chapter 6: systems of equations and inequalities

Chapter 6: Systems of Equations and Inequalities

Algebra I

Table of Contents

• 6.1 - Solving Systems by Graphing• 6.2 - Solving Systems by Substitution• 6.3 - Solving Systems by Elimination• 6.4 - Solving Special Systems• 6.5 - Solving Linear Inequalities• 6.6 - Solving Systems of Linear Inequalities

6.1 - Solving Systems by Graphing

Algebra I

A system of linear equations is a set of two or more linear equations containing two or more variables.

A solution of a system of linear equations with two variables is an ordered pair that satisfies each equation in the system.

-So, if an ordered pair is a solution, it will make both equations true.

6.1 Algebra 1 (bell work)

Just Read

Tell whether the ordered pair is a solution of the given system.

(5, 2);

The ordered pair (5, 2) makes both equations true.

(5, 2) is the solution of the system.

3x – y = 13

2 – 2 0

0 0

03(5) – 2 13

15 – 2 13

13 13

3x – y 13

6.1 Example 1 Identifying Solutions of Systems


(1, 3); 2x + y = 5–2x + y = 1

2x + y = 5

2(1) + 3 5

2 + 3 5

5 5 The ordered pair (1, 3) makes both equations true.

–2x + y = 1

–2(1) + 3 1

–2 + 3 1

1 1

(1, 3) is the solution of the system.

6.1


(2, –1); x – 2y = 43x + y = 6

The ordered pair (2, –1) makes one equation true, but not the other.

(2, –1) is not a solution of the system.

3x + y = 6

3(2) + (–1) 6

6 – 1 6

5 6

x – 2y = 4

2 – 2(–1) 4

2 + 2 4

4 4

6.1 Optional

Math Joke

• Teacher: What is the name of the formula that describes the phases of the moon?

• Student: The lunear (linear) equation

6.1

All solutions of a linear equation are on its graph.

To find a solution of a system of linear equations, you need a point that each line has in common.

In other words, you need their point of intersection.

y = 2x – 1y = –x + 5

The point (2, 3) is where the two lines intersect and is a solution of both equations, so (2, 3) is the solution of the systems.

Just Read6.1

Solve the system by graphing. Check your answer.

y = xy = –2x – 3

CheckSubstitute (–1, –1) into the system.

y = x

(–1) (–1)

–1 –1 y = –2x – 3

(–1) –2(–1) –3

–1 2 – 3–1 – 1

6.1 Example 2 Solving a System of Linear Equations by Graphing

y = x – 6y + x = –1

y + x = –1

− x − x

y =

Solve the system by graphing. Check your answer.6.1

Solve the system by graphing. Check your answer.

y = –2x – 1 y = x + 5

Check Substitute (–2, 3) into the system.

y = x + 5

3 –2 + 5

3 3

y = –2x – 1

3 –2(–2) – 1

3 4 – 1

3 3

6.1 Day 2

2x + y = 4

2x + y = 4

–2x – 2x

y = –2x + 4

6.1 Solve the system by graphing. Check your answer.

Wren and Jenni are reading the same book. Wren is on page 14 and reads 2 pages every night. Jenni is on page 6 and reads 3 pages every night. After how many nights will they have read the same number of pages? How many pages will that be?

Totalpages is

number read

everynight plus

already read.

Wren y = 2 x + 14

Jenni y = 3 x + 6

6.1 Example 3 Problem Solving Application

(8, 30)

Nights

Graph y = 2x + 14 and y = 3x + 6.

The lines appear to intersect at (8, 30).

So, the number of pages read will be the same at 8 nights with a total of 30 pages.

6.1 Method 1 (Graph) Method 2 (Substitution)

6.1 OptionalVideo club A charges $10 for membership and $3 per movie rental. Video club B charges $15 for membership and $2 per movie rental. For how many movie rentals will the cost be the same at both video clubs? What is that cost?

Totalcost is price

for eachrental plus

member-ship fee.

Club A y = 3 x + 10

Club B y = 2 x + 15

6.1

Graph y = 3x + 10 and y = 2x + 15. The lines appear to intersect at (5, 25). So, the cost will be the same for 5 rentals and the total cost will be $25.

