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8-1
Relations and FunctionsRelations and FunctionsPRE-ALGEBRA LESSON 8-1PRE-ALGEBRA LESSON 8-1
Determine if and where Will and Pedro will meet if they startdriving from the same intersection. Pedro travels 2 blocks east, 2blocks south, 2 blocks east, and 2 blocks north. Will starts traveling2 blocks north, 4 blocks east, and 2 blocks south.
Yes, they will meet 4 blocks east.
Relations and FunctionsRelations and FunctionsPRE-ALGEBRA LESSON 8-1PRE-ALGEBRA LESSON 8-1
Graph each point.
1. A(3, 4) 2. B(–3, 1) 3. F(2, 0)
4. D(2, –2) 5. C(–4, –3) 6. E(0, –4)
(For help, go to Lesson 1-10.)
Check Skills You’ll Need
8-1
Relations and FunctionsRelations and FunctionsPRE-ALGEBRA LESSON 8-1PRE-ALGEBRA LESSON 8-1
Solutions
8-1
Relations and FunctionsRelations and FunctionsPRE-ALGEBRA LESSON 8-1PRE-ALGEBRA LESSON 8-1
Is each relation a function? Explain.
a. {(0, 5), (1, 6), (2, 4), (3, 7)}
List the domain values and the range values in order.
Draw arrows from the domain values to their range values.
There is one range value for each domain value. This relation is a function.
8-1
Relations and FunctionsRelations and FunctionsPRE-ALGEBRA LESSON 8-1PRE-ALGEBRA LESSON 8-1
(continued)
b. {(0, 5), (1, 5), (2, 6), (3, 7)}
There is one range value for each domain value. This relation is a function.
8-1
Relations and FunctionsRelations and FunctionsPRE-ALGEBRA LESSON 8-1PRE-ALGEBRA LESSON 8-1
(continued)
c. {(0, 5), (0, 6), (1, 6), (2, 7)}
There are two range values for the domain value 0.
This relation is not a function.
Quick Check
8-1
Relations and FunctionsRelations and FunctionsPRE-ALGEBRA LESSON 8-1PRE-ALGEBRA LESSON 8-1
Is the time needed to mow a lawn a function of the
size of the lawn? Explain.
No; two lawns of the same size (domain value) can require different lengths of time (range values) for mowing.
Quick Check
8-1
Relations and FunctionsRelations and FunctionsPRE-ALGEBRA LESSON 8-1PRE-ALGEBRA LESSON 8-1
a. Graph the relation shown in the table.
Domain Value
–3
–5
3
5
Range Value
5
3
5
3
Graph the ordered pairs(–3, 5), (–5, 3), (3, 5), and (5, 3).
8-1
Relations and FunctionsRelations and FunctionsPRE-ALGEBRA LESSON 8-1PRE-ALGEBRA LESSON 8-1
b. Use the vertical-line test. Is the relation a function? Explain.
(continued)
The pencil does not pass through two points at any one of its positions, so the relation is a function.
Pass a pencil across the graph as shown. Keep the pencil vertical (parallel to the y-axis) to represent a vertical line.
Quick Check
8-1
Relations and FunctionsRelations and FunctionsPRE-ALGEBRA LESSON 8-1PRE-ALGEBRA LESSON 8-1
Is each relation a function? Explain.
1. {(–2, –1), (4, 2), (–8, –4), (6, 3)}
2. {(5, 0), (7, 2), (9, 4), (5, 1)}
3. Graph the relation in the table.
Is the relation a function?
Explain.
x
y
–1
7
0
7
1
7
2
7
Yes; there is only one range value for each domain value.
No; the domain value 5 has two range values, 0 and 1.
Yes; there is one range value for each domain value. Check students’ graphs.
8-1
8-2
Equations With Two VariablesEquations With Two VariablesPRE-ALGEBRA LESSON 8-2PRE-ALGEBRA LESSON 8-2
Identify the data needed: At a fundraiser, $5 was collected fromeach parent and $1 was collected from each child. How muchmoney was raised?
the number of parents and children at the fundraiser
PRE-ALGEBRA LESSON 8-2PRE-ALGEBRA LESSON 8-2
Evaluate each expression for x = 2.
1. 2 + x 2. x – 12 3. 8x – 13 4. 24 ÷ 2x
(For help, go to Lesson 1-3.)
