SECTION Ready To Go On? Skills Intervention 6A 6-1 … 6 Ready to... · Ready To Go On? Skills...

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Find these vocabulary words in Lesson 6-1 and the Multi-Language Visual Glossary.

Vocabulary

system of linear equations solution of a system of linear equations

Identifying Solutions of SystemsTell whether the ordered pair (1, 2) is a solution of the system, { y ! "2x # 4 y ! x # 1 .

To be a solution to the system of equations, the ordered pair (1, 2) has to

make both equations . Let x ! and let y ! .

y ! "2x # 4

! "2( ) # 4

! # 4

!

The point (1, 2) makes this equation .

y ! x # 1

! # 1

!

The point (1, 2) makes this equation .

Since the point (1, 2) makes both equations , it is a

to the given system.

Solving a System of Linear Equations by Graphing

Solve the system of equations { y ! " 1 __ 2 x # 2 2x # y ! "1 by graphing.

Consider y ! " 1 __ 2 x # 2. This equation is

written in form. The

y-intercept is (0, ) and the slope ! .

Start at point (0, ). Plot this point on the

coordinate system. Move down unit

and right units and plot another point. Connect these points.

Consider 2x # y ! "1. This equation is

written in form.

Find the x-intercept: Find the y-intercept:

Plug 0 in for y. Plug 0 in for x.

2x # 0 ! "1 2(0) # y ! "1

x ! y ! "1

( , 0) (0, )

Plot these points and connect the points.

The point where the two lines intersect is at ( , ).

Ready To Go On? Skills Intervention6-1 Solving Systems by Graphing6A

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Copyright © by Holt McDougal. 74 Holt McDougal Algebra 1All rights reserved.

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The solution to a system of linear equations is the point where the graphs of the two lines intersect.

The local county fair charges $7.00 admission and $0.50 per ticket to ride the rides. The state fair charges $5.00 admission and $1.50 per ticket to ride the rides. For how many tickets will the total cost be the same at both fairs? What is that cost?

Understand the Problem

1. How much is the admission for the local county fair?

How much does each ride ticket cost at the local county fair?

2. How much is the admission for the state fair?

How much does each ride ticket cost at the state fair?

Make a PlanLet x represent the number of tickets purchased and y represent the total cost.

Total cost is equal to the price per ticket plus admission

3. Equation for local county fair: ! x " 7

4. Equation for state fair: ! x " 5

Solve

5. Graph the line ! x " 7. The slope ! or 1 __ 2 .

The y-intercept of this line is (0, ).

6. Graph the line ! x " 5. The slope ! or 3 __ 2 .

The y-intercept of this line is (0, ).

7. What is the ordered pair where the two lines intersect?

8. If a person buys tickets, the cost at both fairs is the same: .

Look Back

9. To check, substitute the intersection point, , into both equations.

y ! 0.5x " 7 y ! 1.5x " 5 Does the point make both equations

true? ! 0.5( ) " 7 ! 1.5( ) " 5

! !

6AReady to Go On? Problem Solving Intervention6-1 Solving Systems by Graphing

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Solving a System of Linear Equations by SubstitutionSolve each system by substitution.

A. { y ! "x # 3

y ! 2 __ 3 x " 2

STEP 1 Substitute "x # 3 for in the second equation and solve for x.

STEP 2 y ! 2 __ 3 x " 2 STEP 3 Solve for the other variable, y.

! 2 __ 3 x " 2 You know from Step 2 that x ! .

y ! "x # 3

y ! " ! " # 3

y !

The solution to the system of equations

is ( , 0).

3 ! " ! 3 ! 2 __ 3 x " 2 " "3x # ! " 6

! 5x " 6

! 5x

! x

B. { x # y ! 4 x # 2y ! "2

STEP 1 Solve the equation x # y ! 4 for x.

x # y ! 4

x ! 4 "

STEP 2 Substitute 4 " y for in the second equation.

x # 2y ! "2

# 2y ! "2

STEP 3 Solve for . STEP 4 Solve for the other variable, x.

x # y ! 4

x # ! 4

x !

The solution to the system is (10, ).

4 " y # 2y ! "2

4 # ! "2

y !

6AReady To Go On? Skills Intervention6-2 Solving Systems by Substitution

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Elimination using Addition

Solve { x ! 4y " 16 5x # 4y " 8 by elimination.

STEP 1 Combine the two equations by addition.

x ! 4y " 16 5x # 4y " 8

____

! " 24

Which variable is eliminated?

STEP 2 Solve for . STEP 3 Substitute and solve for .

6x " 24 x ! 4y " 16

! 4y " 16

4y "

y "

6x _____ " 24 _____

x "

STEP 4 The solution to the system is (4, ).

Elimination using Multiplication First

Solve { 3x # y " 4 x ! 4y " 10 by elimination.

STEP 1 To eliminate y, multiply the 4(3x # y ) " 4(4) x # y "

x ! 4y " 10 x ! 4y " 10 first equation by .

STEP 2 Combine the two equations using . x # y "

x ! 4y " 10

____

x ! " 26

STEP 3 Solve for x. STEP 4 Substitute and solve for y.

13x " 26 x ! 4y " 10

! 4y " 10

4y "

y "

x ______ " 26 _____

x "

STEP 5 The solution to the system of equations is (2, ).

6AReady To Go On? Skills Intervention6-3 Solving Systems by Elimination

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Ready to Go On? Problem Solving Intervention6-3 Solving Systems by Elimination

You can multiply one or both of the equations in a system by a number so that when the equations are combined, one of the variables is eliminated.

Jack has 38 animals on his farm. Some of the animals have two legs and some of the animals have 4 legs. If Jack counted 128 legs total, how many of each type of animal does he have?


1. How many animals does Jack have total?

2. What type of animals does Jack have?

3. How many legs did Jack count on his farm?

Make a PlanLet t represent the number of two-legged animals and f represent the number of four-legged animals.

4. Two-legged animals plus four-legged animals equals the total number of animals.

t ! "

5. (2)(number of two-legged animals) ! (4)(number of 4-legged animals) " Total legs

2(t ) ! 4( ) "

Solve

6. To eliminate t multiply the first equation by .

t ! f " 38 (t ! f ) " (38) t ! f "

2t ! 4f " 128

____

2f " 52

f " 26

2t ! 4f " 128 2t ! 4f " 128

7. There are four-legged animals. To find the number of t ! 26 " 38

t " two-legged animals, substitute in for f and solve for t.

8. There are two-legged animals.

Look Back

9. To check, substitute the solution, , into both equations.

2t ! 4f " 128

2( ) ! 4( ) " 128

" 128

t ! f " 38

! " 38

" 38

Does your solution make both

equations true?

