106 CHAPTER 3
Algebra
3.5 GRAPHING EQUATIONS AND INEQUALITIES
Textbook Reference Section 6.1 &6.2
CLAST OBJECTIVE
" Identify regions of the coordinate plane that correspond to specific conditions
and vice-versa
Graphing Equations
Case 1: The graph of x = a, where a is a constant, is a vertical line.
Examples
a) Graph: x = 3
Solutions
y
Locate the point (3,0)
on the x – axis and
draw a vertical line
through the point.
4
3
2
1
-4 -3 -2 -1
0
-1
-2
-3
-4
1 2
3
4
x
y
b) Graph: x = -4
Locate the point (-4,0)
on the x – axis and
draw a vertical line
through the point.
4
3
2
1
-4 -3 -2 -1
0
-1
-2
-3
-4
1 2
3
4
x
c) Graph: x = 0
4
3
2
1
-4 -3 -2 -1
0
-1
-2
-3
-4
1 2
3
4
Locate the point (0, 0) on
the x – axis and draw a
vertical line through the
point. Notice that this line
is the same as the y – axis.
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SECTION 3.5
Graphing Equations and Inequalities 107
Case 2: The graph of y = b, where b is a constant, is a horizontal line.
Solutions
Examples
d) Graph y = 3.
y
Locate the point (0,3)
on the y – axis and
draw a horizontal line
through the point.
4
3
2
1
-4 -3 -2 -1
0
-1
-2
-3
-4
1 2
3
4
x
y
e) Graph y = -2.
Locate the point (0,-2)
on the y – axis and
draw a horizontal line
through the point.
4
3
2
1
-4 -3 -2 -1
0
-1
-2
-3
-4
1 2
3
4
x
Case 3: The graph of y = mx + b, where m ≠ 0, is a line of slope m and y-intercept b.
rise
• Slope (m) =
run
•
Y-intercept (b) is the point (0, b) where the line crosses or intersects the y – axis.
•
To use the slope and the y – intercept to graph a line, be sure to write the equation
of the line in slope-intercept form, y = mx + b, if possible.
Example
Explanation
Note that the equation of the line is already in slope-intercept
f) Graph y = 3x + 2.
3
y
form with m = 3 = and b = 2.
•
1
7
6
5
•
4
3
2•
1
-4 -3 -2 -1
0
1 2
3
4
x
1. Graph the y-intercept (0, 2).
2. Now use the slope to find other points on the line. From the
y-intercept move up three places and one place to the right.
Place a point there.
3. From the 2nd point, use the slope again to get another point
on the line. Move up three places and one place to the right.
Place a point there.
4. Now draw the line using the three points.
5. To obtain more points on the line, proceed as in step 3.
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108 CHAPTER 3
Algebra
Examples
Explanations
First, rewrite the equation in slope-intercept form by solving
for y.
g) Graph 2x + 3y = 3.
y
•
-4 -3 -2 -1
4
3
2
1•
0
-1
-2
-3
-4
1 2
3
•
4
x
1. Graph the y-intercept (0, 1).
2. Now use the slope to find other points on the line. From the
y-intercept move down two places and three places to the
right. Place a point there.
3. Now draw the line using the two points.
4. To obtain more points on the line, proceed as in step 2.
5. To get a point to the left of the y-intercept, start at the yintercept and move up two spaces and to the left three spaces.
Place a point there.
h) What is an equation of the line
below?
y
0 1 2
-1
-2•
-3
-4
From the graph, we can determine that the y-intercept is
(0, -2). Thus b = -2. Next, we must determine how the next
point was obtained. From the y-intercept, if we move up 2
(rise) and to the right 3 (run), we get the point shown on the
graph. Therefore, slope (m) is
4
3
2
1
-4 -3 -2 -1
2x + 3y = 3
-2x
-2x
3y = -2x + 3
3
3 3
2
y =− x + 1
3
−2
Note that m =
and b = 1.
3
2
.
3
We now have the necessary information to write an equation
of the line in slope-intercept form. b = -2 and m =
3• 4
x
2
3
2
x−2
3
All these equations
⎛2 ⎞
are equivalent.
3( y ) = 3⎜ x ⎟ − 2(3)
⎝3 ⎠
3 y = 2x − 6
+6
+6
3y + 6 = 2x
-3y
-3y
6 = 2x – 3y same as 2x – 3y = 6
same as -2x + 3y = -6
y=
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SECTION 3.5
Graphing Equations and Inequalities 109
Check Your Progress 3.5
Graph each of the following lines.
1. x = 5
2. x = -2
3. y = 3
4. y = -6
5. y = 2x –3
6. y =
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1
x+2
3
110 CHAPTER 3
Algebra
8. y – 2x = 4
7. 4x – 5y = 20
For Questions 9 and 10, write an equation of the line shown.
9.
3
2
1
-3 -2 -1 0
-1
1
2 3
4
5
1 2
3
4
-2
-3
-4
-5
-6
10.
5
4
3
2
1
-4 -3 -2 -1 0
-1
5
-2
-3
-4
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SECTION 3.5
Graphing Equations and Inequalities 111
Graphing Inequalities
1. Graph the corresponding equation. (The corresponding equation is formed by replacing
the inequality symbol with an equal sign.)
•
•
If the symbols < or > appear in the inequality, use a dashed line when graphing
the corresponding equation.
If the symbols ≤ or ≥ appear in the inequality, use a solid line when graphing the
corresponding equation.
The graph of the corresponding equation separates the grid into two regions, one of
which will be shaded to satisfy the original inequality.
2. Test the inequality and shade only the region which makes the inequality true. Begin
by choosing a point that doesn’t lie on the graphed line. (If the line does not go through
the origin, the point (0, 0) is the easiest one to test the inequality.)