Check Skills You’ll Need
8-2
Equations With Two VariablesEquations With Two Variables
PRE-ALGEBRA LESSON 8-2PRE-ALGEBRA LESSON 8-2
Solutions
1. 2 + x = 2 + 2 2. x – 12 = 2 – 12 = 4 = –10
3. 8x – 13 = 8(2) – 13 4. 24 ÷ 2x = 24 ÷ 2(2) = 16 – 13 = 24 ÷ 4 = 3 = 6
8-2
Equations With Two VariablesEquations With Two Variables
PRE-ALGEBRA LESSON 8-2PRE-ALGEBRA LESSON 8-2
Find the solution of y = 4x – 3 for x = 2.
y = 4x – 3
y = 4(2) – 3 Replace x with 2.
y = 8 – 3 Multiply.
y = 5 Subtract.
A solution of the equation is (2, 5).
Quick Check
8-2
Equations With Two VariablesEquations With Two Variables
PRE-ALGEBRA LESSON 8-2PRE-ALGEBRA LESSON 8-2
The equation a = 5 + 3p gives the price for
admission to a park. In the equation, a is the admission
price for one car with p people in it. Find the price of
admission for a car with 4 people in it.
a = 5 + 3p
a = 5 + 3(4) Replace p with 4.
a = 5 + 12 Multiply.
a = 17 Add.
A solution of the equation is (4, 17). The admission price for one car with 4 people in it is $17.
Quick Check
8-2
Equations With Two VariablesEquations With Two Variables
x 4x – 2 (x, y)
–2 4(–2) – 2 = –8 – 2 = –10 (–2, –10)
0 4(0) – 2 = 0 – 2 = –2 (0, –2)
2 4(2) – 2 = 8 – 2 = 6 (2, 6)
Graph the ordered pairs.
Draw a line through the points.
PRE-ALGEBRA LESSON 8-2PRE-ALGEBRA LESSON 8-2
Graph y = 4x – 2.
Make a table of values to show ordered-pair solutions.
Quick Check
8-2
Equations With Two VariablesEquations With Two Variables
For every value of x, y = –3.
PRE-ALGEBRA LESSON 8-2PRE-ALGEBRA LESSON 8-2
Graph each equation. Is the equation a function?
This is a horizontal line.
a. y = –3 b. x = 4
The equation y = –3 is a function.
This is a vertical line.The equation y = 4 is not a function.
For every value of y, x = 4.
Quick Check
8-2
Equations With Two VariablesEquations With Two Variables
PRE-ALGEBRA LESSON 8-2PRE-ALGEBRA LESSON 8-2
Solve y – x = 3 for y. Then graph the equation.12
Solve the equation for y.
y – x = 312
y = x + 3 Simplify.12
y – x + x = 3 + x Add x to each side.12
12
12
12
8-2
Equations With Two VariablesEquations With Two Variables
PRE-ALGEBRA LESSON 8-2PRE-ALGEBRA LESSON 8-2
(continued)
Graph.Make a table of values.
x x + 3 (x, y)
–2 (–2) + 3 = –1 + 3 = 2 (–2, 2)
0 (0) + 3 = 0 + 3 = 3 (0, 3)
2 (2) + 3 = 1 + 3 = 4 (2, 4)
12121212
Quick Check
8-2
Equations With Two VariablesEquations With Two Variables
PRE-ALGEBRA LESSON 8-2PRE-ALGEBRA LESSON 8-2
Find the solution for each equation for x = 2.1. y = –2x + 5 2. y = 7x 3. y = 3x – 9
Solve each equation for y. Then graph each equation.4. y – 2x = 3 5. 2x + 2y = 8
(2, 1) (2, 14) (2, –3)
y = 2x + 3 y = –x + 4
8-2
Equations With Two VariablesEquations With Two Variables
8-3
Slope and y-interceptSlope and y-interceptPRE-ALGEBRA LESSON 8-3PRE-ALGEBRA LESSON 8-3
Floyd has $35 in the bank. He writes a check for $52 and makes a deposit of $10. What is his new balance?
– $7
Slope and y-interceptSlope and y-interceptPRE-ALGEBRA LESSON 8-3PRE-ALGEBRA LESSON 8-3
Find each difference.
1. –4 – 5 2. 3 – (–2)
3. 6 – 9 4. –1 – (–1)
(For help, go to Lesson 1-6.)