6ASECTION



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Ready To Go On? Skills Intervention6-4 Solving Special Systems


Systems with No Solution

Solve { y ! 4x " 3 "4x # y ! 8 .

Consider y ! 4x " 3.

What is the slope of this equation?

What is the y-intercept of this equation?

(0, )

Consider "4x # y ! 8.Solve the equation for y."4x # y ! 8

y ! # 8


What is the y-intercept of this equation? (0, )

The slopes of both equations are . The y-intercepts of the equations are

. The lines are so the lines do not . Therefore

there is no .

Classifying Systems of Linear Equations

Classify the system { y ! " 1 __ 3 x # 6 x # 3y ! 18 . Give the number of solutions.

Consider y ! " 1 __ 3 x # 6.



(0, )

Consider x # 3y ! 18.

To identify the slope and y-intercept solve the

equation for .x # 3y ! 18

3y ! 18 "

3y _____ ! 18 " __________

y ! " 1 __ 3 x


What is the y-intercept of this equation? (0, )

The slopes of both equations are . The y-intercepts of equations are

. Since the lines overlap this system has many solutions.

This system is called and .

SECTION

6A

Vocabulary

inconsistent system consistent system independent system dependent system



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6-1 Solving Systems by GraphingTell whether the ordered pair is a solution of the given system.

1. (!1, 4); { y " !3x # 1 y " x # 5 2. (1, 5); { x ! 3y " 2

2x ! 5y " 5 3. (8, !2); { y " ! 1 __ 4 x y # 3x " 22

Solve each system by graphing.

4. { y " x # 4

y " 3 __ 4 x # 2 5. { y " !2x ! 3

3x ! y " 3 6. { 1 __ 3 x # y " 5 4x ! 2y " 4

x

y5

–5

–9

x

y5

–5

–5 5

x

y5

–5

–5 5

7. Bill and Steve both owe their mother money. Bill owes his mother $300 and plans to pay her $25 every week. Steve owes his mother $550 and plans to pay her $75 every week. After how many weeks will they both owe their mother the same amount of money? What amount will that be?

6-2 Solving Systems by SubstitutionSolve each system by substitution.

8. { y " x ! 2 4x # y " 8 9. { !4x # 2y " !6

3x # y " 2 10. { !y " x y " 3x # 4

11. { y " x # 10

!2x ! 10y " !52 12. { y " !x

!y " 4x # 9 13. { y " 6x ! 3 y " x # 5

Ready To Go On? Quiz6A

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6-3 Solving Systems by EliminationSolve each system by elimination.

14. { x ! 2y " 20 3x # 2y " #12 15. { x ! y " 5

3x ! y " #11 16. { 4x ! 3y " 19

#3x ! 4y " #8

17. { x # y " 12 2x ! y " 0 18. { 12x # 6y " 9

6y # 9x " 3 19. { #x # y " 8

#2x # y " 10

20. Chris sold 180 cookies and cupcakes over the weekend at a bake sale. The cupcakes sold for $0.50 each and the cookies sold for $0.25 each. If Chris collected $66 how many of each did Chris sell?

6-4 Solving Special Systems Solve each system of linear equations.

21. { y " #x # 3 x ! y " 4 22. { x # y " 5

3x # 3y " 12 23. { y " #3x ! 5 3x ! y " 5

24. { #x # y " 7 x ! y " 2 25. { y " 6x ! 2

y # 6x " 2 26. { 3x # y " 12 y " #3x ! 1

Classify each system. Give the number of solutions.

27. { 2y " 1 # 1 __ 2 x x " 2 # 4y 28. { y " #3x ! 1

3x ! y " #1 29. { 4x # 3y " 12

y " 3(x # 4)

Ready to Go On? Quiz continued6A

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Name Date Class

Using a Matrix to Represent a System of EquationsA matrix is a rectangular array of numbers enclosed in a single set of brackets. If each equation in a system of equations is written in standard form, you can represent the system with a matrix equation. The matrix equation is made up of three matrices; one for the coefficients on the variables x and y, one for the variables x and y, and one for the constants.

For example, the system of equations { 3x ! y " 6 x # y " !2 is represented by the matrix

equation [ 3 !1 1 1 ] [ x y ] " [ 6 !2 ] .

Determine which system of equations represents the correct matrix equation.

1. { x # y " 8 x ! y " 2 a. [ 3 !1 6 2 ] [ x y ] " [ 4

!8 ]

2. { 3x ! y " 4 6x # 2y " !8 b. [ 1 !1 2 3 ] [ x y ] " [ 2 9 ]

3. { x ! y " 2 2x # 3y " 9 c. [ !5 8 10 3 ] [ x y ] " [ 21 15 ]

4. { !5x # 8y " 21 10x # 3y " 15 d. [ 1 1 1 !1 ] [ x y ] " [ 8 2 ]

Create a matrix equation for each system of equations.

5. { x ! 5y " 0 2x ! 3y " 7 6. { 4x # 3y " 19

3x ! 4y " 8 7. { 5x # 3y " 12 4x ! 5y " 17

8. { x # y " 7 x ! y " 9 9. { 12x ! 9y " 114

12x # 7y " 82 10. { 2x ! 3y " !4 x # 3y " 7

11. { 1 __ 2 x # y " 12

x # 1 __ 5 y " 10 12. { 2 __ 3 x # 1 __ 3 y " !9

1 __ 4 x # 3 __ 4 y " 16

13. { 1.2x ! 1.6y " 2.4

!0.8x # 0.2y " !1.2

Ready to Go On? Enrichment6A

SECTION



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11ASECTION


Identifying Solutions of InequalitiesTell whether the ordered pair, (!4, !2), is a solution of the inequality, y " !3x # 6.

To be a solution to the inequality, the ordered pair must make the inequality .

Substitute the ordered pair (!4, !2) into the inequality. Let x " and y " .y # !3x $ 6

# ? ! 3( ) $ 6

# ? $ 6

#

Graphing Linear Inequalities in Two VariablesGraph the solutions of the linear inequality, y $ !2x ! 3.

The inequality, y % !2x ! 3, is in form.

The line is given by the equation y !2x ! 3.

The slope of this line is . The y-intercept of this line is (0, ).

Use the slope and y-intercept to plot the boundary line.

Start at the point (0, ) and go up units and left

unit. Plot the point.

Make the boundary line a line because the

inequality symbol is not “or equal to.”

The inequality symbol is , so should the solutions be located

in the half-plane above or below the line? Shade this region to represent the solutions.

Check: Choose the point (0, 0) and substitute it into the inequality.

y % !2x ! 3 0 % !2(0) ! 3 0 % !3

Does the point make the inequality true? Did you shade correctly?


