Pre-AlgebraPre-Algebra
Pre-AlgebraPre-Algebra
The Coordinate PlaneThe Coordinate Plane
Lesson 1-10
Write the coordinates of point G. In which
quadrant is point G located?
Point G is located 2 units to the left of the y-axis.So the x-coordinate is –2.
The point is 3 units below the x-axis.
So the y-coordinate is –3.
The coordinates of point G are (–2, –3). Point G is located in Quadrant III.
Pre-AlgebraPre-Algebra
The Coordinate PlaneThe Coordinate Plane
Lesson 1-10
Graph point M(–3, 3).
Pre-AlgebraPre-Algebra
Relations and FunctionsRelations and Functions
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.
Pre-AlgebraPre-Algebra
Relations and FunctionsRelations and Functions
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.
Pre-AlgebraPre-Algebra
Relations and FunctionsRelations and Functions
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.
Pre-AlgebraPre-Algebra
Relations and FunctionsRelations and Functions
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.
Pre-AlgebraPre-Algebra
Relations and FunctionsRelations and Functions
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).
Pre-AlgebraPre-Algebra
Relations and FunctionsRelations and Functions
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.
Pre-AlgebraPre-AlgebraLesson 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).
Equations With Two VariablesEquations With Two Variables
Pre-AlgebraPre-AlgebraLesson 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.
Equations With Two VariablesEquations With Two Variables
Pre-AlgebraPre-Algebra
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.
Lesson 8-2
Graph y = 4x – 2.
Make a table of values to show ordered-pair solutions.
Equations With Two VariablesEquations With Two Variables
Pre-AlgebraPre-Algebra
For every value of x, y = –3.
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.
Equations With Two VariablesEquations With Two Variables
Pre-AlgebraPre-AlgebraLesson 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
Equations With Two VariablesEquations With Two Variables
Pre-AlgebraPre-AlgebraLesson 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
Equations With Two VariablesEquations With Two Variables
Pre-AlgebraPre-Algebra
Slope and y-interceptSlope and y-intercept
Lesson 8-3
Pre-AlgebraPre-Algebra
Slope and y-interceptSlope and y-intercept
Lesson 8-3
Find the slope of each line.
a. b.
slope = = = 4riserun
41 slope = = = –2
riserun
–63
Pre-AlgebraPre-Algebra
Slope and y-interceptSlope and y-intercept
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
Pre-AlgebraPre-Algebra
Slope and y-interceptSlope and y-intercept
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.
Pre-AlgebraPre-Algebra
Slope and y-interceptSlope and y-intercept
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
Pre-AlgebraPre-Algebra
Writing Rules for Linear FunctionsWriting Rules for Linear Functions
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
Pre-AlgebraPre-Algebra
Writing Rules for Linear FunctionsWriting Rules for Linear Functions
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.
Pre-AlgebraPre-Algebra
Writing Rules for Linear FunctionsWriting Rules for Linear Functions
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.
Pre-AlgebraPre-Algebra
Writing Rules for Linear FunctionsWriting Rules for Linear Functions
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.
Pre-AlgebraPre-Algebra
Scatter PlotsScatter Plots
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.
Pre-AlgebraPre-Algebra
Scatter PlotsScatter Plots
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
29
Pre-AlgebraPre-Algebra
Scatter PlotsScatter Plots
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.
Pre-AlgebraPre-Algebra
Problem Solving Strategy: Solve by Graphing Problem Solving Strategy: Solve by Graphing
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
Pre-AlgebraPre-Algebra
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
Lesson 8-6
(continued)
Pre-AlgebraPre-Algebra
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
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.
Pre-AlgebraPre-Algebra
Solve the system y = x – 7 and y = 4x + 2 by
graphing.
Solving Systems of Linear EquationsSolving Systems of Linear Equations
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.
Pre-AlgebraPre-Algebra
Solve each system of equations by graphing.
Solving Systems of Linear EquationsSolving Systems of Linear Equations
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.
Pre-AlgebraPre-Algebra
Find two numbers with a sum of 10 and a
difference of 2.
Solving Systems of Linear EquationsSolving Systems of Linear Equations
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.
Pre-AlgebraPre-Algebra
(continued)
Solving Systems of Linear EquationsSolving Systems of Linear Equations
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.
Pre-AlgebraPre-Algebra
Graph each inequality on a coordinate plane.
Graphing Linear InequalitiesGraphing Linear Inequalities
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.
Pre-AlgebraPre-Algebra
(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
>?
>?
Lesson 8-8
Pre-AlgebraPre-Algebra
(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.<–
Lesson 8-8
Pre-AlgebraPre-Algebra
(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
<–
<–
<–?
<–?
Lesson 8-8
Pre-AlgebraPre-Algebra
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
Lesson 8-8
Pre-AlgebraPre-Algebra
(continued)
Graphing Linear InequalitiesGraphing Linear Inequalities
Step 3: Graph y = –2x + 10 in Quadrant I since weight is not negative.
Lesson 8-8
Step 2: Write the equation of the boundary line in slope-intercept form.
2y + 4x 20
y –2x + 10
<–
y = –2x + 10
<–
Pre-AlgebraPre-Algebra
(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.
<–<–?
<–
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.
Lesson 8-8
Pre-AlgebraPre-Algebra
Step 1: Graph y x + 1 on a coordinate plane. Shade in red.
>–
Solve the system y x + 1 and y < 2x + 3 by graphing.
Graphing Linear InequalitiesGraphing Linear Inequalities
>–
Step 2: Graph y < 2x + 3 on the same coordinate plane. Shade in blue.
Lesson 8-8
Pre-AlgebraPre-Algebra
(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.
>–>–?
>–
<?
Lesson 8-8