Check Skills You’ll Need
8-3
Slope and y-interceptSlope and y-interceptPRE-ALGEBRA LESSON 8-3PRE-ALGEBRA LESSON 8-3
Solutions
1. –4 – 5 = –4 + (–5) 2. 3 – (–2) = 3 + 2 = –9 = 5
3. 6 – 9 = 6 + (–9) 4. –1 – (–1) = –1 + 1 = – 3 = 0
8-3
Slope and y-interceptSlope and y-interceptPRE-ALGEBRA LESSON 8-3PRE-ALGEBRA LESSON 8-3
Find the slope of each line.
a. b.
slope = = = 4riserun
41 slope = = = –2
riserun
–63
Quick Check
8-3
Slope and y-interceptSlope and y-interceptPRE-ALGEBRA LESSON 8-3PRE-ALGEBRA LESSON 8-3
Find the slope of the line through E(7, 5) and F(–2, 0).
slope = = = =difference in y-coordinatesdifference in x-coordinates
0 – 5–2 – 7
–5–9
59
Quick Check
8-3
Slope and y-interceptSlope and y-interceptPRE-ALGEBRA LESSON 8-3PRE-ALGEBRA LESSON 8-3
Find the slope of each line.
a. b.
slope = –3 – (–3)4 – (–2) = = 0
06
Slope is 0 for a horizontal line.
slope = –1 – 3
–2 – (–2) =–40
Division by zero is undefined. Slope is undefined for a vertical line.
Quick Check
8-3
Slope and y-interceptSlope and y-interceptPRE-ALGEBRA LESSON 8-3PRE-ALGEBRA LESSON 8-3
A ramp slopes from a warehouse door down to a
street. The function y = – x + 4 models the ramp, where x is
the distance in feet from the bottom of the door and y is the
height in feet above the street. Graph the equation.
15
Step 1 Since the y-intercept is 4, graph (0, 4).
Step 3 Draw a line through the points.
Then move 5 units right to graph a second point.
Step 2 Since the slope is – , move
1 unit down from (0, 4).
15
Quick Check
8-3
Slope and y-interceptSlope and y-interceptPRE-ALGEBRA LESSON 8-3PRE-ALGEBRA LESSON 8-3
Find the slope of the line through each pair of points.
1. A(2, 4), B(–2, –4) 2. F(–5, 1), G(0, –9)
3. Identify the slope and y-intercept of y = – x + 3. Then graph the line.
2 –2
43
– ; 343
8-3
8-4
Writing Rules for Linear FunctionsWriting Rules for Linear FunctionsPRE-ALGEBRA LESSON 8-4PRE-ALGEBRA LESSON 8-4
Find the slope of each line.
a. 10x + 5y = 28
– 2
b. 3x +3y = 5
–1
Writing Rules for Linear FunctionsWriting Rules for Linear FunctionsPRE-ALGEBRA LESSON 8-4PRE-ALGEBRA LESSON 8-4
Find the slope of the line through each pair of points.
1. A(3, 1), B(2, 1) 2. S(3, 4), T(1, 2)
3. P(0, –2), Q(0, 2) 4. C(–5, 2), D(4, –1)
(For help, go to Lesson 8-3.)
Check Skills You’ll Need
8-4
Writing Rules for Linear FunctionsWriting Rules for Linear FunctionsPRE-ALGEBRA LESSON 8-4PRE-ALGEBRA LESSON 8-4
Solutions
1. slope = = = 0
2. slope = = = 1
3. slope = = ; slope is undefined.
4. slope = = = –
1 – 12 – 3
0–1
2 – 41 – 3
–2–2
2 – (–2)0 – 0
40
–1 – 24 – (–5)
–39
13
8-4
Writing Rules for Linear FunctionsWriting Rules for Linear FunctionsPRE-ALGEBRA LESSON 8-4PRE-ALGEBRA LESSON 8-4
A long-distance phone company charges its
customers a monthly fee of $4.95 plus 9¢ for each minute of
a long-distance call.
a. Write a function rule that relates the total monthly bill to the number of minutes a customer spent on long-distance calls.
A rule for the function is t(m) = 4.95 + 0.09m.
