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2. Graph of Linear Equations11
Graph of Linear Equations
11.1 Graphs and Applications of Linear Equations
11.2 More with Graphing and Intercepts
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3. GRAPHS and APPLICATIONS OF LINEAR EQUATIONS
11.1
GRAPHS and APPLICATIONS OF LINEAR EQUATIONS a
Plot points associated with ordered pairs of numbers, determine the quadrant in which a point lies.
Find the coordinates of a point on a graph.
Determine whether an ordered pair is a solution of an equation with two variables.
Graph linear equations of the type y = mx + b and Ax + By = C, identifying the y-intercept.
Solve applied problems involving graphs of linear equations. b c d e
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4. Objective a
Plot points associated with ordered pairs of numbers, determine the quadrant in which a point lies.
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5. Points and Ordered Pairs
To graph, or plot, points we use two perpendicular number lines called axes. The point at which the axes cross is called the origin. Arrows on the axes indicate the positive directions.
Consider the pair (2, 3). The numbers in such a pair are called the coordinates. The first coordinate in this case is 2 and the second coordinate is 3.
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6. Points and Ordered Pairs continued
To plot the point (2, 3) we start at the origin.
Move 2 units in the horizontal direction.
The second number 3, is positive. We move 3 units in the vertical direction (up).
Make a “dot” and label the point.
(2, 3)
(2, 3)
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7. Example A Plot the point (4, 3).
Solution
Starting at the origin, we move 4 units in the negative horizontal direction.
The second number, 3, is positive, so we move 3 units in the positive vertical direction (up).
(4, 3)
3 units up
4 units left
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8. The horizontal and vertical axes divide the plane into four regions, or quadrants.
In which quadrant is the point (3, 4) located? IV
In which quadrant is the point (3, 4) located?
III
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9. Objective Find the coordinates of a point on a graph. b
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10. Example B Find the coordinates of points A, B, C, D, E, F, and G.
Solution
Point A is 5 units to the right of the origin and 3 units above the origin. Its coordinates are (5, 3). The other coordinates are as follows:
B: (2, 4)
C: (3, 4)
D: (3, 2)
E: (2, 3)
F: (3, 0)
G: (0, 2) A B C D E F G
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11. Objective c
Determine whether an ordered pair is a solution of an equation with two variables.
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12. Example C Determine whether each of the following pairs is a solution of 4y + 3x = 18: a) (2, 3); b)(1, 5).
Solution
a) We substitute 2 for x and 3 for y.
4y + 3x = 18
4•3 + 3•2 | 18
| 18
18 = 18 True
b) We substitute 1 for x and 5 for y.
4•5 + 3•1 | 18
| 18
23 = 18 False
Since 18 = 18 is true, the pair (2, 3) is a solution.
Since 23 = 18 is false, the pair (1, 5) is not a solution.
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13. Objective d
Graph linear equations of the type y = mx + b and Ax + By = C, identifying the y-intercept.
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14. Graph of An Equation
The graph of an equation is a drawing that represents all its solutions.
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15. To Graph a Linear Equation
1. Select a value for one variable and calculate the corresponding value of the other value. Form an ordered pair using alphabetical order as indicated by the variables.
2. Repeat step (1) to obtain at least two other ordered pairs. Two points are essential to determine a straight line. A third ordered point serves as a check.
3. Plot the ordered pairs and draw a straight line passing through the points.
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16. Example D Graph y = 3x
Solution Find some ordered pairs that are solutions. We choose any number for x and then determine y by substitution. y x
y = 3x
(x, y) 2 1 1 2
(2, 6) 6
(2, 6) 3
(1, 3)
(1, 3)
(0, 0)
(0, 0) 3
(1, 3) 6
(2, 6)
1. Choose x.
2. Compute y.
4. Plot the points.
3. Form the ordered pair (x, y).
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17. Example E Graph y = 4x + 1 Solution
We select convenient values for x and compute y, and form an ordered pair.
If x = 2, then y = 4(2) + 1 = 7 and (2, 7) is a solution.
If x = 0, then y = 4(0) + 1 = 1 and (0, 1) is a solution.
If x = 2, then y = 4(2) + 1 = 9 and (2, 9) is a solution.
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18. Solution (continued) Results are often listed in a table.
(1) Choose x.
(2) Compute y.
(3) Form the pair (x, y).
(4) Plot the points. x y
(x, y) 2 7
(2, 7) 1
(0, 1) 2 9
(2, 9)
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19. Solution (continued)
Note that all three points line up. If they didn’t we would know that we had made a mistake.
Finally, use a ruler or other straightedge to draw a line.
Every point on the line represents a solution of
y = 4x + 1
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20. y-Intercept
The graph of the equation y = mx + b passes through the y-intercept (0, b).
