Download 6.2 | Graphing Linear Equations

198
Chapter 6 | Linear systems
7. Three vertices of a square, ABCD, have points A (-3, 3), B (1, 3), and C (1, -1). Find the coordinates of the 4th
vertex, D.
8. Three vertices of a rectangle, PQRS, have points P (-3, 4), Q (6, 4), and R (6, -1). Find the coordinates of the 4th
vertex, S.
9. A vertical line has a length of 7 units and the coordinates of one end of the line is (1, 5). Find the possible
coordinates of the other end of the line.
10. A horizontal line has a length of 6 units and the coordinates of one end of the line is (-1, 3). Find the possible
coordinates of the other end of the line.
6.2 |
Graphing Linear Equations
Consider a simple linear equation with two variables.
2x - y + 3 = 0 This equation is in the form of Ax + By + C = 0 and is called the
equation in 'standard' form, where A = 2, B = -1, and C = 3.
This equation in standard form, can be rearranged to:
y = 2x + 3 This equation is in the form of y = mx + b and is called the equation
in 'slope-intercept' form, where the slope, m = 2, and the y-intercept,
b = 3.
It is not possible to solve the equation because it has two variables, x and y. However, it is possible
to solve for a set of values by replacing one variable (either x or y) with any number and then
solving for the value of the other variable. Therefore, if an equation has two variables, then the
solutions are pairs of values.
Consider the same equation
y = 2x + 3 Substituting for x = 1 in the equation, we obtain,
y = 2(1) + 3 = 5.
Therefore, x = 1 and y = 5 is one of the solutions to the equation; i.e., (1, 5) is a point on the
line that represents the equation.
Similarly, we can obtain any number of points on the line by using different values for x.
y = 2x + 3 Substituting for x = 2 in the equation, we obtain,
y = 2(2) + 3 = 7
Therefore, x = 2 and y = 7 is another solution to the equation; i.e., (2, 7) is another point on the line.
Graphing a linear equation means representing the solution of the equation on a graph.
Conversely, any point on the line is a solution to the linear equation.
­­­­­­
Graphing Linear Equations Using a Table of Values
Steps in drawing the graph of a linear equation using a table of values:
Drawing a linear
graph requires only
two points. However,
at least 3 points will
ensure that the formed
line truly represents
the given linear
equation.
Create a table of values by choosing a few different values for the variable x.
Compute the corresponding value for the variable y.
Form the ordered pairs (x, y).
Repeat 2 more times to create 3 ordered pairs.
Plot the ordered pairs (points) on the coordinate system, choosing a scale.
5. Join the points in a straight line.
1.
2.
3.
4.
6. Label the graph with the equation of the line.
Chapter 6 | Linear Systems
For example, let x = 3,
y = 2x + 3
x
y
(x, y)
3
9
(3, 9)
2
7
(2, 7)
1
5
(1, 5)
0
3
(0, 3)
y = 2x + 3
= 2(3) + 3
=6+3=9
(3, 9) is a point on the line.
Exhibit 6.2(a): Graphing a Linear Equation
Using a Table of Values
Graphing Linear Equations Using the x-intercept and
the y-intercept
We may use the x-intercept and y-intercept as two points to draw a linear graph and use a 3rd point to
test the drawn line.
Let us consider another equation y = 3x + 9, as shown in Exhibit
6.2(b).
x-intercept: it is the point at which the line crosses the X-axis and
where the y-coordinate is zero.
Substituting y = 0 in the above equation and solving for x, we obtain,
0 = 3x + 9, thus, x = -3.
Therefore, (-3, 0) is the x-intercept.
y-intercept: it is the point at which the line crosses the Y-axis and
where the x-coordinate is zero. Substituting x = 0 in the above
equation and solving for y, we obtain,
y = 3(0) + 9, thus, y = 9.
Therefore, (0, 9) is the y-intercept.
To test the line, let us find another point on this line.
Let x = -1.
Substituting this in the equation, we obtain,
Exhibit 6.2(b): Graphing a Linear
Equation Using the x-intercept and
the y-intercept.