Words total bill is $4.95 plus 9¢ timesnumber of minutes
Let = the number of minutes.m
Let = total bill, a function of the number of minutes.t( m )
Rule t( m ) = 4.95 + • m0.09
8-4
Writing Rules for Linear FunctionsWriting Rules for Linear FunctionsPRE-ALGEBRA LESSON 8-4PRE-ALGEBRA LESSON 8-4
(continued)
b. Find the total monthly bill if the customer made 90 minutes of long-distance calls.
t(m) = 4.95 + 0.09m
The total monthly bill with 90 minutes of long-distance calls is $13.05.
t(90) = 4.95 + 0.09(90) Replace m with 90.
t(90) = 4.95 + 8.10 Multiply.
t(90) = 13.05 Add.
Quick Check
8-4
Writing Rules for Linear FunctionsWriting Rules for Linear FunctionsPRE-ALGEBRA LESSON 8-4PRE-ALGEBRA LESSON 8-4
Write a rule for the linear function in the table below.
x
2
0
–2
–4
f(x)
3
–5
–13
–21
–8
–8
–8
–2
–2
–2
As the x values decrease by 2, the f(x) values decrease by 8.
So m = = 4.–8–2
When x = 0, f(x) = –5. So b = –5.
A rule for the function is f(x) = 4x – 5.
Quick Check
8-4
Writing Rules for Linear FunctionsWriting Rules for Linear FunctionsPRE-ALGEBRA LESSON 8-4PRE-ALGEBRA LESSON 8-4
Write a rule for the linear function graphed below.
slope = –2 – 20 – 2 =
–4–2 = 2
y-intercept = –2
A rule for the function is f(x) = 2x – 2.
Quick Check
8-4
Writing Rules for Linear FunctionsWriting Rules for Linear FunctionsPRE-ALGEBRA LESSON 8-4PRE-ALGEBRA LESSON 8-4
Write a rule for each function.
1. Sarena earns a salary of $150 a week plus a 10% commission on each sale.
2. 3.
f(x) = 0.10x + 150
x
f(x)
0
0
1
1
2
2
f(x) = x
f(x) = 5x – 3
8-4
8-5
Scatter PlotsScatter PlotsPRE-ALGEBRA LESSON 8-5PRE-ALGEBRA LESSON 8-5
Use a graph with 100 squares to draw and shade squares that spell a 3-letter word. Estimate what fraction of the grid is unshaded.
Check students’ answers.
Scatter PlotsScatter PlotsPRE-ALGEBRA LESSON 8-5PRE-ALGEBRA LESSON 8-5
Write the coordinates of each point.
1. A 2. B 3. C 4. D
(For help, go to Lesson 1-10.)
Check Skills You’ll Need
8-5
Scatter PlotsScatter PlotsPRE-ALGEBRA LESSON 8-5PRE-ALGEBRA LESSON 8-5
Solutions
1. A(–2, 2) 2. B(0, 3)
3. C(–3, 0) 4. D(2, 3)
8-5
Scatter PlotsScatter PlotsPRE-ALGEBRA LESSON 8-5PRE-ALGEBRA LESSON 8-5
The scatter plot shows education and income data.
a. Describe the person represented by the point with coordinates (10, 30).This person has 10 years of education and earns $30,000 each year.
b. How many people have exactly 14 years of education? What are their incomes?The points (14, 50), (14, 80), and (14, 90) have education coordinate 14.The three people they represent earn $50,000, $80,000, and $90,000, respectively.
Quick Check
8-5
Scatter PlotsScatter PlotsPRE-ALGEBRA LESSON 8-5PRE-ALGEBRA LESSON 8-5
Use the table to make a scatter plot of the
elevation and precipitation data.
Atlanta, GA
Boston, MA
Chicago, IL
Honolulu, HI
Miami, FL
Phoenix, AZ
Portland, ME
San Diego, CA
Wichita, KS
Elevation Above Sea Level (ft)
1,050
20
596
18
11
1,072
75
40
1,305
CityMean Annual
Precipitation (in.)
51
42
36
22
56
8
44
10
29Quick Check
8-5
Scatter PlotsScatter PlotsPRE-ALGEBRA LESSON 8-5PRE-ALGEBRA LESSON 8-5
Use the scatter plot below. Is there a positive
correlation, a negative correlation, or no correlation between
temperatures and amounts of precipitation? Explain.
The values show no relationship.
There is no correlation.
Quick Check
8-5
Scatter PlotsScatter PlotsPRE-ALGEBRA LESSON 8-5PRE-ALGEBRA LESSON 8-5
Answer the following questions based on the graph.