(0, b)
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21. Example F Graph Solution Complete a table of values.y
(x, y) 4
(4, 4) 3
(0, 3) 4 2
(4, 2)
y-intercept
(4, 4)
(4, 2)
We see that (0, 3) is a solution. It is the y-intercept.
(0, 3) is the y-intercept.
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22. Example G Graph and identify the y-intercept. Solution
To find an equivalent equation in the form y = mx + b, we solve for y:
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23. continued Complete a table of values. x y (x, y) (0, 0) 2 3 (2, 3)
(0, 0) 2 3
(2, 3) 2 3
(2, 3)
 y-intercept
y-intercept
(2, 3)
(2, 3)
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24. Objective Solve applied problems involving graphs of linear equations.
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25. Example H
The cost c, in dollars, of shipping a FedEx Priority Overnight package weighing 1 lb or more a distance of 1001 to 1400 mi is given by c = 2.8w where w is the package’s weight in pounds.
Graph the equation and then use the graph to estimate the cost of shipping a 10 ½-pound package.
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26. Solution Select values for w and then calculate c. c = 2.8w + 21.05
If w = 2, then c = 2.8(2) = 26.65
If w = 4, then c = 2.8(4) = 32.25
If w = 8, then c = 2.8(8) = 43.45 w c 2
26.65 4
32.25 8
43.45
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27. Solution (continued) Plot the points.
To estimate an 10 ½ pound package, we locate the point on the line that is above 10 ½ and then find the value on the c-axis that corresponds to that point.
The cost of shipping an 10 ½ pound package is about $51.00.
Weight (in pounds)
Mail cost (in dollars)
10 ½ pounds
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28. More with Graphing and Intercepts
11.2
More with Graphing and Intercepts a
Find the intercepts of a linear equation, and graph using intercepts.
Graph equations equivalent to those of the type x = a and y = b. b
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29. Objective a
Find the intercepts of a linear equation, and graph using intercepts.
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30. Intercepts
The y-intercept is (0, b). To find b, let x = 0 and solve the original equation for y.
The x-intercept is (a, 0). To find a, let y = 0 and solve the original equation for x.
(0, b)
(a, 0)
y-intercept
x-intercept
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31. Example A Consider 5x + 2y = 10. Find the intercepts
Example A Consider 5x + 2y = 10. Find the intercepts. Then graph the equation using the intercepts.
Solution
To find the y-intercept, we let x = 0 and solve for y:
5 • 0 + 2y = 10
2y = 10
y = 5
The y-intercept is (0, 5).
To find the x-intercept, we let y = 0 and solve for x.
5x + 2• 0 = 10
5x = 10
x = 2
The x-intercept is (2, 0).
Replacing x with 0
Replacing y with 0
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32. continued
We plot these points and draw the line, or graph. A third point should be used as a check. We substitute any convenient value for x and solve for y. If we let x = 4, then
5 • 4 + 2y = 10
y = 10
2y = 10
y = 5
5x + 2y = 10
x-intercept (2, 0)
y-intercept (0, 5) x y 5 2 4 5
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33. Example B Graph y = 4x Solution
We know that (0, 0) is both the x-intercept and y-intercept. We calculate two other points and complete the graph, knowing it passes through the origin.
y = 4x
(1, 4)
(1, 4)
(0, 0) x y 1 4 1 4
x-intercept y-intercept
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34. Objective b
Graph equations equivalent to those of the type x = a and y = b.
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35. Example C Graph y = 2 y = 2 Solution
We regard the equation y = 2 as 0 • x + y = 2. No matter what number we choose for x, we find that y must equal 2.
y = 2
Choose any number for x.
y must be 2. x y
(x, y) 2
(0, 2) 4
(4, 2) 4
(4 , 2)
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36. continued y = 2
Solution
When we plot the ordered pairs (0, 2), (4, 2) and (4, 2) and connect the points, we obtain a horizontal line.
Any ordered pair of the form (x, 2) is a solution, so the line is parallel to the x-axis with y-intercept (0, 2).
y = 2
(4, 2)
(0, 2)
(4, 2)
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37. Example D Graph x = 2 Solution
We regard the equation x = 2 as x + 0 • y = 2. We make up a table with all 2 in the x-column.
x = 2
x must be 2. x y
(x, y) 2 4
(2, 4) 1
(2, 1) 4
(2, 4)
Any number can be used for y.
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38. continued x = 2 Solution
When we plot the ordered pairs (2, 4), (2, 1), and (2, 4) and connect them, we obtain a vertical line.
Any ordered pair of the form (2, y) is a solution. The line is parallel to the y-axis with x-intercept (2, 0).
x = 2
(2, 4)
(2, 4)
(2, 1)
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39. Horizontal and Vertical Lines
The graph of y = b is a horizontal line. The y-intercept is (0, b).
The graph of x = a is a vertical line. The x-intercept is (a, 0).
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