199
y = 3(-1) + 9 = 6
Since the point (-1, 6), when plotted on the graph, falls on the line, it
verifies that the plotted line represents the equation.
­­­ raphing Linear Equations Using the Slope and
G
the y-intercept
As noted earlier, a linear equation in the form of y = mx + b is known as the equation in
slope-intercept form, where 'm' is the slope and 'b' is the y-intercept.
If the equation is in the standard form Ax + By + C = 0, it can be rearranged to represent the
slope-intercept form as follows:
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Chapter 6 | Linear systems
Ax + By + C = 0
This is of the form
Where, By = -Ax - C
y = - Ax- C
B
B
y = mx + b
C
m = - A and b = - B
B
The Slope and y-intercept of a Line
The Slope (m) is the steepness of the line relative to the X-axis. It is the ratio of the change in value
of y (called 'rise') to the corresponding change in value of x (called 'run').
If P (x1, y1) and Q (x2, y2) are two different points on a line, then the slope of the line between the
points PQ is:
y2 - y1
Change in y value
3y
Rise
m = Change in x value = 3 x =
= x -x
Run
2
1
as illustrated below, in Exhibit 6.2(c) and Exhibit 6.2(d):
Exhibit 6.2(c): Coordinates of Points P and Q
Exhibit 6.2(d): Rise and Run Between Points P and Q
As defined earlier, the y-intercept (b) is the point where the line crosses the Y-axis and is
calculated by substituting zero for the x-coordinate of the equation.
Example 6.2(a)
Finding the Slope and y-intercept of an Equation and Graphing the Equation
Find the slope and y-intercept of the linear equation -2x + 3y - 12 = 0 and graph the equation.
Solution
Rearranging, -2x + 3y - 12 = 0
3y = 2x + 12
Therefore,
y= 2x+ 4
3
This is in the form y = mx + b,
2
where, the y-intercept, b = 4, and the slope m = .
3
Therefore, (0, 4) is a point on the line and slope,
Change in y value
m = Change in x value
Rise
Run
2
= 3
=
Chapter 6 | Linear Systems
Solution
continued
Representing this on a graph:
(i) First, plot the y-intercept (0, 4).
(ii) From this point, move 3 units to the right and
then move 2 units up to locate the new point (3, 6). This is the same as moving 2 units
up, then 3 units to the right to locate the new point (3, 6).
(iii) Similarly, from the point (3, 6) move 3 units to the right
and 2 units up to locate another point (6, 8).
(iv) Draw the line through these points to graph the equation.
Or,
(i) From the y-intercept (0,4), move 3 units to the left and then move 2 units down to locate
a point (-3, 2).
(ii) Similarly, by using the same order for both the rise and the run find another point (-6,0).
(iii) Draw the line through these points to graph the equation.
Note: All the points will lie on the same line.
Example 6.2(b)
Graphing a Linear Equation in the Slope-Intercept Form
Graph the equation y = - 3 x - 2 .
4
Solution
The equation is in the form y = mx + b.
Change in y value
Therefore, the slope, m = Change in x value
=
Rise
Run
= -
3
4
3
m = - or = 3
4
-4
The y-intercept = -2, which gives the point (0, -2).
First plot the point (0, -2). Then, using the slope,
3
m = - , from the point (0, -2), move 4 units to the
4
right and 3 units down to locate the new point, (4, -5).
Alternatively, first plot the point (0, -2). Then, using
the slope, m = 3 , from the point (0, -2), move 4
-4
units to the left and 3 units up to get another point on
the line,(-4, 1).
Draw a line through these points to graph the equation.
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Chapter 6 | Linear systems
Note: The sign of the coefficient 'm' of the equation y = mx + b indicates the direction of the line.
If the sign of the coefficient 'm' is positive, then the
line slopes upwards to the right, as illustrated in Exhibit
6.2(e).
Exhibit 6.2(e): Slope of a Line When 'm' is Positive
■■ The slope of a line parallel to the X-axis is zero; i.e.,
If the sign of the coefficient of 'm' is negative,
then the line slopes downwards to the right, as
illustrated in Exhibit 6.2(f).