1. What do you know about the student at point A?
2. How many students read 3 books per month?
3. Is there a positive correlation, a negative correlation, or no correlation between books read and semester grades?
read 2 books per month, grade: 80
positive correlation
3
8-5
8-6
Problem Solving Strategy: Solve by GraphingProblem Solving Strategy: Solve by GraphingPRE-ALGEBRA LESSON 8-6PRE-ALGEBRA LESSON 8-6
Draw 6 nonlinear points. How many different segments can you draw connecting two points?
15
Problem Solving Strategy: Solve by GraphingProblem Solving Strategy: Solve by GraphingPRE-ALGEBRA LESSON 8-6PRE-ALGEBRA LESSON 8-6
Write a rule for each linear function.
1. 2.
(For help, go to Lesson 8-4.)
8-6
Check Skills You’ll Need
Problem Solving Strategy: Solve by Graphing Problem Solving Strategy: Solve by Graphing PRE-ALGEBRA LESSON 8-6PRE-ALGEBRA LESSON 8-6
Solutions
1. slope = = = 2
y-intercept = 2
A rule for the function is f(x) = 2x + 2.
2. slope = = = –
y-intercept = –3
A rule for the function is f(x) = – x – 3.
6 – 22 – 0
42
–2 – (– 3)–4 – 0
1–4
14
14
8-6
Problem Solving Strategy: Solve by Graphing Problem Solving Strategy: Solve by Graphing PRE-ALGEBRA LESSON 8-6PRE-ALGEBRA LESSON 8-6
Use the data in the table below. Suppose this year
there are 16 wolves on the island. Predict how many moose
are on the island.
Isle Royale Populations
Wolf Moose
14
23
24
22
20
16
700
900
811
1,062
1,025
1,380
Year
1982
1983
1984
1985
1986
1987
Wolf Moose
12
11
15
12
12
13
1,653
1,397
1,216
1,313
1,600
1,880
Year
1988
1989
1990
1991
1992
1993
Wolf Moose
15
16
22
24
14
25
1,800
2,400
1,200
500
700
750
Year
1994
1995
1996
1997
1998
1999
8-6
Step 1 Make a scatter plot by graphing the (wolf, moose) ordered pairs. Use the x-axis for wolves and the y-axis for moose.
Step 2 Sketch a trend line. The line should be as close as possible to each data point. There should be about as many points above the trend line as below it.
Problem Solving Strategy: Solve by Graphing Problem Solving Strategy: Solve by Graphing PRE-ALGEBRA LESSON 8-6PRE-ALGEBRA LESSON 8-6
(continued)
8-6
Look up to find the point on the trend line that corresponds to 16 wolves.
Then look across to the value on the vertical axis, which is about 1,300.
Problem Solving Strategy: Solve by Graphing Problem Solving Strategy: Solve by Graphing PRE-ALGEBRA LESSON 8-6PRE-ALGEBRA LESSON 8-6
(continued)
Step 3 To predict the number of moose when there are 16 wolves, find 16 along the horizontal axis.
There are about 1,300 moose on the island.
Quick Check
8-6
Problem Solving Strategy: Solve by Graphing Problem Solving Strategy: Solve by Graphing PRE-ALGEBRA LESSON 8-6PRE-ALGEBRA LESSON 8-6
Answer each question.
1. When is there no trend line for a scatter plot?
2. What type of correlation will the data have for a scatter plot of ages and heights of all students in your school?
3. Why is it important to say “about” when making a prediction?
when there is no correlation
The prediction is based on a trend line that approximates the locations of the related points.
positive
8-6
8-7
Solving Systems of Linear EquationsSolving Systems of Linear EquationsPRE-ALGEBRA LESSON 8-7PRE-ALGEBRA LESSON 8-7
Find the slope and y-intercept for the equation 1.3x – y + 5 = 0.
Slope is 1.3; y-intercept is 5.
Solving Systems of Linear EquationsSolving Systems of Linear EquationsPRE-ALGEBRA LESSON 8-7PRE-ALGEBRA LESSON 8-7
Graph each equation.
1. y = –x – 4 2. y = 2x – 1 3. –4x = 6y 4. 3x – 2y = 5
(For help, go to Lesson 8-2.)