Name Date Class

Ready To Go On? Skills Intervention6-5 Solving Linear Inequalities6B

SECTION

Vocabulary

linear inequality solution of a linear inequality

Does the ordered pair (!4, !2) make the inequality true?

Therefore the ordered pair (!4, !2) a solution to the inequality y # !3x $ 6.

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5

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11ASECTION

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A solution to a linear equality lies in a half-plane with other ordered pairs that make the inequality true.

Janel has $10 to buy pencils and paper. Pencils sell for $1.50 per dozen and paper sells for $0.50 per package. Write a linear inequality to represent the situation. Then graph the linear inequality and give two possible combinations of pencils and paper that Janel can buy.


1. How much money does Janel have?

2. A dozen pencils cost and a package of paper cost .

Make a Plan

Let x represent the dozens of pencils and y represent the number of packages of paper.

3. (1.50)(dozens of pencils) ! (0.50)(packages of paper) " $10

# x ! (0.50) # "

4. Write the inequality in slope-intercept form:

1.5x ! y " 10 The slope of the boundary line is .

0.5y " 10 $ The y-intercept of the boundary line is (0, ).

0.5y ____ 0.5 " 10 ___ 0.5 $ ______ 0.5 The symbol is “"” so the boundary line is .

y " 20 $ x Shade the half-plane the boundary line.

Solve

5. Graph the inequality and shade the correct half-plane.

6. Choose two combinations: (2, ) and ( , 5).

Look Back

7. Substitute your combinations into the inequality from Exercise 3.

y " 20 $ 3x y " 20 $ 3x

" 20 $ 3( ) " 20 $ 3( )

" "

So Janel can buy 2 dozen pencils and packages of paper or 1 dozen pencils and

packages of paper.

Ready to Go On? Problem Solving Intervention6-5 Solving Linear Inequalities6B

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11ASECTION

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Ready To Go On? Skills Intervention6-6 Solving Systems of Linear Inequalities6B

SECTION


Identifying Solutions of Systems of Linear InequalitiesTell whether the ordered pair (!3, 1) is a solution of the system, { y " !x # 3 y $ x ! 2 .

To be a solution to the system of inequalities, the ordered pair (!3, 1) has to make

both inequalities . Let x " and let y " .

Substitute x " !3 and y " 1 into y # !x $ 3.

# $ 3

# $ 3

#

The point (!3, 1) makes this inequality .

Substitute x " !3 and y " 1 into y % x ! 2.

% ! 2

%

The point (!3, 1) makes this inequality .

Since the point (!3, 1) makes both inequalities , it is a to the system.

Solving a System of Linear Inequalities by Graphing

Graph the system of inequalities { y % !x # 3 y & 2x . Give two ordered

pairs that are solutions and two that are not solutions.

y & !x $ 3 The y-intercept is (0, )

and the slope " . Use this information

to draw a line. Because the symbol

is than, shade the half-plane

the line.

y ' 2x The

y-intercept is (0, )

and the slope " . Use this

information to draw a line. Because

the symbol is than or equal to,

shade the half-plane the line.

Look at the graph. Are (2, 3) and (4, 1) solutions?

Are (0, 0) and (1, !1) solutions?

Did you shade the graph correctly?

Vocabulary

system of linear inequalities solutions of a system of linear inequalities

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–5–5

55

5555



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11ASECTION

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Ready to Go On? Problem Solving Intervention6-6 Solving Systems of Linear Inequalities6B

SECTION

The solutions to a system of linear inequalities are located in the intersection of the half-planes.

As a fundraiser, the chess club sells hotdogs and hamburgers. They make $1 for every hotdog and $2 for every hamburger they sell. The club cannot sell more than 100 hotdogs or 200 hamburgers. The club’s goal is to make at least $200 in total profit. Show and describe possible combinations of hotdogs and hamburgers that can be sold to meet the goal.


1. How much does the club earn from each hotdog? Each hamburger?

2. How many hotdogs do they have to sell? Hamburgers?

3. How many inequalities need to be written?

Make a PlanLet x represent the number of hotdogs and y represent the number of hamburgers.

4. The club cannot sell more than 100 hotdogs.

The club cannot sell more than 200 hamburgers.

Profit from hotdogs ! Profit from hamburgers " 200

x # y # 1x ! y "

5. Graph each inequality on the coordinate system.

Inequality Boundary LineDashed or Solid Line Shaded Region

x # Vertical line at x $ line Shade to the of the line.

y # Horizontal line at y $ line Shade the line.

1x ! y " x-intercept: (200, 0) y-intercept: (0, 100)

line Shade above the line.

Solve

6. Graph each inequality and shade the correct half-planes.

7. Are the points (50, 100), (75, 150), and (80, 175) located in

the intersection of all three regions?

Look Back

8. Check that the x-coordinates of each point are less than 100 and that the y-coordinates

are less than 200. Does each point make the inequality x ! 2y " 200 true?

x

y250

0225




Name Date ClassName Date Class


6-5 Solving Linear InequalitiesTell whether the ordered pair is a solution of the inequality.

1. (4, !3); y " !3x # 2 2. (5, 18); y $ 4x ! 6 3. (2, !4); y % 5x ! 12

Graph the solutions of each linear inequality.

4. y $ 5x ! 4 5. 4x ! y " 3 6. 3x # 4y " 12

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y5

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x

y5

–5

–5 5

x

y5

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7. Barbara has no more than $144 to buy jewelry. Earrings cost $12 each and necklaces cost $24 each. How many of each can she buy? Write a linear inequality to describe the situation. Graph the linear inequality and give three possible combinations of earrings and necklaces Barbara can buy.

Write an inequality to represent each graph.

8. 9. 10.

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x

y5

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x

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Ready To Go On? Quiz6B

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6-6 Solving Systems of Linear InequalitiesTell whether the ordered pair is a solution of the given system.

11. (!1, !4); { y " !3x y # x ! 4 12. (3, 2); { y $ x % 2

y & !3x ! 2 13. (0, 0); { y & 2x 2x % y " !4

Graph each system of linear inequalities. Give two ordered pairs that are solutions and two that are not solutions.

14. { y # 3 y " x % 1 15. { x % y $ 4

3x ! y & !3 16. { 3x ! 2y " 6 2x % 3y & !6

x

y5

–5

–5 5

x

y5

–5

–5 5

x

y5

–5

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Graph each system of linear inequalities and describe the solutions.