Exhibit 6.2(f): Slope of a Line When 'm' is Negative
■■The slope of a line parallel to the Y-axis is undefined;
if the slope of a line is zero, then the line is horizontal,
as shown in Exhibit 6.2(g).
i.e., if the slope of a line is undefined, then the line is
vertical, as shown in Exhibit 6.2(h).
For example, in the equation, y = 3, (y = 0x + 3) the
slope is zero and the value of the y-coordinate is 3 for all
values of x. Therefore, the line is horizontal and passes
through (0, 3), as illustrated in Exhibit 6.2(g).
For example, in the equation, x = 2, the value of the
x-coordinate is 2 for all values of y. Therefore, the line
is vertical and passes through (2, 0), as illustrated in
Exhibit 6.2(h).
Exhibit 6.2(g): Slope of a Line Parallel
to the X-axis
Exhibit 6.2(h): Slope of a Line Parallel
to the Y-axis
Chapter 6 | Linear Systems
An equation with a y-intercept equal to 0 will have the
graph passing through the origin.
For example, in the equation, y = 2x,
2
the slope (m) = 2 = , and the y-intercept = 0.
1
Therefore, the line passes through (0, 0), as illustrated
in Exhibit 6.2(i).
Exhibit 6.2(i): Slope of a Line When the y-intercept is 0
Example 6.2(c)
Finding the Slope and the Equation of a Line Given Two Points
Find the equation of a line that passes through points (3, 2) and (4, 5).
Solution
Step 1: Calculate the slope
m=
y2 - y1
Change in y value
3
= x - x = 5- 2 =
1
4
3
Change in x value
2
1
Step 2: Replace m with the calculated slope
Substituting for m in the slope-intercept equation y = mx + b, we obtain, y = 3x + b.
Step 3: Substitute one ordered pair into the equation to solve for b
Substituting the coordinate (3,2) into the above equation to solve for b,
2 = 3(3) + b, b = -7.
Step 4: Write the equation y = mx + b with the calculated values of m and b
Therefore, the equation of the line is y = 3x - 7.
Example 6.2(d)
Finding the Equation of a Line Given the Slope and One Point
Find the equation of a line having a slope of -2 and passing through (3, 5).
Solution
m = -2,
point = (3, 5)
Substituting for m in the slope-intercept equation y = mx + b, we obtain, y = -2x + b.
Substituting the coordinates of the given point (3, 5) in the above equation to solve for b,
5 = -2(3) + b, b = 5 + 6 = 11.
Therefore, the equation of the line is y = -2x + 11.
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Chapter 6 | Linear systems
Example 6.2(e)
Finding the Equation of a Line in the
Standard Form Given a Graph
Find the equation of the line that is plotted in
the graph shown:
Solution
Start by choosing any 2 points (with integer
coordinates) on the line: e.g., (1, 6) and (4, 10).
The slope of the line is,
y -y
4
10 - 6
m = x 2 x1 =
= .
3
4- 1
2- 1
Let the equation be y = mx + b.
4
Therefore, y = 3 x + b .
Substituting the coordinates of one of the points
(1, 6) into the above equation, we obtain,
4 ^1 h + b
3
14
Solving for b,
b = 6 - 4 ^1 h =
3
3
Therefore, the equation of the line is
6=
4
14
y= 3 x+ 3 Multiplying both sides by 3, we obtain,
This is in the slope-intercept form.
3y = 4x + 14.
Rearranging,4x - 3y + 14 = 0
Therefore, the equation of the line in standard form is 4x - 3y + 14 = 0.
A higher value for a slope indicates a steeper incline.