Check Skills You’ll Need
8-7
Solving Systems of Linear EquationsSolving Systems of Linear EquationsPRE-ALGEBRA LESSON 8-7PRE-ALGEBRA LESSON 8-7
Solutions
1. y = –x + 4 2. y = 2x – 1
8-7
Solving Systems of Linear EquationsSolving Systems of Linear EquationsPRE-ALGEBRA LESSON 8-7PRE-ALGEBRA LESSON 8-7
Solutions (continued)
3. –4x = 6y 4. 3x – 2y = 5
– x = y 3x – 2y – 3x = 5 – 3x
y = – x –2y = –3x + 5
y = x –
23 3
252
46
8-7
Solve the system y = x – 7 and y = 4x + 2 by
graphing.
Solving Systems of Linear EquationsSolving Systems of Linear EquationsPRE-ALGEBRA LESSON 8-7PRE-ALGEBRA LESSON 8-7
Step 1 Graph each line.Step 2 Find the point of intersection.
The lines intersect at one point, (–3, –10). The solution is (–3, –10).
y = x – 7
Check See whether (–3, –10) makes both equations true.
–10 –3 – 7
–10 = –10
y = 4x + 2
–10 4(–3) + 2
–10 = –10
Replace x with – 3and y with –10.
The solution checks.
Quick Check
8-7
Solve each system of equations by graphing.
Solving Systems of Linear EquationsSolving Systems of Linear EquationsPRE-ALGEBRA LESSON 8-7PRE-ALGEBRA LESSON 8-7
a. 27x + 9y = 36; y = 4 – 3x b. 8 = 4x + 2y; 2x + y = 5
The lines are the same line.There are infinitely many solutions.
The lines are parallel.They do not intersect.There is no solution.
Quick Check
8-7
Find two numbers with a sum of 10 and a
difference of 2.
Solving Systems of Linear EquationsSolving Systems of Linear EquationsPRE-ALGEBRA LESSON 8-7PRE-ALGEBRA LESSON 8-7
Step 1 Write equations.Let x = the greater number.Let y = the lesser number.
Equation 1 Sum is 10.x + y = 10
Equation 2 Difference is 2.x – y = 2
Step 2 Graph the equations.The lines intersect at (6, 4).The numbers are 6 and 4.
8-7
(continued)
Solving Systems of Linear EquationsSolving Systems of Linear EquationsPRE-ALGEBRA LESSON 8-7PRE-ALGEBRA LESSON 8-7
Check Since the sum of 6 and 4 is 10 and the difference of 6 and 4 is 2, the answer is correct.
Quick Check
8-7
Solving Systems of Linear EquationsSolving Systems of Linear EquationsPRE-ALGEBRA LESSON 8-7PRE-ALGEBRA LESSON 8-7
Solve each system by graphing.
1. y = –x – 4 and y = 4x + 1
2. y = –3x + 12 and 18x + 6y = 42
3. Find two numbers with a sum of 15 and a
difference of 1. Show your work.
(–1, –3)
(8, 7)
no solution
8-7
8-8
Graphing Linear InequalitiesGraphing Linear InequalitiesPRE-ALGEBRA LESSON 8-8PRE-ALGEBRA LESSON 8-8
Use two of the given digits to make the equation true. 2, 3, 4 2 3
8 ÷ = 10
3 4
?
?
Graphing Linear InequalitiesGraphing Linear InequalitiesPRE-ALGEBRA LESSON 8-8PRE-ALGEBRA LESSON 8-8
Is the given value of x a solution of the inequality? Explain.
1. x + 3 –2; x = –5 2. 5 – x < 4; x = 1
3. –2 – 2x 6; x = –4 4. 4x + 1 > –7; x = –2
(For help, go to Lesson 2-9.)
<–
<–
Check Skills You’ll Need
8-8
Graphing Linear InequalitiesGraphing Linear InequalitiesPRE-ALGEBRA LESSON 8-8PRE-ALGEBRA LESSON 8-8
Solutions
1. yes; x + 3 –2; x = –5 2. no; 5 – x < 4; x = 1 –5 + 3 –2 5 – 1 < 4 –2 –2 true 4 < 4 false
3. yes; –2 – 2x 6; x = –4 4. no; 4x + 1 > –7; x = –2 –2 – 2(–4) 6 4(–2) + 1 > – 7 –2 + 8 6 –8 + 1 > –7 6 6 true –7 > –7 false
>–
>–
>–
<–<–<–<–
8-8
Graph each inequality on a coordinate plane.
