Unit 3 – Chapter 7. Unit 3 Section 7.1 – Solve Linear Equations by graphing Section 7.2 – Solve Linear Equations by substitution.Section 7.2 – Solve Linear

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.

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

ANSWER2 lawns and 5 hedges, or4 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:1. Read and highlight key words2. DEFINE THE VARIABLES (MOST

CRITICAL STEP!!)3. Write Mathlish sentence left to right (careful

with subtraction and division!)•3 Step Process To Solve Linear Eqns by graphing 1.Write equations in slope-intercept form2.Graph them on calculator3.Find intersection of lines•CHECK ANSWERS!

Examples 7.1

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.

EXAMPLE 2

STEP 2

Use the graph-and-check method

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– x + y = – 7

– 7 = – 7–(4) + (– 3) – 7=

?

Equation 2x + 4y = – 8

– 8 = – 84 + 4(– 3) – 8=

?

ANSWER

Because (4, – 3) is a solution of each equation, it is a solution of the linear system.



Solve the linear system by graphing. Check your solution.

y = 5x

5x + y = -5x + 10

GUIDED PRACTICE for Examples 1 and 2

– 5x + y = 01.5x + y = 10

The Intersection point is at (1,5).

Put eqns in slope-Intercept form

EXAMPLE 3 Standardized Test Practice

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.

The parks and recreation department in your town offers a season pass for $90.

• Write a system of equations to model these situations and find out where they are equal.

SOLUTION


Write a system of equations where y is the total cost (in dollars) for x sessions.

EQUATION 1

y = 13 x


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


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

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.

RENTAL BUSINESS

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

15x + 30y = 450

Equation for number of rentals

Equation for money collected from rentals

STEP 2Graph both equations.


STEP 3

Estimate 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 = 45025 = 25

ANSWER

The business rented 20 pairs of skates and 5 bicycles.

EXAMPLE 4 Solve a multi-step problemGUIDED PRACTICE for Example 4

SOLUTION

In Example 4, suppose the business has a total of 20 rentals and collects $ 420. Find the number of bicycles rented.

6.

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

15x + 30y = 420

Equation for number of rentals.

Equation for money collected from rentals.


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.

15(12) + 30(8) 420=420 = 420

12 + 8 = 2020 = 20

15x + 30y 420=x + y 20=

EXAMPLE 4

ANSWER

The business rented number of bicycle is 8.


Warm-Up – 7.2


Solve the equation.

1. 6a – 3 + 2a = 13

2. 4(n + 2) – n = 11

ANSWER a = 2

ANSWER n = 1

3. You burned 8 calories per minute on a treadmilland 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?


Solve the equation.

HINTDefine the variablesCreate two equations (one for total minutesAnd one for total calories)

ANSWEREQN 1 – 8x + 10y = 560EQN 2 – x + y = 60treadmill: 20 min,elliptical trainer: 40 min


Solve the system of equations BUT YOU CAN’T USE THE CALCULATOR OR GRAPH IT!!! WORK WITH YOUR GROUP!!.

1. 2x + 3y = 402. y = x + 5

HINT! Do the Dance!!!!

ANSWER X = 5Y = 10

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

1. Get ONE of the variables in ONE of the equations by itself.

2. Substitute that variable into the OTHER equation.

3. Solve the equation from #2.4. Plug the answer back into the original equation.

Examples 7.2

EXAMPLE 2 Use the substitution method

Solve the linear system:x – 2y = – 6 Equation 1

4x + 6y = 4 Equation 2

SOLUTION

Solve Equation 1 for x.

x – 2y = – 6 Write original Equation 1.

x =2y – 6 Revised Equation 1

STEP 1


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.

Distributive property8y – 24 + 6y = 4

14y – 24 = 4 Simplify.

14y = 28 Add 24 to each side.

y = 2 Divide each side by 14.

STEP 2


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).

STEP 3

4( –2 )+ 6 (2 ) = 4 ?

GUIDED PRACTICE

CHECK

–2 –2 (2)= – 6?

– 6 = – 6

Substitute –2 for x and 2 for y in each of the original equations.

4x + 6y = 4

4 = 4

Equation 1 Equation 2

x – 2y = – 6



Solve the linear system using the substitution method.

Equation 2

Equation 1

3x + y = 10

Solve for y. Equation 1 is already solved for y.

SOLUTION

STEP 1

y = 2x + 51.



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

STEP 2


EXAMPLE 2

Substitute 1 for x in the revised Equation 1 to find the value of y.

