2x – 4) for y. – 12 = – 12 Simplify.

GUIDED PRACTICE for Examples 1 and 2 ANSWER The variables are eliminated and you are left with a true statement regardless of the values of x and y. This tells you that the system has infinitely many solution.

Warm-Up – 7.6

Lesson 7.6, For use with pages 466-472 1. Graph y < x – 1. 2 3 ANSWER 1. Solve -3x >= 12 ANSWER: x <= -4

Lesson 7.6, For use with pages 466-472 2. You are running one ad that costs $6 per day and another that costs $8 per day. You can spend no more than $120. Graph this inequality. HINT: WRITE THE EQUATION FIRST! Graph the following on a number line: x <= 5 and x >= 0 ANSWER: -2 -1 0 1 2 3 4 5 6 7

Vocabulary – 7.6 System of Linear Inequalities Two or more linear inequalities in the same variables. Solution of a system of linear inequalities An ordered pair that makes ALL the inequalities true at the same time. Graph of a system of linear inequalitites Graph of all the solutions of the system.

Notes–7.6–Solve Systems of linear inequalities. REVIEW Graphing inequalities - similar to graphing lin.eqns Play the pretend game and let equation be =. Dotted line is < or > Solid line is <= or >= Pick a point NOT ON THE LINE, check the answer, and shade the correct side of the line. TO GRAPH SYSTEMS OF INEQUALITIES Graph each inequality Find the AREA where solutions intersect. Pick a point and check your solution.

Examples 7.6

Graph a system of two linear inequalities EXAMPLE 1 Graph a system of two linear inequalities Graph the system of inequalities. y > – x – 2 y < 3x + 6 Inequality 1 Inequality 2 SOLUTION Graph 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.

Graph a system of two linear inequalities EXAMPLE 1 Graph a system of two linear inequalities CHECK Choose 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 > 0 + 6 ? 1 > – 2 1 > 6

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

GUIDED PRACTICE for Examples 1 and 2 Graph the system of linear inequalities. 1. y < x – 4 y > – x + 3 ANSWER

GUIDED PRACTICE for Examples 1 and 2 Graph the system of linear inequalities. 2. y > – x + 2 y < 4 x < 3 ANSWER

GUIDED PRACTICE for Examples 1 and 2 Graph the system of linear inequalities. 3. y > – x y > x – 4 y < 5 ANSWER

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

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

EXAMPLE 4 Write and solve a system of linear inequalities BASEBALL The 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.

Write and solve a system of linear inequalities EXAMPLE 4 Write and solve a system of linear inequalities SOLUTION a. 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: x – y < 3 The difference of the bat’s length and weight can be at most 3. x ≥ 26 The length of the bat must be at least 26 inches. x ≤ 34 The length of the bat can be at most 34 inches. y ≥ 0 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 4 Write and solve a system of linear inequalities b. Graph the point that represents a bat that is 31 inches long and weighs 25 ounces. ANSWER Because the point falls outside the solution region, the bat cannot be used by a player on an NCAA team.

GUIDED PRACTICE for Examples 3 and 4 Write a system of inequalities that defines the shaded region. 4. ANSWER x ≤ 3, y > x 1 3 2

GUIDED PRACTICE for Examples 3 and 4 Write a system of inequalities that defines the shaded region. 5. ANSWER y ≤ 4, x < 2

GUIDED PRACTICE for Examples 3 and 4 6. WHAT IF? In Example 4, suppose a Senior League (ages 10–14) 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. x y ≤ 8, x ≥ 26, x ≤ 34, y ≥ 0 ANSWER

Review – Ch. 7 – PUT HW QUIZZES HERE

Daily Homework Quiz For use after Lesson 7.1 Use the graphing to solve the linear system 1. 3x – y = 5 – x + 3y = 5 ANSWER (2.5,2.5)

Daily Homework Quiz For use after Lesson 7.1 2. Solve the linear system by graphing. 2x + y = – 3 – 6x + 3y = 3 ANSWER (–1, –1)

Daily Homework Quiz For use after Lesson 7.1 A pet store sells angel fish for $6 each and clown loaches for $4 each . If the pet store sold 8 fish for $36, how many of each type of fish did it sell? 3. ANSWER 2 angel fish and 6 clown loaches

Daily Homework Quiz For use after Lesson 7.2 Solve the linear system using substitution 1. –5x – y = 12 3x – 5y = 4 ANSWER (–2, –2) 2. 2x + 9y = –4 x – 2y = 11 ANSWER (7, –2 )

Daily Homework Quiz For use after Lesson 7.2 3. You are making 6 quarts of fruit punch for a party. You want the punch to contain 80% fruit juice. You have bottles of 100% fruit juice and 20% fruit juice. How many quarts of 20% fruit juice should you mix to make 6 quarts of 80% fruit juice? ANSWER 4.5 quarts of 100% fruit juice and 1.5 quarts of 20% fruit juice

Daily Homework Quiz For use after Lesson 7.3 Solve the linear system using elimination. 1. –5x +y = 18 3x – y = –10 ANSWER (–4, –2) 2. 4x + 2y = 14 4x – 3y = –11 ANSWER (1, 5) 3. 2x – y = –14 y = 3x + 6 ANSWER (8, 30)

Daily Homework Quiz For use after Lesson 7.3 4. x + 4y = 15 2y = x – 9 ANSWER (11, 1) A business center charges a flat fee to send faxes plus a fee per page.You send one fax with 4 pages for $5.36 and another fax with 7 pages for $7.88. Find the flat fee and the cost per page to send a fax. HINT: Find two points and then utilize the slope-intercept form of a line. 5. ANSWER flat fee: $2, price per page: $.84

Daily Homework Quiz For use after Lesson 7.4 Solve the linear system using elimination. 1. 8x + 3y = 12 – 2x + y = 4 ANSWER (0,4) 2. – 3x + 2y = 7 5x – 4y = – 15 ANSWER (1,5) 3. – 7x – 3y = 11 4x – 2y = 16 ANSWER (1, – 6)

Daily Homework Quiz For use after Lesson 7.4 A recreation center charges nonmembers $3 to use the pool and $5 to use the basketball courts. A person pays $42 to use the recreation facilities 12 times. How many times did the person use the pool. 4. ANSWER 9 times (-4,-2)

Daily Homework Quiz For use after Lesson 7.5 Without solving the linear system, tell whether the linear system has one solution, no solution, or infinitely many solutions. 1. 4x + 2y = 12 y = –2x + 8 ANSWER no solution 2. – 2x + 5y = 5 y = x + 1 2 5 ANSWER infinitely many solutions

Daily Homework Quiz For use after Lesson 7.3 A group of 12 students and 3 teachers pays $57 for admission to a primate research center. Another group of 14 students and 4 teachers pays $69. Find the cost of one student ticket. 3. ANSWER $3.50

Daily Homework Quiz For use after Lesson 7.6 Write a system of inequalities for the shaded region. 1. ANSWER x < 2, y > x =1

Daily Homework Quiz For use after Lesson 7.6 A bibliography can refer to at most 8 articles, at most 4 books, and at most 8 references in all. Write and graph a system of inequalities that models the situation. 2. ANSWER x = articles, y = books; x 8, y 4, x + y 8, x 0, and y 0 > <

Warm-Up – X.X

Vocabulary – X.X Holder Holder 2 Holder 3 Holder 4

Notes – X.X – LESSON TITLE. Holder

Examples X.X