Method 1 (Graph) Method 2 (Substitution)

HW pg. 400

• 6.1-– Day 1: 1-7, 27, 33-42– Day 2: 12-15, 8, 16, 17, 25, 26, 28– For the story problems you can use method 1 or 2– Ch: 18, 23, 24, 31, 32

– HW Guidelines or ½ off– Start a new HW sheet for Ch 6

6.2 – Solving Systems by Substitution

Algebra I

1. Solve one equation for x= or y=2. Substitute that equation into the other

equation3. Use the new equation to solve for the value

of x or y4. Plug the value back in to find other variable

(Back Substition)


Steps for solving by Substitution

Solve the system by substitution.

y = 3x y = x – 2

Step 1 y = 3x

y = x – 2

Step 2 y = x – 23x = x – 2

Step 3 –x –x

2x = –22x = –22 2x = –1

Step 4 y = 3xy = 3(–1)y = –3

Step 5 (–1, –3)

6.2

Solution is always an ordered point or pair

Example 1 Solving a System of Linear Equations by Substitution


y = x + 14x + y = 6

Step 1 y = x + 1

Step 2 4x + y = 64x + (x + 1) = 6

Step 3 –1 –1

5x = 5 5 5x = 1

5x = 5

5x + 1 = 6

Step 4 y = x + 1

y = 1 + 1y = 2

Step 5 (1, 2)

6.2

Math Joke

• Q: Why did the chef throw the unsolved system of equations out of the restaurant?

• A: The menu said, “No substitution.”

6.2


x + 2y = –1x – y = 5

Step 1 x + 2y = –1

Step 2 x – y = 5(–2y – 1) – y = 5

–3y – 1 = 5

−2y −2yx = –2y – 1

Step 3 –3y – 1 = 5+1 +1

–3y = 6

–3y = 6–3 –3y = –2

Step 4 x – y = 5 x – (–2) = 5

x + 2 = 5 –2 –2

x = 3

Step 5 (3, –2)

6.2 Optional

y + 6x = 11 3x + 2y = –5

Step 1 y + 6x = 11– 6x – 6xy = –6x + 11

3x + 2(–6x + 11) = –5

3x + 2y = –5 Step 2

3x + 2(–6x + 11) = –5

Step 3 3x + 2(–6x) + 2(11) = –5

–9x + 22 = –5

–9x = –27 – 22 –22

–9x = –27–9 –9

x = 3

3x – 12x + 22 = –5

6.2Solve the system by substitution.

Step 4 y + 6x = 11 y + 6(3) = 11

y + 18 = 11 –18 –18

y = –7Step 5 (3, –7)

Example 2 Using the Distributive Property

–2x + y = 8 3x + 2y = 9

Step 1 –2x + y = 8

+ 2x +2x

y = 2x + 8

3x + 2(2x + 8) = 9

3x + 2y = 9 Step 2

3x + 2(2x + 8) = 9

Step 3 3x + 2(2x) + 2(8) = 9

7x + 16 = 9

7x = –7

–16 –16

7x = –77 7

x = –1

3x + 4x + 16 = 9

6.2 Solve the system by substitution.

Step 4 –2x + y = 8

–2(–1) + y = 8

y + 2 = 8–2 –2

y = 6Step 5 (–1, 6)

Day 2

Jenna is deciding between two cell-phone plans. The first plan has a $50 sign-up fee and costs $20 per month.The second plan has a $30 sign-up fee and costs $25 per month.

After how many months will the total costs be the same? What will the costs be? If Jenna has to sign a one-year contract, which plan will be cheaper? Explain.

Total paid is

sign-up fee plus

paymentamount

for eachmonth.

Option 1 t = $50 + $20 m

Option 2 t = $30 + $25 m

Step 1 t = 50 + 20mt = 30 + 25m

Step 2 50 + 20m = 30 + 25m

6.2 Example 3 Application

Step 3 50 + 20m = 30 + 25m–20m – 20m50 = 30 + 5m

–30 –30

20 = 5m

Step 4 t = 30 + 25m

t = 30 + 25(4)

t = 30 + 100

t = 130

5 5m = 4

20 = 5m

6.2Total paid is

sign-up fee plus

paymentamount

for eachmonth.

Option 1 t = $50 + $20 mOption 2 t = $30 + $25 m

Step 5 (4, 130)

In 4 months, the total cost for each option would be the same $130.

Jenna should choose the first plan because it costs $290 for the year and the second plan costs $330.

Option 1: t = 50 + 20(12) = 290

Option 2: t = 30 + 25(12) = 330

If Jenna has to sign a one-year contract, which plan will be cheaper? Explain.

After how many months will the total costs be the same? What will the costs be?