17. { y & x % 2 y & x ! 3 18. { y # !3x ! 2

y " !3x % 3 19. { y # !2x % 4 y " !2x ! 1

x

y5

–5

–5 5

x

y5

–5

–5 5

x

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–5

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20. A vendor sells hotdogs for $2.00 and sausage for $3.00. The vendor begins each day with 150 hotdogs and 200 sausages. The vendor wants to make at least $600. Graph and describe all possible combinations of sandwiches that could be sold to meet the goal. List two possible combinations.

Ready to Go On? Quiz continued6B

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Describing Geometric Regions with a System of Inequalities

Describe the shaded region of the graph by writing a system of inequalities consisting of three different linear inequalities. To write this system, follow these steps:

1. Determine the slope and y-intercept of each line.2. Write inequalities in slope-intercept form.3. If the line is solid use either ! or ". If the line is dashed, use either # or $.4. If the shaded region is “above” the line use the symbol $, and if the shaded

region is “below” the line, use the symbol #.

For example, the system of inequalities that describes the region below is { x " %2y " %4

y ! %3 ___ 2 x % 1

.

x

y5

–5

–5 5

Write a system of inequalities that describes each region.

1.

x

y5

–5

–5 5

2.

x

y5

–5

–5 5

3.

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–5

–5 5

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Ready to Go On? Enrichment6B

SECTION




Ready To Go On? Skills Intervention5-10 Transforming Linear Functions5B

SECTION

Every linear function is a transformation of the function f (x ) ! x.

A pre-school charges a one-time fee of $180 for supplies and then $275 per month. The school’s total bill after teaching a child for x months is f (x ) ! 275x " 180. How will the graph of this function change if the fee for supplies is increased to $250? if the monthly rate is decreased to $200?


1. How much does it cost for supplies? $180 2. How much does it cost for a child to attend this pre-school each month? $275

Make a Plan

3. What is the slope of f (x ) ! 275x " 180? 275 This represents the cost

per month . What is the y-intercept? 180 This represents the

fee for supplies.

Solve

4. If a linear equation is translated , the y-intercept is increased or decreased.

5. If the fee for supplies is increased to $250, the graph of the function will be

translated up 70 units.

6. If a linear equation is rotated, the slope of the equation is changed.

7. If the monthly rate is decreased to $200, the graph of the function will be

rotated about the point (0, 180) . Will the graph of the function with a monthly fee of $200 be less steep or more steep than the original

function? Less steep

Look Back

8. Compare the functions f (x ) ! 275x " 180 to f (x ) ! 275x " 250. If you plot these on a coordinate grid, which function has the greatest y-intercept?

f (x ) ! 275x " 250 9. Compare the slopes of the functions f (x ) ! 275x " 180 to f (x ) ! 200x " 180.

Which has the smallest slope? f (x ) ! 200x " 180

Which has the steepest line? f (x ) ! 275x " 180

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Vocabulary

system of linear equations solution of a system of linear equations

Identifying Solutions of SystemsTell whether the ordered pair (1, 2) is a solution of the system, { y ! #2x " 4 y ! x " 1 .

To be a solution to the system of equations, the ordered pair (1, 2) has to

make both equations true . Let x ! 1 and let y ! 2 .

y ! #2x " 4

2 ! #2( 1 ) " 4

2 ! #2 " 4

2 ! 2

The point (1, 2) makes this equation true .

y ! x " 1

2 ! 1 " 1

2 ! 2

The point (1, 2) makes this equation true .

Since the point (1, 2) makes both equations true , it is a

solution to the given system.

Solving a System of Linear Equations by Graphing

Solve the system of equations { y ! # 1 __ 2 x " 2 2x " y ! #1 by graphing.

Consider y ! # 1 __ 2 x " 2. This equation is

written in slope-intercept form. The

y-intercept is (0, 2 ) and the slope ! # 1 _ 2

.

Start at point (0, 2 ). Plot this point on the

coordinate system. Move down 1 unit

and right 2 units and plot another point. Connect these points.

Consider 2x " y ! #1. This equation is

written in standard form.

Find the x-intercept: Find the y-intercept:

Plug 0 in for y. Plug 0 in for x.

2x " 0 ! #1 2(0) " y ! #1

x ! # 1 _ 2

y ! #1

(# 1 _ 2

, 0) (0, #1 )

Plot these points and connect the points.

The point where the two lines intersect is at ( #2 , 3 ).

Ready To Go On? Skills Intervention6-1 Solving Systems by Graphing6A

SECTION

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The solution to a system of linear equations is the point where the graphs of the two lines intersect.

The local county fair charges $7.00 admission and $0.50 per ticket to ride the rides. The state fair charges $5.00 admission and $1.50 per ticket to ride the rides. For how many tickets will the total cost be the same at both fairs? What is that cost?


1. How much is the admission for the local county fair? $7.00

How much does each ride ticket cost at the local county fair? $0.50

2. How much is the admission for the state fair? $5.00

How much does each ride ticket cost at the state fair? $1.50

Make a PlanLet x represent the number of tickets purchased and y represent the total cost.

Total cost is equal to the price per ticket plus admission

3. Equation for local county fair: y ! 0.5 x " 7

4. Equation for state fair: y ! 1.5 x " 5

Solve

5. Graph the line y ! 0.5 x " 7. The slope ! 0.5 or 1 __ 2 .

The y-intercept of this line is (0, 7 ).

6. Graph the line y ! 1.5 x " 5. The slope ! 1.5 or 3 __ 2 .

The y-intercept of this line is (0, 5 ).

7. What is the ordered pair where the two lines intersect? (2, 8) 8. If a person buys 2 tickets, the cost at both fairs is the same: $8.00 .

Look Back

9. To check, substitute the intersection point, (2, 8) , into both equations.

y ! 0.5x " 7 y ! 1.5x " 5 Does the point make both equations

true? yes8 ! 0.5( 2 ) " 7 8 ! 1.5( 2 ) " 5

8 ! 8 8 ! 8

6AReady to Go On? Problem Solving Intervention6-1 Solving Systems by Graphing

SECTION

x

y

8

6

4

2

0–1 8642

44444444



Solving a System of Linear Equations by SubstitutionSolve each system by substitution.

A. { y ! #x " 3

y ! 2 __ 3 x # 2

STEP 1 Substitute #x " 3 for y in the second equation and solve for x.

STEP 2 y ! 2 __ 3 x # 2 STEP 3 Solve for the other variable, y.

#x " 3 ! 2 __ 3 x # 2 You know from Step 2 that x ! 3 .

y ! #x " 3

y ! # ! 3 " " 3

y ! 0The solution to the system of equations

is ( 3 , 0).