ANSWER The solution is ( 1, 7).

y = 2x + 5 = 2(1) + 5 = 7

STEP 3



CHECK

y = 2x + 5

7 = 2(1) + 5?

7 = 7


Substitute 1 for x and 7 for y in each of the original equations.

3x + y = 10

3 ( 1 ) + 7 = 11?

10 = 10




Equation 1

x + 2y = – 6 Equation 2

SOLUTION

Solve Equation 1 for x.

x – y = 3 Write original Equation 1.

x = y + 3 Revised Equation 1

STEP 1


x – y = 32.


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.

y = – 3 Divide each side by 3.

STEP 2

for Examples 1 and 2GUIDED PRACTICE

3y = – 9 Add 3 to each side.


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 ).

STEP 3

GUIDED PRACTICE for Examples 1and 2

0 + 2 (–3 ) = – 6?

GUIDED PRACTICE

CHECK

0 – (–3) = 3?

3 = 3


Substitute 0 for x and – 3 for y in each of the original equations.

x + 2y = – 6

– 6 = – 6


x – y = 3



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.

WEBSITES

Solve a multi-step problem

EXAMPLE 3

SOLUTION

Write a system of equations. Let y be the totalcost after x months.

Equation 1: Internet service provider

y = 10 + 21.95 x

STEP 1


EXAMPLE 3

Equation 2: Website hosting company

y = 22.45 x

The system of equations is:

y = 22.45x

Equation 1y = 10 + 21.95x

Equation 2


EXAMPLE 3

Substitute 22.45x for y in Equation 1 and solvefor x.

y = 10 + 21.95x

22.45x = 10 + 21.95x

0.5x = 10

x = 20

The total cost will be the same for both companies after 20 months.

ANSWER

STEP 2

Write Equation 1.

Substitute 22.45x for y.

Subtract 21.95x from each side.

Divide each side by 0.5.


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


The total cost for website hosting for each company after 20 months is $ 449.

ANSWER

SOLUTION

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.

STEP 1

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?

ANTIFREEZE

Solve a mixture problemEXAMPLE 4

Equation 1: Total number of quarts

x + 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 = 7.7



x + 0.5y = 7.7

Solve Equation 1 for x.x + y = 11

x = 11 – y

Substitute 11 – y for x in Equation 2 and solvefor y.

x + 0.5y = 7.7

STEP 2

STEP 3

Equation 1x + y =11

Equation 2

Write Equation 1.

Revised Equation 1

Write Equation 2.


Solve a mixture problem

EXAMPLE 4

(11 – y) = 0.5y = 7.7

y = 6.6

Substitute 6.6 for y in the revised Equation 1 tofind the value of x.

STEP 4

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.

Substitute 11 – y for x.

Solve for y.


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

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.

STEP 1


Equation 1: Total number of quarts

x + y = 16

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(16)x + 0.5y = 11.2



x + 0.5y = 11.2

Solve Equation 1 for x.x + y = 16

x = 16 – y

STEP 2

Equation 1x + y =16

Equation 2.

Write Equation 1.

Revised Equation 1


(16 – y) + 0.5y = 7.7

y = 9.6

Substitute 16 – y for x.

Solve for y.

Substitute 16 – y for x in Equation 1 and solvefor x.

x + 0.5y = 7.7

STEP 3

Write Equation 2.


Substitute 9.6 for y in the revised Equation 1 tofind the value of x.

STEP 4

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


Solve the linear systems by GRAPHING!!!

1. x + y = -22. -x + y = 6

2. x – y = 03. 5x + 2y = -7

ANSWER x = -4y = 2

ANSWER x = -1y = -1


Solve the linear systems by SUBSTITUTION!!!!!!

1. y = x – 42. -2x + y = 18

2. 5x – 4y = 273. -2x + y = 3

ANSWER x = -22y = -26

ANSWER x= -13y = -23


1. Solve the linear system using substitution.2x + y = 123x – 2y = 11

2. One auto repair shop charges $30 for a diagnosisand $25 per hour for labor. Another auto repair shopcharges $35 per hour for labor. For how many hoursare the total charges for both of the shops the same?

HINT: FIND EQUATIONS FOR TOTAL AND SUBSTITUTE!

ANSWER (5, 2)

ANSWER 3 h


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!

1. Remember the goal: You are trying to eliminate one variable!

2. Line up like terms under each other.3. NEVER subtract. Add the negative instead!

Examples 7.3

Use addition to eliminate a variable EXAMPLE 1


2x + 3y = 11 – 2x + 5y = 13

Equation 1

Equation 2

SOLUTION

Add the equations toeliminate one variable.