If Jenna has to sign a one-year contract, which plan will be cheaper? Explain.

6.2

One cable television provider has a $60 setup fee and $80 per month, and the second has a $160 equipment fee and $70 per month.a. In how many months will the cost be the

same? What will that cost be.

6.2

Total paid is fee plus

paymentamount

for eachmonth.

Option 1 t = $60 + $80 mOption 2 t = $160 + $70 m

Step 1 t = 60 + 80mt = 160 + 70m

Step 2 60 + 80m = 160 + 70m

Optional

Step 3 60 + 80m = 160 + 70m –70m –70m

60 + 10m = 160 –60 –60

10m = 10010 10

m = 10

Step 4 t = 160 + 70mt = 160 + 70(10)t = 160 + 700t = 860

6.2

Step 5 (10, 860)In 10 months, the total cost for each option would be the same, $860.

The first option is cheaper for the first six months.

Option 1: t = 60 + 80(6) = 540Option 2: t = 160 + 270(6) = 580

b. If you plan to move in 6 months, which is the cheaper option? Explain.

6.2

HW pg. 408

• 6.2- – Day 1: 1-5 (Odd), 9-15 (Odd), 45-47– Day 2: 7, 17, 19, 26, 32, 33– Ch: 24, 25, 27

– HW Guidelines or ½ off

6.3 - Solving Systems by Elimination

Algebra I

1. Line x and y up vertically

2. Eliminate one variable and solve for the other

3. Take solved x or y and use back substitution

4. Check your answer

6.3

Steps for solving by EliminationAlgebra 1 (bell work)

3x – 4y = 10x + 4y = –2Solve by elimination.

Step 1 3x – 4y = 10

4x = 8

x + 4y = –24x + 0 = 8Step 2

4x = 84 4

x = 2

Step 3 x + 4y = –2

2 + 4y = –2

–2 –2

4y = –44y –44 4

y = –1

Step 4 (2, –1)

6.3 Example 1 Elimination Using Addition

y + 3x = –22y – 3x = 14Solve by elimination.

Step 1 2y – 3x = 14

y + 3x = –2

Step 2 3y + 0 = 12

3y = 12

y = 4

Step 3 y + 3x = –2

4 + 3x = –2 –4 –4

3x = –6

3x = –63 3

x = –2

Step 4 (–2, 4)

6.3

2x + y = –52x – 5y = 13Solve by elimination.

Step 1–(2x – 5y = 13)

2x + y = –5

2x + y = –5–2x + 5y = –13

0 + 6y = –18 Step 2

6y = –18y = –3

Step 3 2x + y = –5

2x + (–3) = –5

2x – 3 = –5

+3 +3

2x = –2

x = –1

Step 4 (–1, –3)

6.3 Example 2 Elimination Using Subtraction

Math Joke

• Teacher: What’s your solution of the system of equations?

• Student: Nothing – I used elimination and got rid of the whole thing!

6.3

x + 2y = 11–3x + y = –5

Solve the system by elimination.

x + 2y = 11Step 1 –2(–3x + y = –5)

x + 2y = 11+(6x –2y = +10)

7x + 0 = 21

Step 2

7x = 21

x = 3

Step 3 x + 2y = 11

3 + 2y = 11–3 –3

2y = 8y = 4

Step 4 (3, 4)

6.3

–5x + 2y = 32 2x + 3y = 10


Step 1 2(–5x + 2y = 32) 5(2x + 3y = 10)

–10x + 4y = 64

+(10x + 15y = 50)

19y = 114y = 6

Step 2

Step 3 2x + 3y = 10

2x + 3(6) = 10

–18 –18

2x = –8

2x + 18 = 10

x = –4

Step 4 (–4, 6)

6.3 Day 2Example 3 Elimination Using Multiplication First


2x + 5y = 26 –3x – 4y = –25

Step 1 3(2x + 5y = 26) +(2)(–3x – 4y = –25)

6x + 15y = 78

+(–6x – 8y = –50)

y = 40 + 7y = 28Step 2

Step 3 2x + 5y = 26

2x + 5(4) = 26

Step 4 (3, 4) x = 3

2x + 20 = 26–20 –20

2X = 6

6.3

Sally spent $14.85 to buy 13 flowers. She bought lilies, which cost $1.25 each, and tulips, which cost $0.90 each. How many of each flower did Sally buy?

Write a system. Use l for the number of lilies and t for the number of tulips.