3 ! #x " 3 " ! 3 ! 2 __ 3 x # 2 " #3x " 9 ! 2x # 6

9 ! 5x # 6

15 ! 5x

3 ! x

B. { x " y ! 4 x " 2y ! #2

STEP 1 Solve the equation x " y ! 4 for x.

x " y ! 4

x ! 4 # y

STEP 2 Substitute 4 # y for x in the second equation.

x " 2y ! #2

4 # y " 2y ! #2

STEP 3 Solve for y . STEP 4 Solve for the other variable, x.

x " y ! 4

x " #6 ! 4

x ! 10

The solution to the system is (10, #6 ).

4 # y " 2y ! #2

4 " y ! #2

y ! #6

6AReady To Go On? Skills Intervention6-2 Solving Systems by Substitution

SECTION



BUA1_11NL_RTGO_AK_220-256.indd 238BUA1_11NL_RTGO_AK_220-256.indd 238 9/9/09 8:05:18 AM9/9/09 8:05:18 AM


Elimination using Addition

Solve { x ! 4y " 16 5x # 4y " 8 by elimination.

STEP 1 Combine the two equations by addition.

x ! 4y " 16 5x # 4y " 8

____

6x ! 0 " 24

Which variable is eliminated? y

STEP 2 Solve for x . STEP 3 Substitute and solve for y .

6x " 24 x ! 4y " 16

4 ! 4y " 16

4y " 12

y " 3

6x _____

6 " 24 _____

6

x " 4

STEP 4 The solution to the system is (4, 3 ).

Elimination using Multiplication First

Solve { 3x # y " 4 x ! 4y " 10 by elimination.

STEP 1 To eliminate y, multiply the 4(3x # y ) " 4(4) 12 x # 4 y " 16 x ! 4y " 10 x ! 4y " 10 first equation by 4 .

STEP 2 Combine the two equations using addition . 12 x # 4 y " 16

x ! 4y " 10

____

13 x ! 0 " 26

STEP 3 Solve for x. STEP 4 Substitute and solve for y.

13x " 26 x ! 4y " 10

2 ! 4y " 10

4y " 8

y " 2

13 x

______

13 " 26 _____

13

x " 2

STEP 5 The solution to the system of equations is (2, 2 ).

6AReady To Go On? Skills Intervention6-3 Solving Systems by Elimination

SECTION



Ready to Go On? Problem Solving Intervention6-3 Solving Systems by Elimination

You can multiply one or both of the equations in a system by a number so that when the equations are combined, one of the variables is eliminated.

Jack has 38 animals on his farm. Some of the animals have two legs and some of the animals have 4 legs. If Jack counted 128 legs total, how many of each type of animal does he have?


1. How many animals does Jack have total? 38

2. What type of animals does Jack have? 2-legged and 4-legged

3. How many legs did Jack count on his farm? 128 legs

Make a PlanLet t represent the number of two-legged animals and f represent the number of four-legged animals.

4. Two-legged animals plus four-legged animals equals the total number of animals.

t ! f " 38

5. (2)(number of two-legged animals) ! (4)(number of 4-legged animals) " Total legs

2(t ) ! 4( f ) " 128

Solve

6. To eliminate t multiply the first equation by #2 .

t ! f " 38 #2 (t ! f ) " #2 (38) #2 t ! #2 f " #76

2t ! 4f " 128

____

2f " 52

f " 26

2t ! 4f " 128 2t ! 4f " 128

7. There are 26 four-legged animals. To find the number of t ! 26 " 38

t " 12two-legged animals, substitute 26 in for f and solve for t.

8. There are 12 two-legged animals.

Look Back

9. To check, substitute the solution, (12, 26), into both equations.

2t ! 4f " 128

2( 12 ) ! 4( 26 ) " 128

128 " 128

t ! f " 38

12 ! 26 " 38

38 " 38

Does your solution make both

equations true? Yes

6ASECTION



Ready To Go On? Skills Intervention6-4 Solving Special Systems


Systems with No Solution

Solve { y " 4x # 3 #4x ! y " 8 .

Consider y " 4x # 3.

What is the slope of this equation? 4


(0, #3 )

Consider #4x ! y " 8.Solve the equation for y.#4x ! y " 8

y " 4x ! 8

What is the slope of this equation? 4What is the y-intercept of this equation? (0, 8 )

The slopes of both equations are equal . The y-intercepts of the equations are different . The lines are parallel so the lines do not intersect . Therefore

there is no solution .

Classifying Systems of Linear Equations

Classify the system { y " # 1 __ 3 x ! 6 x ! 3y " 18 . Give the number of solutions.

Consider y " # 1 __ 3 x ! 6.

What is the slope of this equation? # 1 __

3


(0, 6 )

Consider x ! 3y " 18.

To identify the slope and y-intercept solve the

equation for y .x ! 3y " 18

3y " 18 # x

3y _____

3 " 18 # x __________

3

y " 6 # 1 __ 3 x

What is the slope of this equation? # 1 __

3

What is the y-intercept of this equation? (0, 6 )

The slopes of both equations are equal . The y-intercepts of equations are

equal . Since the lines overlap this system has infinitely many solutions.

This system is called dependent and consistent .

SECTION

6A

Vocabulary

inconsistent system consistent system independent system dependent system




11ASECTION


Identifying Solutions of InequalitiesTell whether the ordered pair, (#4, #2), is a solution of the inequality, y $ #3x ! 6.

To be a solution to the inequality, the ordered pair must make the inequality true .

Substitute the ordered pair (#4, #2) into the inequality. Let x " #4 and y " #2 .y $ #3x ! 6

#2 $ ? # 3( #4 ) ! 6

#2 $ ? 12 ! 6

#2 $ 18

Graphing Linear Inequalities in Two VariablesGraph the solutions of the linear inequality, y % #2x # 3.

The inequality, y % #2x # 3, is in slope-intercept form.

The boundary line is given by the equation y " #2x # 3.

The slope of this line is #2 . The y-intercept of this line is (0, #3 ).

Use the slope and y-intercept to plot the boundary line.

Start at the point (0, #3 ) and go up 2 units and left

1 unit. Plot the point.

Make the boundary line a dashed line because the

inequality symbol is not “or equal to.”

The inequality symbol is % , so should the solutions be located

in the half-plane above or below the line? Above Shade this region to represent the solutions.

Check: Choose the point (0, 0) and substitute it into the inequality.

y % #2x # 3 0 % #2(0) # 3 0 % #3

Does the point make the inequality true? Yes Did you shade correctly? Yes


Ready To Go On? Skills Intervention6-5 Solving Linear Inequalities6B

SECTION

Vocabulary

linear inequality solution of a linear inequality

Does the ordered pair (#4, #2) make the inequality true? Yes

Therefore the ordered pair (#4, #2) is a solution to the inequality y $ #3x ! 6.

x

y5

–5

–5 5

55

55555555





11ASECTION

A solution to a linear equality lies in a half-plane with other ordered pairs that make the inequality true.