2x + 3y = 11– 2x +5y = 13

Solve for y. 8y = 24y = 3

Substitute 3 for y in either equation and Solve for x.

STEP 1

STEP 2

STEP 3

Use addition to eliminate a variable EXAMPLE 1

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

Solve the linear system:4x + 3y = 2 Equation 1

5x + 3y = – 2 Equation 2

SOLUTION

Subtract the equations toeliminate one variable.

4x + 3y = 25x + 3y = – 2

Solve for x. – x = 4

Substitute 4 for x in either equation and solvefor y.

STEP 1

STEP 2

STEP 3

x = 4

Use subtraction to eliminate a variable EXAMPLE 2


4(– 4) + 3y = 2 Substitute – 4 for x.

y = 2 Solve for y.

ANSWERThe solution is (– 4, 6).

Arrange like terms EXAMPLE 3

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 = –44y = 3x + 14

8x – 4y = –43x + 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.

Arrange like terms EXAMPLE 3

4y = 3x + 14 Write Equation 2.

4y = 3(2) + 14 Substitute 2 for x.

y = 5 Solve for y.

ANSWERThe solution is (2, 5).

GUIDED PRACTICE for Example 1,2 and 3


4x – 3y = 5 – 2x + 3y = – 7

Equation 1

Equation 2

SOLUTION

Add the equations toeliminate one variable.

4x – 3y = 5– 2x +3y = – 7

Solve for x. 2x = – 2x = – 1

Substitute – 1 for y in either equation and Solve for x.

STEP 1

STEP 2

STEP 3

`

1.


4x – 3y = 5 Write Equation 1.

2(– 1) – 3y = 5 Substitute – 1 for x.

y = – 3 Solve for x.

ANSWER

The solution is (– 1, – 3).


4x – 3y = 5

4(– 1) – 3(– 3) = 5?

CHECK Substitute 1 for x and 3 for y in each of the. original equation

– 2x + 3y = – 7

– 2(– 1) + 5(– 3) – 7?=

5 = 5– 7 = – 7


Solve the linear system:Equation 1

Equation 2

SOLUTION

Subtract the equations toeliminate one variable.Solve for y.

Substitute 1 for y in either and solve for x.

STEP 1

STEP 2

STEP 3

7x – 2y = 57x – 3y = 4

4.

7x – 2y = 57x – 3y = 4

y = 1


Write Equation 1.

Substitute 1 for y.

x = 1 Solve for x.


7x – 2y = 57x – 2(1) = 5



Equation 2

SOLUTION

STEP 1 Rewrite Equation 2 so that the like terms are arranged in columns.

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.

3x + 4y = – 65. = 3x + 62y

3x + 4y = – 6= 3x + 62y 3x – 2y = – 6

3x + 4y = – 6


Write Equation 2.

Substitute 0 for y.

x = – 2 Solve for x .

ANSWERThe solution is (– 2, 0).

= 3x + 62y= 3x + 62(0)



Equation 2

SOLUTION

STEP 1 Rewrite Equation 2 so that the like terms are arranged in columns.

STEP 2 Subtract the equations.

STEP 3 Solve for x. x = 1

STEP 4 Substitute 1 for x in either equation and solve for y.

2x + 5y = 126. = 4x + 65y

2x + 5y = 12– 4x + 5y = 6

2x + 5y = 12= 4x + 65y

6x = 6


Write Equation 2.

Substitute 1 for x.

x = 2 Solve for y .


+ 5y = 62(1)2x + 5y = 12

KAYAKING

EXAMPLE 4 Write and solve a linear system

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.


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

12 = r 3

4 = r

Downstream: d = rt

12 = r 2

6 = r


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.

x y 4=–

Equation 1: Going upstream


Equation 2: Going downstream

x y 6=+


STEP 2

Solve the system of equations.

x – y= 4

2x = 10

x = 5

Write Equation 1.

Write Equation 2.

Add equations.

Solve for x.

Substitute 5 for x in Equation 2 and solve for y.

x + y = 6


5 + y = 6

y = 1

Substitute 5 for x in Equation 2.

Subtract 5 from each side.


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.

STEP 1

Write a system of equations. First find the speed of the kayak going downstream.

SOLUTION


Upstream: d = rt

10 = r 5

2 = r

Downstream: d = rt

10 = r 2

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.


x y 2=–

Equation 2: Going downstream

x y 5=+

Going upstreamEquation 1:


STEP 2

Solve the system of equations.

x – y= 2

2x = 7

x = 3.5

Equation 1.

Equation 2.

Add equations.

Solve for x.