1.25l + 0.90t = 14.85l + t = 13


Step 4 (9, 4)

Sally bought 9 lilies and 4 tulips.Step 3

9 + t = 13–9 –9

t = 4

l + t = 13

Step 1 1.25l + .90t = 14.85

+ (–.90)(l + t) = 13

1.25l + 0.90t = 14.85 + (–0.90l – 0.90t = –11.70)

Step 2

0.35l = 3.15

l = 9

6.3

HW pg. 415

• 6.3-– Day 1: 1-6, 15, 25, 27, 41-47 (Odd)– Day 2: 7-10, 17, 19, 20– Ch: 21, 30, 31


6.4 – Solving Special Systems

Algebra I

1. Systems with at least one solution are called consistent.

2. A system that has no solution is an inconsistent system. (Parallel Lines)

3. An independent system has exactly one solution.

4. A dependent system has infinitely many solutions. (Same Line, which means same slope and y-intercept)


Write down the following definitions

Pg. 421

6.4 1. One solution, consistent, independent

2. All real, consistent, dependent.

3. No solution, inconsistent

Solve y = x – 4

Method 1 Compare slopes and y-intercepts.

y = x – 4 y = 1x – 4 Write both equations in slope-intercept form.

–x + y = 3 y = 1x + 3

–x + y = 3

The lines are parallel because they have the same slope and different y-intercepts.

This system has no solution so it is an inconsistent system.

6.4

Method 2 Solve the system algebraically.

–x + (x – 4) = 3

–4 = 3


Solve y = x – 4 –x + y = 3

6.4

Solve y = –2x + 5


2x + y = 1

y = –2x + 5 y = –2x + 5 2x + y = 1 y = –2x + 1

Write both equations in slope-intercept form.

The lines are parallel because they have the same slope and different y-intercepts.


6.4

Method 2 Solve the system algebraically.

2x + (–2x + 5) = 1


5 = 1

Solve y = –2x + 5 2x + y = 1

6.4

Solve y = 3x + 2 3x – y + 2= 0


y = 3x + 2 y = 3x + 2 Write both equations in slope-

intercept form. The lines have the same slope and the same y-intercept.

3x – y + 2= 0 y = 3x + 2

If this system were graphed, the graphs would be the same line. There are infinitely many solutions, consistent, dependent.

6.4

Solve y = 3x + 2 3x – y + 2= 0

Method 2 Solve the system algebraically. y = 3x + 2 y − 3x = 2

3x − y + 2= 0 −y + 3x = −20 = 0

There are infinitely many solutions, consistent, dependent.

6.4

Solve3y = x + 3 x + y = 1

Classify the system. Give the number of solutions.

The system is consistent and dependent. It has infinitely many solutions.

Day 26.4

Solvex + y = 5

4 + y = –x

The system is inconsistent. It has no solutions.

Classify the system. Give the number of solutions.6.4

Solvey = 4(x + 1)

y – 3 = x

The system is consistent and independent. It has one solution.

Classify the system. Give the number of solutions.6.4

Jared and David both started a savings account in January.

If the pattern of savings in the table continues, when will the amount in Jared’s account equal the amount in David’s account?


Total saved is start

amount plusamountsaved

for eachmonth.

Jared y = $25 + $5 xDavid y = $40 + $5 x

y = 5x + 25y = 5x + 40

y = 5x + 25y = 5x + 40

The graphs of the two equations are parallel lines, so there is no solution. If the patterns continue, the amount in Jared’s account will never be equal to the amount in David’s account.

6.4

Matt has $100 in a checking account and deposits $20 per month. Ben has $80 in a checking account and deposits $30 per month. Will the accounts ever have the same balance? Explain.

y = 20x + 100y = 30x + 80

y = 20x + 100y = 30x + 80

The accounts will have the same balance.

The graphs of the two equations have different slopes so they intersect.

6.4 Optional

HW pg. 423

• 6.4-– Day 1: 2-7, 15, 19, 39-43 (Odd)– Day 2: 8-11, 21, 23, 24, 29, 31– Ch: 28, 30


6.5 - Solving Linear Inequalities

Algebra I

A linear inequality is similar to a linear equation, but the equal sign is replaced with an inequality symbol.

A solution of a linear inequality is any ordered pair that makes the inequality true.


Copy Definitions

Tell whether the ordered pair is a solution of the inequality.

(–2, 4); y < 2x + 1

y < 2x + 14 2(–2) + 14 –4 + 14 –3

<

(–2, 4) is not a solution.