Janel has $10 to buy pencils and paper. Pencils sell for $1.50 per dozen and paper sells for $0.50 per package. Write a linear inequality to represent the situation. Then graph the linear inequality and give two possible combinations of pencils and paper that Janel can buy.


1. How much money does Janel have? $10 2. A dozen pencils cost $1.50 and a package of paper cost $0.50 .

Make a Plan

Let x represent the dozens of pencils and y represent the number of packages of paper.

3. (1.50)(dozens of pencils) ! (0.50)(packages of paper) " $10

(1.5) # x ! (0.50) # y " 10 4. Write the inequality in slope-intercept form:

1.5x ! 0.5 y " 10 The slope of the boundary line is $3 .

0.5y " 10 $ 1.5x The y-intercept of the boundary line is (0, 20 ).

0.5y ____ 0.5 " 10 ___ 0.5 $

1.5x ______ 0.5 The symbol is “"” so the boundary line is solid .

y " 20 $ 3 x Shade the half-plane under the boundary line.

Solve

5. Graph the inequality and shade the correct half-plane.

6. Choose two combinations: (2, 2 ) and ( 1 , 5).

Look Back

7. Substitute your combinations into the inequality from Exercise 3.

y " 20 $ 3x y " 20 $ 3x

2 " 20 $ 3( 2 ) 5 " 20 $ 3( 1 )

2 " 14 5 " 17

So Janel can buy 2 dozen pencils and 2 packages of paper or 1 dozen pencils and

5 packages of paper.

Ready to Go On? Problem Solving Intervention6-5 Solving Linear Inequalities6B

SECTION

x

y20

16

12

8

4

0–1 8642

00

6666

2222

8888

4444

00–1–1 666644442222

2020

–1–1 664422


11ASECTION Ready To Go On? Skills Intervention

6-6 Solving Systems of Linear Inequalities6BSECTION


Identifying Solutions of Systems of Linear InequalitiesTell whether the ordered pair (!3, 1) is a solution of the system, { y " !x # 3 y $ x ! 2 .

To be a solution to the system of inequalities, the ordered pair ($3, 1) has to make

both inequalities true . Let x % !3 and let y % 1 .

Substitute x % $3 and y % 1 into y & $x ! 3.

1 & !(!3) ! 3

1 & 3 ! 3

1 & 6

The point ($3, 1) makes this inequality true .

Substitute x % $3 and y % 1 into y ' x $ 2.

1 ' !3 $ 2

1 ' !5

The point ($3, 1) makes this inequality true .

Since the point ($3, 1) makes both inequalities true , it is a solution to the system.

Solving a System of Linear Inequalities by Graphing

Graph the system of inequalities { y % !x # 3 y & 2x . Give two ordered

pairs that are solutions and two that are not solutions.

y ( $x ! 3 The y-intercept is (0, 3 )

and the slope % !1 . Use this information

to draw a dashed line. Because the symbol

is greater than, shade the half-plane

above the line.

y " 2x The

y-intercept is (0, 0 )

and the slope % 2 . Use this

information to draw a solid line. Because

the symbol is less than or equal to,

shade the half-plane below the line.

Look at the graph. Are (2, 3) and (4, 1) solutions? Yes

Are (0, 0) and (1, $1) solutions? No

Did you shade the graph correctly? Yes

Vocabulary

system of linear inequalities solutions of a system of linear inequalities

x

y5

–5

–5 5

–5–5–5–5

5555

55555555


BUA1_11NL_RTGO_CH06_074-083.indd 82 8/26/09 12:33:56 PM11ASECTION Ready to Go On? Problem Solving Intervention

6-6 Solving Systems of Linear Inequalities6BSECTION

The solutions to a system of linear inequalities are located in the intersection of the half-planes.

As a fundraiser, the chess club sells hotdogs and hamburgers. They make $1 for every hotdog and $2 for every hamburger they sell. The club cannot sell more than 100 hotdogs or 200 hamburgers. The club’s goal is to make at least $200 in total profit. Show and describe possible combinations of hotdogs and hamburgers that can be sold to meet the goal.


1. How much does the club earn from each hotdog? $1 Each hamburger? $2 2. How many hotdogs do they have to sell? 100 Hamburgers? 200 3. How many inequalities need to be written? 3

Make a PlanLet x represent the number of hotdogs and y represent the number of hamburgers.

4. The club cannot sell more than 100 hotdogs.

The club cannot sell more than 200 hamburgers.

Profit from hotdogs ! Profit from hamburgers ' 200

x " 100 y " 200 1x ! 2 y ' 200 5. Graph each inequality on the coordinate system.

Inequality Boundary LineDashed or Solid Line Shaded Region

x " 100 Vertical line at x % 100 Solid line Shade to the left of the line.

y " 200 Horizontal line at y % 200 Solid line Shade under the line.

1x ! 2 y ' 200 x-intercept: (200, 0) y-intercept: (0, 100)

Solid line Shade above the line.

Solve

6. Graph each inequality and shade the correct half-planes.

7. Are the points (50, 100), (75, 150), and (80, 175) located in

the intersection of all three regions? Yes

Look Back

8. Check that the x-coordinates of each point are less than 100 and that the y-coordinates

are less than 200. Does each point make the inequality x ! 2y ' 200 true? Yes

x

y250

0225



Evaluating Expressions with Zero and Negative Exponents

Zero Exponent: Any nonzero number raised to the zero power is 1 .

4 0 % 1Negative Exponent: A nonzero number raised to a negative exponent is

equal to one divided by that number raised to the opposite exponent.

3 $4 % 1 __ 3 4

Evaluate each expression for the given value(s) of the variable(s).

A. a $3 b 2 for a % 2 and b % 3

2 $3

# 3 2 Substitute 2 for a and 3 for b.

% 1 __ 2 3

# 9 Use the definition x $n % 1 __ x n

and simplify.

% 1 _________ 2 # 2 # 2

# 9 Write the power in the denominator as a product.

% 1 ___ 8

# 9 Evaluate the power in the denominator.

% 9 ___ 8

Simplify.

B. x 0 y $3 for x % 7 and y % $5

7 0 # ($5)

!3 Substitute 7 for x and !5 for y.

% 1 # 1 _____ ($5 ) 3

Any nonzero number raised to the zero power is 1

and x $n % 1 __ x n

. % 1 # 1 ______

!125

Evaluate the power.