Graphing Linear InequalitiesGraphing Linear InequalitiesPRE-ALGEBRA LESSON 8-8PRE-ALGEBRA LESSON 8-8
a. y > 2x + 1
Step 1 Graph the boundary line.Points on the boundary line do not make y > 2x + 1 true. Use a dashed line.
8-8
(continued)
Graphing Linear InequalitiesGraphing Linear Inequalities
Step 2 Test a point not on the boundary line.Test (0, 0) in the inequality.
Since the inequality is false for (0, 0), shade the region that does not contain (0, 0).
y > 2x + 10 2(0) + 1 Substitute.0 0 + 1
0 > 1 false
>?
>?
PRE-ALGEBRA LESSON 8-8PRE-ALGEBRA LESSON 8-8
8-8
(continued)
Graphing Linear InequalitiesGraphing Linear Inequalities
b. y 3x – 2<–Step 1 Graph the boundary line.
Points on the boundary line make y 3x – 2 true. Use a solid line.
PRE-ALGEBRA LESSON 8-8PRE-ALGEBRA LESSON 8-8
8-8
(continued)
Graphing Linear InequalitiesGraphing Linear Inequalities
Step 2 Test a point not on the boundary line.Test (3, 0) in the inequality.
Since the inequality is true for (3, 0), shade the region containing (3, 0).
y 3x – 20 3(3) – 2 Substitute.0 9 – 20 7 true
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PRE-ALGEBRA LESSON 8-8PRE-ALGEBRA LESSON 8-8
Quick Check
8-8
Cashews cost $2/lb. Pecans cost $4/lb. You plan
to spend no more than $20. How many pounds of each can
you buy?
Graphing Linear InequalitiesGraphing Linear Inequalities
Step 1 Write an inequality.
Wordscost of
cashewsplus
cost of pecans
is at most
twenty dollars
Let = number of pounds of pecans.x
+Inequality 2y 4x <– 20
Let = number of pounds of cashews.y
PRE-ALGEBRA LESSON 8-8PRE-ALGEBRA LESSON 8-8
8-8
(continued)
Graphing Linear InequalitiesGraphing Linear Inequalities
Step 3 Graph y = –2x + 10 in Quadrant I since weight is not negative.
PRE-ALGEBRA LESSON 8-8PRE-ALGEBRA LESSON 8-8
Step 2 Write the equation of the boundary line in slope-intercept form.
2y + 4x 20
y –2x + 10
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y = –2x + 10
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8-8
(continued)
Graphing Linear InequalitiesGraphing Linear Inequalities
Step 4 Test (1, 1).y –2x + 101 –2(1) + 101 8The inequality is true. (1, 1) is a solution.
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Step 5 Shade the region containing (1, 1).
The graph shows the possible solutions. For example, you could buy 1 pound of pecans and 5 pounds of cashews.
PRE-ALGEBRA LESSON 8-8PRE-ALGEBRA LESSON 8-8
Quick Check
8-8
Step 1 Graph y x + 1 on a coordinate plane. Shade in red.
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Solve the system y x + 1 and y < 2x + 3 by graphing.
Graphing Linear InequalitiesGraphing Linear Inequalities
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Step 2 Graph y < 2x + 3 on the same coordinate plane. Shade in blue.
PRE-ALGEBRA LESSON 8-8PRE-ALGEBRA LESSON 8-8
8-8
(continued)
Graphing Linear InequalitiesGraphing Linear Inequalities
The solutions are the coordinates of all the points in the region that is shaded in both colors.
Check See whether the solution (2, 5) makes both of the inequalities true.y x + 1 5 2 + 1 Replace x with 2 and y with 5.5 3y The solution checks.
y < 2x + 3 5 2(2) + 3 Replace x with 2 and y with 5.5 < 7y The solution checks.
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PRE-ALGEBRA LESSON 8-8PRE-ALGEBRA LESSON 8-8
Quick Check
8-8
Graphing Linear InequalitiesGraphing Linear InequalitiesPRE-ALGEBRA LESSON 8-8PRE-ALGEBRA LESSON 8-8
Tell whether the ordered pair is a solution of the inequality. Justify your response.
1. y < 4x; (3, 0) 2. y > –x – 1; (2, –3)
3. y 4x – 2; (–1, 1) 4. Solve the system y 3x + 1 and y > –2x + 2 by graphing.
Yes; 0 < 12.
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No; 1 6.–<
No; –3 > –3.
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