Substitute 3.5 for x in Equation 2 and solve for y.

x + y = 5


3.5 + y = 6

y = 1.5

Substitute 3.5 for x in Equation 2.

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


Solve the linear system.

1. 4x – 3y = 152x – 3y = 9

2. –2x + y = – 82x – 2y = 8

ANSWER (3, – 1)

ANSWER (4, 0)

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?


Solve the linear system.

ANSWER 4.5 mi/h

Multiply the second equation by -2 and rewrite it. Then use it to solve the system.

1. 8x – 6y = 302. 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:

1. Graphing• Easiest when I can get y by itself and have a calculator!

2. Substitution• Easiest when I can get one variable by itself.

3. 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

SOLUTION

EXAMPLE 1 Multiply one equation, then add


6x +5y = 19 Equation 1

2x +3y = 5 Equation 2

STEP 1 Multiply: Equation 2 by –3 so that the coefficients of x are opposites.

6x + 5y = 19

2x + 3y = 5

6x + 5y = 19

STEP 2 Add: the equations. –4y = 4

–6x – 9y = –15

EXAMPLE 1 Multiply one equation, then add

STEP 3

STEP 4

Solve: for y.

Substitute: –1 for y in either of the original equations and solve for x.

y = –1

ANSWER

The solution is (4, –1).

EXAMPLE 2 Multiply both equations, then subtract



2y = 3x – 9 Equation 2

SOLUTION

STEP 1


–3x + 2y = –9 Rewrite Equation 2.

Arrange: the equations so that like terms are in columns.

EXAMPLE 2 Multiply both equations, then subtract

STEP 2

4x + 5y = 35

–3x + 2y = –9

23x = 115STEP 3

STEP 4

8x + 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.

ANSWER The solution is (5, 3).


Solve the linear system using elimination:

Equation 1

–2x + 3y = –5 Equation 2

SOLUTION

6x – 2y = 11.

y = –2

ANSWER The solution is (–0.5, –2).


Equation 1

3x + 10y = –3 Equation 2

SOLUTION

2x + 5y = 32.

ANSWER The solution is (9, –3).


Equation 1

9y = 5x +5 Equation 2

SOLUTION

3x – 7y = 53.

ANSWER The solution is (–10, –5).

Standardized Test PracticeEXAMPLE 3

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?


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


Equation 2: Cost of fabric

The system of equations is:x + y = 164x +6y = 76

Equation 1

Equation 2

ANSWER

The correct answer is x = 10 and y = 6.

4 766+ =yx


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.


Equation 1:

10x + 2y = 155

Cost of soccer ball Total Cost

Cost of soccer bag+ =


Equation 2:

12x + 3y = 189

Cost of soccer ball

Total CostCost of soccer bag+ =


10x +2y = 155

12x +3y = 189

Equation 1

Equation 2


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 .

– 30x – 6y = –465

STEP 3 Add the equation

24x + 6y = 378

6x = 87

STEP 4 Solve for x x = 14.50

STEP 2 10x +2y = 155

12x +3y = 189


10x + 2y = 155

10(14.50) + 2y = 155

Write Equation 1.

Substitute 14.50 for x.

Substitute 14.50 for x in either of the original equations and solve for y.

y = 5

STEP 5

solve for y

ANSWER

Cost of soccer ball is $ 14.50 and soccer ball bag is $5.

Warm-Up – 7.5


1. Solve the linear system.2x + 3y = – 9 x – 2y = 6

2. You buy 8 pencils for $8 at the bookstore. Standardpencils cost $.85 and specialty pencils cost $1.25.How many specialty pencils did you buy?

ANSWER 3 specialty pencils

ANSWER (0, – 3)


1. Solve the linear system by GRAPHING. Describe the lines on your whiteboard.

x + y = -2 y = -x+5

ANSWER No solution.

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

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:

1. Graph them (put them in slope-intercept form)1. Intersect = 1 solution2. Parallel = No solution3. Same line = Infinite solutions

2. Check if equations are multiples of each other.1. Yes = infinite solutions2. No = Check some more!

3. Eliminate the variables (using Add. or Mult.)1. Always False statement = No solutions2. Always True statement = Infinite solutions

Notes – 7.5–Special Types of Systems

Examples 7.4

SOLUTION

EXAMPLE 1 A linear system with no solution

Solve the linear system by graphing and by elimination!



Graph the linear system.

METHOD 1 Graphing

Answer: NO SOLUTION.

EXAMPLE 1 A linear system with no solution

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.

Subtract the equation.

METHOD 2 Elimination

3x + 2y = 10

3x + 2y = 2

0 = 8 This is a false statement.