% $ 1 ______

125 Simplify.

Simplifying Expressions with Zero and Negative Exponents

A. 9 a $7

9 a $7 % 9 # a !7

% 9 # 1 _____ a 7

% 9 _____ a 7

Ready To Go On? Skills Intervention7-1 Integer Exponents7A

SECTION

B. x $6 ____ y $4

x $6 ____ y $4

% x $6 # 1 ______ y !4

% 1 _____ x 6

# y 4

__ 1 % y 4 _____

x 6

Copyright © by Holt, McDougal. 84 Holt McDougal Algebra 1All rights reserved.





6-1 Solving Systems by GraphingTell whether the ordered pair is a solution of the given system.

1. (!1, 4); { y " !3x # 1 y " x # 5 2. (1, 5); { x ! 3y " 2

2x ! 5y " 5 3. (8, !2); { y " ! 1 __ 4 x y # 3x " 22

Yes No Yes

Solve each system by graphing.

4. { y " x # 4

y " 3 __ 4 x # 2 5. { y " !2x ! 3

3x ! y " 3 6. { 1 __ 3 x # y " 5 4x ! 2y " 4

x

y5

–5

–9

x

y5

–5

–5 5

x

y5

–5

–5 5

(!8, !4) (0, !3) (3, 4)

7. Bill and Steve both owe their mother money. Bill owes his mother $300 and plans to pay her $25 every week. Steve owes his mother $550 and plans to pay her $75 every week. After how many weeks will they both owe their mother the same amount of money? What amount will that be?

5 weeks; They will both owe $175.

6-2 Solving Systems by SubstitutionSolve each system by substitution.

8. { y " x ! 2 4x # y " 8 9. { !4x # 2y " !6

3x # y " 2 10. { !y " x y " 3x # 4

(2, 0) (1, !1) (!1, 1)

11. { y " x # 10

!2x ! 10y " !52 12. { y " !x

!y " 4x # 9 13. { y " 6x ! 3 y " x # 5

(!4, 6) (!3, 3)

Ready To Go On? Quiz6A

SECTION

55

! 8 __ 5 , 33 ___

5 "

BUA1_11NL_RTGO_QUIZE_168-179.indd 168 8/26/09 2:11:23 PM


6-3 Solving Systems by EliminationSolve each system by elimination.

14. { x # 2y " 20 3x ! 2y " !12 15. { x # y " 5

3x # y " !11 16. { 4x # 3y " 19

!3x # 4y " !8

(2, 9) (!8, 13) (4, 1)

17. { x ! y " 12 2x # y " 0 18. { 12x ! 6y " 9

6y ! 9x " 3 19. { !x ! y " 8

!2x ! y " 10

(4, !8) (4, 6.5) (!2, !6)

20. Chris sold 180 cookies and cupcakes over the weekend at a bake sale. The cupcakes sold for $0.50 each and the cookies sold for $0.25 each. If Chris collected $66 how many of each did Chris sell?

96 cookies and 84 cupcakes

6-4 Solving Special Systems Solve each system of linear equations.

21. { y " !x ! 3 x # y " 4 22. { x ! y " 5

3x ! 3y " 12 23. { y " !3x # 5 3x # y " 5

No solution No solution Infinitely many solutions

24. { !x ! y " 7 x # y " 2 25. { y " 6x # 2

y ! 6x " 2 26. { 3x ! y " 12 y " !3x # 1

No solution Infinitely many solutions No solution

Classify each system. Give the number of solutions.

27. { 2y " 1 ! 1 __ 2 x x " 2 ! 4y 28. { y " !3x # 1

3x # y " !1 29. { 4x ! 3y " 12

y " 3(x ! 4)

Consistent, dependent Inconsistent Consistent, independent

infinitely many solutions no solution one solution

Ready to Go On? Quiz continued6A

SECTION


Copyright © by Holt McDougal. 170 Holt McDougal Algebra 1All rights reserved.Copyright © by Holt McDougal. 170 Holt McDougal Algebra 1All rights reserved.

6-5 Solving Linear InequalitiesTell whether the ordered pair is a solution of the inequality.

1. (4, !3); y $ !3x # 2 2. (5, 18); y % 4x ! 6 3. (2, !4); y & 5x ! 12

No Yes Yes

Graph the solutions of each linear inequality.

4. y % 5x ! 4 5. 4x ! y $ 3 6. 3x # 4y $ 12

x

y5

–5

–5 5

x

y5

–5

–5 5

x

y5

–5

–5 5

7. Barbara has no more than $144 to buy jewelry. Earrings cost $12 each and necklaces cost $24 each. How many of each can she buy? Write a linear inequality to describe the situation. Graph the linear inequality and give three possible combinations of earrings and necklaces Barbara can buy.

12x " 24y # 144; Sample Answers: 10 pairs of

earrings and 1 necklace; 7 pairs of earrings and

2 necklaces; 1 pair of earrings and 4 necklaces

Write an inequality to represent each graph.

8. y $ x ! 1 9. y % !4 10. y & !2x ! 4

x

y5

–5

–5 5

x

y5

–5

–5 5

x

y5

–5

–5 5

Ready To Go On? Quiz6B

SECTION

5

–5–5

5

–5

5

5

–5–5

5

–5

55

5

5 55555555

x

y

16

12

8

4

0–2–2

161284

44

02 11288442 161284 116

–5

55

–5

5

5 5555

5

55

5

55


Copyright © by Holt McDougal. 171 Holt McDougal Algebra 1All rights reserved.Copyright © by Holt McDougal. 171 Holt McDougal Algebra 1All rights reserved.

6-6 Solving Systems of Linear InequalitiesTell whether the ordered pair is a solution of the given system.

11. (!1, !4); { y $ !3x y ' x ! 4 12. (3, 2); { y & x # 2

y % !3x ! 2 13. (0, 0); { y % 2x 2x # y $ !4

Yes Yes No

Graph each system of linear inequalities. Give two ordered pairs that are solutions and two that are not solutions.

14. { y ' 3 y $ x # 1 15. { x # y & 4

3x ! y % !3 16. { 3x ! 2y $ 6 2x # 3y % !6

x

y5

–5

–5 5

x

y5

–5

–5 5

x

y5

–5

–5 5

Sample Answers: solutions: (4, 4), (5, 5);

Sample Answers: solutions: (0, 0), (1, 1);

Sample Answers: solutions: (0, 0), (!1, 1);

not solutions: (0, 0), (!1, 2)

not solutions: (!2, 1), (5, 1)

not solutions: (5, 1), (3, !2)

Graph each system of linear inequalities and describe the solutions.