< <

6.5 Identifying Solutions of Inequalities

(3, 1); y > x – 4

y > x − 41 3 – 4

1 – 1>

(3, 1) is a solution.

>

A linear inequality describes a region of a coordinate plane called a half-plane.

All points in the region are solutions of the linear inequality.

The boundary line of the region is the graph of the related equation.

6.5 Don’t Copy Down

Don’t Copy Down6.5

Graphing Linear Inequalities

Step 1 Solve the inequality for y (slope-intercept form).

Step 2 Graph the boundary line. Use a solid line for ≤ or ≥. Use a dashed line for < or >.

Step 3Shade the half-plane above the line for y > or ≥. Shade the half-plane below the line for y < or y ≤. Check your answer.

6.5 Steps to Graphing Linear Inequalities

Graph the solutions of the linear inequality.

y 2x – 3

6.5 Example 2 Graphing Linear Inequalities in Two Variables

Check ( 0, 0 )

Math Joke

• Q: Why did the math student name her boundary line Hope?

• A: Because it was dashed.

6.5

5x + 2y > –8

6.5


Check ( 0, 0 )

4x – 3y > 12

6.5


Check ( 0, 0 )

2x – y – 4 > 0


Check ( 0, 0 )

6.5 Optional

4x – y + 2 ≤ 0


Check ( 0, 0 )

6.5 Day 2

Ada has at most 285 beads to make jewelry. A necklace requires 40 beads, and a bracelet requires 15 beads.

Write a linear inequality to describe the situation. Let x represent the number of necklaces and y the number of bracelets.

Write an inequality. Use ≤ for “at most.”

Necklacebeads

braceletbeadsplus

is atmost

285beads.

40x + 15y ≤ 285

Solve the inequality for y.

40x + 15y ≤ 285–40x –40x

15y ≤ –40x + 285

Example 4 Application6.5

b. Graph the solutions.

Step 1 Since Ada cannot make a negative amount of jewelry, the system is graphed only in Quadrant I.

Graph the boundary line

Use a solid line for ≤.

6.5

b. Graph the solutions.Step 2 Shade below the line. Ada can only make whole numbers of jewelry. All points on or below the line with whole number coordinates are the different combinations of bracelets and necklaces that Ada can make.

6.5

c. Give two combinations of necklaces and bracelets that Ada could make.

Two different combinations of jewelry that Ada could make with 285 beads could be 2 necklaces and 8 bracelets or 5 necklaces and 3 bracelets. (2, 8)

(5, 3)

6.5

Dirk is going to bring two types of olives to the Honor Society induction and can spend no more than $6. Green olives cost $2 per pound and black olives cost $2.50 per pound.

a. Write a linear inequality to describe the situation.


c. Give two combinations of olives that Dirk could buy.


Greenolives

black olivesplus

is no more than

totalcost.

2x + 2.50y ≤ 6

Write an inequality. Use ≤ for “no more than.”

Solve the inequality for y.

2.50y ≤ –2x + 62.50 2.50

2x + 2.50y ≤ 6

2.50y ≤ –2x + 6–2x –2x

6.5


Step 1 Since Dirk cannot buy negative amounts of olive, the system is graphed only in Quadrant I. Graph the boundary line for y = –0.80x + 2.4. Use a solid line for≤.

y ≤ –0.80x + 2.4

Green Olives

Blac

k O

lives

6.5

c. Give two combinations of olives that Dirk could buy.

Two different combinations of olives that Dirk could purchase with $6 could be 1 pound of green olives and 1 pound of black olives or 0.5 pound of green olives and 2 pounds of black olives.

(1, 1)

(0.5, 2)Blac

k O

lives

Green Olives

6.5

Write an inequality to represent the graph.

y-intercept: 0 slope: –1

Write an equation in slope-intercept form.

y = mx + b y= – 1x

The graph is shaded below a dashed boundary line.

Replace = with < to write the inequality y < –x.

6.5 Example 4 Writing an Inequality from a Graph

Write an inequality to represent the graph.

y-intercept: 1; slope:

Write an equation in slope-intercept form.

The graph is shaded above a dashed boundary line.

Replace = with > to write the inequality

6.5

HW pg. 432

• 6.5-– Day 1: 2-8, 15-18, 51-65 (Odd)– Day 2: 9-11, 19-21, 37 (Equation Only), 38– Ch: 22, 27, 28, 40-42

6.6 - Solving Systems of Linear Inequalities

Algebra I

A system of linear inequalities is a set of two or more linear inequalities containing two or more variables.