EXAMPLE 2 A linear system with infinitely many solutions

Solve the system by graphing and substitution.

x – 2y = – 4 Equation 1

Equation 2y = x + 212

SOLUTION

GraphingMETHOD 1

Graph the linear system.

EXAMPLE 2 A linear system with infinitely many solutions

Substitute x + 2 for y in Equation 1 and solve for x.12

x – 2y = – 4 Write Equation 1

2Substitute x + 2 for y.1x – 2 x + 2 =1

2– 4

METHOD 2 Substitution

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

– 4 = – 4 Simplify.


1. 5x + 3y = 6


5x + 3y = 6

– 5x – 3y = 3

– 5x – 3y = 30 = 9 This is a false statement.

Subtract the equations.

Equation 1

Equation 2

Tell whether the linear system has no solution or infinitely many solutions. Explain.


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.

– 6x + 3y = – 12 Write Equation 2

– 6x + 3(2x – 4) = – 12 Substitute (2x – 4) for y.

– 12 = – 12 Simplify.

2. y = 2x – 4

– 6x + 3y = – 12

Equation 1

Equation 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


ANSWER

1. Graph y < x – 1.23

1. Solve -3x >= 12

ANSWER: x <= -4


2. You are running one ad that costs $6 per day andanother that costs $8 per day. You can spend nomore than $120. Graph this inequality. HINT:

WRITE THE EQUATION FIRST!

2. 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.eqns1. Play the pretend game and let equation be =.2. Dotted line is < or >3. Solid line is <= or >=4. Pick a point NOT ON THE LINE, check the

answer, and shade the correct side of the line.•TO GRAPH SYSTEMS OF INEQUALITIES

1.Graph each inequality2.Find the AREA where solutions intersect.3.Pick a point and check your solution.

Examples 7.6

SOLUTION

EXAMPLE 1 Graph a system of two linear inequalities

Graph the system of inequalities. y > – x – 2 y < 3x + 6

Inequality 1

Inequality 2

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.

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 > – 2 1 > 6 1 > 0 + 6 ?

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 3SOLUTION

Graph all three inequalities in the same coordinate plane. The graph of the system is the triangular region shown.


ANSWER

Graph the system of linear inequalities.

1. y < x – 4 y > – x + 3

GUIDED PRACTICE

ANSWER


2. y > – x + 2 y < 4x < 3

for Examples 1 and 2

GUIDED PRACTICE

ANSWER


3. y > – x y > x – 4 y < 5


SOLUTION

EXAMPLE 3 Write a system of linear inequalities

Write a system of inequalities for the shaded region.

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.

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.


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:

The difference of the bat’s length and weight can be at most 3.

x – y < 3

x ≥ 26 The length of the bat must be at least 26 inches.

x ≤ 34

y ≥ 0

The 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.


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

Write a system of inequalities that defines the shaded region.


4.

ANSWER x ≤ 3, y > x 132

GUIDED PRACTICE

Write a system of inequalities that defines the shaded region.


5.

ANSWER y ≤ 4, x < 2

GUIDED PRACTICE

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.


ANSWER

x y ≤ 8, x ≥ 26, x ≤ 34, y ≥ 0

Review – Ch. 7 – PUT HW QUIZZES HERE

Daily Homework Quiz For use after Lesson 7.1

Use the graphing to solve the linear system1.

3x – y = 5

– x + 3y = 5

ANSWER (2.5,2.5)


2. Solve the linear system by graphing.

2x + y = – 3 – 6x + 3y = 3

ANSWER (–1, –1)


ANSWER

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.

2 angel fish and 6 clown loaches


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 )


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


Solve the linear system using elimination.

1. –5x +y = 183x – y = –10

ANSWER (–4, –2)

2. 4x + 2y = 144x – 3y = –11

ANSWER (1, 5)

3. 2x – y = –14y = 3x + 6

ANSWER (8, 30)


4. x + 4y = 152y = 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


Solve the linear system using elimination.

1. 8x + 3y = 12 – 2x + y = 4

ANSWER (0,4)

2. – 3x + 2y = 75x – 4y = – 15

ANSWER (1,5)

3. – 7x – 3y = 11 4x – 2y = 16

ANSWER (1, – 6)


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


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

ANSWER infinitely many solutions

2. – 2x + 5y = 5

y = x + 125


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


Write a system of inequalities for the shaded region.

1.

ANSWER x < 2, y > x =1


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>

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Unit 3 – Chapter 7. Unit 3 Section 7.1 – Solve Linear Equations by graphing Section 7.2 – Solve Linear Equations by substitution.Section 7.2 – Solve Linear