17. { y % x # 2 y % x ! 3 18. { y ' !3x ! 2

y $ !3x # 3 19. { y ' !2x # 4 y $ !2x ! 1

x

y5

–5

–5 5

x

y5

–5

–5 5

x

y5

–5

–5 5

The same as the solutions of y & x " 2. All points between the lines. No solution

20. A vendor sells hotdogs for $2.00 and sausage for $3.00. The vendor begins each day with 150 hotdogs and 200 sausages. The vendor wants to make at least $600. Graph and describe all possible combinations of sandwiches that could be sold to meet the goal. List two possible combinations.

Sample Answers: (100, 150) and (150, 200)

Ready to Go On? Quiz continued6B

SECTION

–5

55

–5

555555

5

55

55

––

5

5

5 55 5

x

y

200

0300200100

100




Onto FunctionsA relation is a function if for every number in the domain, there is one and only one number in the range. Another way to describe a function is to determine if it is onto. A function is onto if and only if all the numbers of the range are paired with all of the numbers in the domain.

For example, the function in Table 1 is described as onto because each number in the range has a number from the domain assigned to it.

The function in the Table 2 is not onto because the number 7 in the range does not have a number in the domain assigned to it.

Determine if each function can be described as onto.

1. Not onto 2. Onto

3. Onto 4. Not onto


Ready To Go On? Enrichment5A

SECTION

Domain

Range 6 7 8 9 10

1 2 3 4 5

Domain

Range 6 7 8 9 11

1 2 3 4 5

Domain

Range 0 2 4 6 8

2 4 6 8 10 Domain

Range 12

1 2 4 8

0 3 6 9 12

Domain

Range 6 7 8 9 10

5 10 15 20 25 Domain

Range 3 4 5 6 7

1 3 5 7 9

Table 1

Table 2



The Parent Function for Quadratic FunctionsA quadratic function is a function described by an equation of the form y ! ax 2 " bx " c, when a # 0. The parent function for all quadratic functions is y ! x 2. The graph and table of this function is shown to the right. Like the family of linear functions, transformations can also be made on quadratic functions.

The graph of f (x ) ! x 2 is drawn on each coordinate grid. Graph the function g (x ) by completing the table of values and plotting the points. Then describe the transformation from the graph of f (x ) to g (x ).

1. g (x ) ! x 2 $ 3 2. g (x ) ! 1 __ 2 x 2

x $2 $1 0 1 2

y 1 "2 "3 "2 1

The graph is translated down

3 units.

The shape of the graph is wider.

3. g (x ) ! $x 2 4. g (x ) ! 2x 2

x $2 $1 0 1 2

y "4 "1 0 "1 "4

The graph is reflected over the

x-axis.

The shape of the graph is

narrower.


SECTION

x $2 $1 0 1 2y 4 1 0 1 4

x

y

5

–5

–5 5

x $2 $1 0 1 2

y 2 1 __ 2

0 1 __ 2

2

x $2 $1 0 1 2

y 8 2 0 2 8

x

y5

–5

–5 5

x

y5

–5

–5 5

x

y5

–5

–5 5

5

x

y5

–5

–5 5



Using a Matrix to Represent a System of EquationsA matrix is a rectangular array of numbers enclosed in a single set of brackets. If each equation in a system of equations is written in standard form, you can represent the system with a matrix equation. The matrix equation is made up of three matrices; one for the coefficients on the variables x and y, one for the variables x and y, and one for the constants.

For example, the system of equations { 3x $ y ! 6 x " y ! $2 is represented by the matrix

equation [ 3 $1 1 1 ] [ x y ] ! [ 6 $2 ] .

Determine which system of equations represents the correct matrix equation.

1. d { x " y ! 8 x $ y ! 2 a. [ 3 $1 6 2 ] [ x y ] ! [ 4

$8 ]

2. a { 3x $ y ! 4 6x " 2y ! $8 b. [ 1 $1 2 3 ] [ x y ] ! [ 2 9 ]

3. b { x $ y ! 2 2x " 3y ! 9 c. [ $5 8 10 3 ] [ x y ] ! [ 21 15 ]

4. c { $5x " 8y ! 21 10x " 3y ! 15 d. [ 1 1 1 $1 ] [ x y ] ! [ 8 2 ]

Create a matrix equation for each system of equations.

5. { x $ 5y ! 0 2x $ 3y ! 7 6. { 4x " 3y ! 19

3x $ 4y ! 8 7. { 5x " 3y ! 12 4x $ 5y ! 17

[ 1 "5

2 "3 ] [ x y ] ! [ 0

7 ]

[ 4 3 3 "4

] [ x y ] ! [ 19 8

] [ 5 3

4 "5 ] [ x y ] ! [ 12

17 ]

8. { x " y ! 7 x $ y ! 9 9. { 12x $ 9y ! 114

12x " 7y ! 82 10. { 2x $ 3y ! $4 x " 3y ! 7

[ 1 1

1 "1 ] [ x y ] ! [ 7

9 ]

[ 12 "9 12 7

] [ x y ] ! [ 114 82

] [ 2 "3

1 3 ] [ x y ] ! [ "4

7 ]

11. { 1 __ 2 x " y ! 12

x " 1 __ 5 y ! 10 12. { 2 __ 3 x " 1 __ 3 y ! $9

1 __ 4 x " 3 __ 4 y ! 16

13. { 1.2x $ 1.6y ! 2.4

$0.8x " 0.2y ! $1.2

! 1 __ 2

1

1 1 __ 5

" [ x y ] ! ! 12

10 "

! 2 __ 3 1 __

3

1 __ 4 3 __

4 " [ x y ] ! ! "9

16 "

Ready to Go On? Enrichment6A

SECTION

! 1.2 "1.6 "0.8 0.2

" [ x y ] !

! 2.4 "1.2

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Describing Geometric Regions with a System of Inequalities

Describe the shaded region of the graph by writing a system of inequalities consisting of three different linear inequalities. To write this system, follow these steps:

1. Determine the slope and y-intercept of each line.2. Write inequalities in slope-intercept form.3. If the line is solid use either % or &. If the line is dashed, use either ' or (.4. If the shaded region is “above” the line use the symbol (, and if the shaded

region is “below” the line, use the symbol '.

For example, the system of inequalities that describes the region below is { x & $2y & $4

y % $3 ___ 2 x $ 1

.

x

y5

–5

–5 5

Write a system of inequalities that describes each region.

1.

x

y5

–5

–5 5

{ y # 3 __ 2 x $ 3

y # " 7 __ 4 x $ 3 y % "4

2.

x

y5

–5

–5 5

{ y # 3 y % "3

y % x " 4 y % "x " 4

y # "x $ 4 y # x $ 4

3.

x

y5

–5

–5 5

{ x % "4 x # 3 y % "x " 5

y # 4 y % "3 y # x " 4

y % "x $ 5 y # x $ 6


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