Chapter 6 | Linear Systems
6.2 |
Exercises Answers to the odd-numbered problems are available at the end of the textbook
1. For the equation 2x + 3y = 18, find the missing value of the following ordered pairs:
a. (3, ?)
b. (-6, ?)
c. (0, ?)
d. (?, 0)
e. (?, -4)
f. (?, 2)
2. For the equation x + 5y = 20, find the missing value of the following ordered pairs:
a. (0, ?)
b (-15, ?)
c. (5, ?)
d. (?, 6)
e. (?,-3)
f. (?, 0)
Graph the following equations using a table of values:
3. a. y = x + 3
b. y = -5x + 1
4. a. y = 3x + 2
b. y = -2x + 3
5. a. 2x + y + 1 = 0
b. 2x - y - 3 = 0
6. a. 4x + y + 2 = 0
b. x - y - 1 = 0
7. Find the x-intercepts and y-intercepts for the following equations:
a. 3x + y = -2
b. x + y - 3 = 4
c. y = 2x + 4
8. Find the x-intercepts and y-intercepts for the following equations:
a. 5x + y = -3
b. x + y - 4 = 7
c. y = 4x + 1
9. Point 'A' is in the 3rd quadrant and Point 'B' is in the 1st quadrant.
Find the sign of the slope of the line AB.
10. Point 'C' is in the 4th quadrant and Point 'D' is in the 2nd quadrant.
Find the sign of the slope of the line CD.
Find the slope and y-intercept of the following equations and graph the equations:
11. 2x - 3y - 18 = 0
12. 5x - 2y + 10 = 0
13. -4x + 7y - 21 = 0
14. -7x + 8y - 32 = 0
Find the equation of the line that passes through the following points:
15. (1, 2) and (5, 2)
16. (5, 0) and (4, 5)
17. (-3, -5) and (3, 1)
18. (-4, -7) and (5, 2)
Find the equation of the following lines that have:
19. Slope = 1 and passing through (2, 6).
20. Slope = -5 and passing through (3, -2).
21. Slope = 2 and passing through the origin.
22. Slope = 4 and passing through the origin.
23. x-intercept = 4 and y-intercept = -5
24. x-intercept = -3 and y-intercept = 3.
Find the slope of the line passing through:
25. (2, 1) and (6, 1)
26. (-5, 4) and (3, -1)
27. (-6, 4) and (2, 4)
28. (5, 6) and (5, -4)
29. The slope of a line is 3. The line passes through A (4, y) and B (6, 8). Find y.
30. The slope of a line is 2. The line passes through A (x, 8) and B (2, 4). Find x.
31. Points A (2, 3), B (6, 5), and C (10, y) are on a line. Find y.
32. Points D (3, 2), E (6, 5), and F (x, 1) are on a line. Find x.
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Chapter 6 | Linear systems
Find the equation of the lines (in standard form) for the graphs shown in Problems 33 and 34.
33. 34.
35. Hanna charges $20 for the first shirt that you purchase at her store. However, for every additional shirt, she charges
$15. Write an equation that shows the relationship between her total revenue (y) and number of shirts sold (x). If
you plot a graph of y vs. x, what is the y-intercept and slope of the line that represents the equation?
36. An online music store charges $2 for the first song that you download. For every additional song you download, you
will be charged only $1.50. Write an equation that shows the relationship between the total revenue (y) and the number
of songs sold (x). If you plot a graph of y vs. x, what is the y-intercept and slope of the line that represents the equation?
6.3 |
Solving Systems of Linear Equations with
Two Variables Graphically
Two or more equations analyzed together are called systems of equations. In this section, we will
be analyzing two linear equations with two variables.
The solution to a system of two equations with two variables is an ordered pair of numbers
(coordinates) that satisfy both equations.
If the graph of an equation is linear, then every point on the line is a solution to that equation. If
we graph a system of two linear equations, then the point at which the two lines intersect will be
a solution to both lines.
of linear equations may not be parallel and intersect, be parallel and distinct, or
■■ Graphs
be parallel and coincident.
■■ If they intersect once (lines are not parallel), it indicates that there is one solution, as in
Exhibit 6.3(a).
■■ If they do not intersect (lines parallel and distinct), it indicates that there is no solution,
as in Exhibit 6.3(b).
■■ If they intersect at every point (lines are parallel and coincident), it indicates that there
are infinite solutions, as in Exhibit 6.3(c).
­­­­Consistent and Inconsistent Systems
A system of linear equations that has one or many solutions is called a consistent linear system.
A system of linear equations that has no solution is called an inconsistent linear system.
­­­Dependent and Independent Equations
If the system of linear equations has many solutions then the equations are dependent.
If the system of linear equations has one or no solution then the equations are independent.