The solutions of a system of linear inequalities consists of all the ordered pairs that satisfy all the linear inequalities in the system.



(–1, –3); y ≤ –3x + 1 y < 2x + 2

y ≤ –3x + 1–3 –3(–1) + 1–3 3 + 1–3 4≤

(–1, –3) (–1, –3)

–3 –2 + 2–3 0< –3 2(–1) + 2 y < 2x + 2

(–1, –3) is a solution to the system because it satisfies both inequalities.

6.6 Example 1 Identifying Solutions of Systems of Linear Inequalities


(–1, 5); y < –2x – 1 y ≥ x + 3

y < –2x – 1 5 –2(–1) – 1

5 2 – 15 1<

(–1, 5) (–1, 5)

5 2≥ 5 –1 + 3 y ≥ x + 3

(–1, 5) is not a solution to the system because it does not satisfy both inequalities.

Optional6.6

6.5

y ≤ 3 y > –x + 5

Example 2 Solving a System of Linear Inequalities by Graphing

Math Joke

• Q: Why did the math student wear two pairs of sunglasses?

• A: He wanted to have overlapping shades

6.5

y > x – 7 3x + 6y ≤ 12

6.5

–3x + 2y ≥ 2 y < 4x + 3

6.5Graph and Describe theSolutions

y ≤ –2x – 4 y > –2x + 5

Example 3 Graphing Systems with Parallel Boundary Lines Day 2

6.5Graph and Describe theSolutions

y > 3x – 2 y < 3x + 6

In one week, Ed can mow at most 9 times and rake at most 7 times. He charges $20 for mowing and $10 for raking. He needs to make more than $125 in one week.

Show and describe all the possible combinations of mowing and raking that Ed can do to meet his goal. List two possible combinations.

Earnings per Job ($)

Mowing

Raking

20

10

Step 1 Write a system of inequalities.Let x represent the number of mowing jobs and y represent the number of raking jobs.


Step 1 Write a system of inequalities.

Let x represent the number of mowing jobs and y represent the number of raking jobs.

x ≤ 9

y ≤ 7

20x + 10y > 125

He can do at most 9 mowing jobs.

He can do at most 7 raking jobs.

He wants to earn more than $125.

In one week, Ed can mow at most 9 times and rake at most 7 times. He charges $20 for mowing and $10 for raking. He needs to make more than $125 in one week. Show and describe all the possible combinations of mowing and raking that Ed can do to meet his goal. List two possible combinations.

6.6

Step 2 Graph the system.

The graph should be in only the first quadrant because the number of jobs cannot be negative.

Solutions

6.6

Step 3 Describe all possible combinations.

All possible combinations represented by ordered pairs of whole numbers in the solution region will meet Ed’s requirement of mowing, raking, and earning more than $125 in one week. Answers must be whole numbers because he cannot work a portion of a job.

Step 4 List the two possible combinations.

Two possible combinations are: 7 mowing and 4 raking jobs 8 mowing and 1 raking jobs

Solutions

6.6

At her party, Alice is serving pepper jack cheese and cheddar cheese. She wants to have at least 2 pounds of each. Alice wants to spend at most $20 on cheese. Show and describe all possible combinations of the two cheeses Alice could buy. List two possible combinations.

Price per Pound ($)

Pepper Jack

Cheddar

4

2

Step 1 Write a system of inequalities.

Let x represent the pounds of pepper jack and y represent the pounds of cheddar.

x ≥ 2

y ≥ 2

4x + 2y ≤ 20

She wants at least 2 pounds of pepper jack.

She wants to spend no more than $20.

She wants at least 2 pounds of cheddar.

6.6

Step 2 Graph the system.

The graph should be in only the first quadrant because the amount of cheese cannot be negative.

Solutions

6.6

Step 3 Describe all possible combinations. All possible combinations within the gray region will

meet Alice’s requirement of at most $20 for cheese and no less than 2 pounds of either type of cheese. Answers need not be whole numbers as she can buy fractions of a pound of cheese.

Step 4 Two possible combinations are (3, 2) and (2.5, 4).

3 pepper jack, 2 cheddar or 2.5 pepper jack, 4 cheddar.

6.6

HW pg. 438

• 6.6-– Day 1: 2-8, 19, 21, 55, 57– Day 2: 9-15, 25, 29-30 (Equations Only)– Ch: 38-42

chapter 6: systems of equations and inequalities

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