•
•
If the inequality is satisfied, shade the region that includes the point.
If the inequality is not satisfied, shade the other region.
Examples
i) Graph the inequality: x ≤ 3
Solution
y
4
3
2
1
-4 -3 -2 -1
0
-1
-2
-3
-4
1 2
3
4
x
1. First, draw the graph of the equation y = 2. Use a dashed line
because the inequality symbol is <.
y
2. Notice that the line y = 2 divides the grid into two regions.
One region lies above the line. The other region lies below the
line. Now choose a point to test the inequality. We can use (0,
3). In testing this inequality, we use only the y-value since the
inequality does not contain x.
4
3
2
1
-4 -3 -2 -1
0
-1
-2
-3
-4
2. Notice that the line x = 3 divides the grid into two regions.
One region lies to the right of the line. The other region lies to
the left of the line. Now choose a point to test the inequality. We
can use (0, 0). In testing this inequality, we use only the x-value
since the inequality does not contain y.
Substitute 0 for x. Is 0 ≤ 3 a true statement? The answer is yes.
Thus, the region to the left of the line satisfies the inequality and
will be shaded.
To show that the region to the right of the line x = 3 does not
satisfy the inequality, let’s test the point ( 4 , 1 ). Substitute 4 for
x. Is 4 ≤ 3 a true statement? The answer is no. This means
that no point lying in the region to the right of the line x = 3 will
satisfy the inequality.
j) Graph: y < 2
Solution
Explanations
1. First, draw the graph of the equation x = 3. Use a solid line
because the inequality symbol is ≤ .
1 2
3
4
x
Substitute 3 for y. Is 3 < 2 a true statement? The answer is no,
which indicates that the region above the line does not satisfy the
inequality. Thus, the region below the line satisfies the
inequality and will be shaded.
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112 CHAPTER 3
Algebra
Examples
Explanations
1. First, draw the graphs of the equations x = -2 and y = 1. Use
a solid line because the inequality symbols are ≤ and ≥.
k) Graph: x ≤ -2 and y ≥ 1
Solution
y
2. Notice that the lines divide the grid into four regions. We will
test all four regions. Note that both inequalities must be satisfied.
We can use the points: (0, 0), (2, 4), (-3, -1), and (-3, 4).
4
3
2
1
-4 -3 -2 -1
0
-1
-2
-3
-4
1 2
3
4
x
Test Point x ≤ -2
(0, 0)
0 ≤ -2 False
(2, 4)
2 ≤ -2 False
(-3, -1) -3 ≤ -2 True
(-3, 4) -3 ≤ -2 True
y≥3
0≥3
4≥3
-1 ≥ 3
4≥3
False
True
False
True
Shade the region for which both inequalities are true.
Note that we can also use the intersection of sets to determine
the solution. In which case, we would shade only the region that
includes all points where x ≤ 0 and y ≥ 1.
1. First, draw the graph of the equation y + x = 3. Use a dashed
line because the inequality symbol is >.
l) Graph: y + x > 3
Solution
y
2. Now choose a point to test the inequality. We can use (0, 0).
4
3
2
1
-4 -3 -2 -1
0
-1
-2
-3
-4
1 2
3
4
x
m) Graph: x ≥ 0; y ≥ 0; and
2x + 3y < 6
Solution
y
0
-1
-2
-3
-4
1. First, draw the graphs of the corresponding equations. Use a
dashed line when graphing 2x + 3y =6 because the inequality
symbol is <.
2. Note that we can use the intersection of sets to determine the
solution. The inequality x ≥ 0 is true for all points to the right of
the y-axis. The inequality y ≥ 0 is true for all points above the xaxis. The inequality 2x + 3y < 6 is true for all points below the
line 2x + 3y = 6. The intersection is shaded.
4
3
2
1
-4 -3 -2 -1
Substitute 0 for x and 0 for y. Is 0 + 0 > 3 a true statement? The
answer is no, which indicates that this region containing the
point (0, 0) does not satisfy the inequality. Thus, the other region
satisfies the inequality and will be shaded.
1 2
3
4
x
When using test points, note that there are seven regions to test.
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SECTION 3.5
Example
Explanation
n) Give the conditions for the
shaded region.
y
4
3
2
1
-4 -3 -2 -1
0
-1
-2
-3
-4
Graphing Equations and Inequalities 113
1. Find all the boundaries of the shaded region.
• The vertical bound is x = 0 and its shading is x ≤ 0.
• The horizontal bound is y = 0 and its shading is y ≥ 0.
• The dashed line has y-intercept at (0, 4) and slope ¾.
Thus, the equation of the line is y = ¾ x + 4.
Rewriting the equation in standard form we have
3x – 4y = -16 and its shading is 3x – 4y > -16.
1 2
Solution
3
4
3
x+4
4
⎛3 ⎞
4( y ) = 4⎜ x ⎟ + 4(4 )
⎝4 ⎠
4 y = 3 x + 16
− 16 − 16
4 y − 16 = 3x
− 4y
− 4y
− 16 = 3x − 4 y
y=
x
-
x ≤ 0; y ≥ 0; 3x – 4y > 16
2. List the conditions: x ≤ 0; y ≥ 0; 3x – 4y > -16
Check Your Progress 3.5
11. Graph: x ≥ 3
12. Graph: x < -2
13. Graph: y > -3
14. Graph: y ≤ 5
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114 CHAPTER 3
Algebra
15. Graph: x + y ≤ 5
16. Graph: x – y > -4
17. Graph: 2x – 3y > -6
18. Graph: x + 2y ≥ -1
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SECTION 3.5
Graphing Equations and Inequalities 115
For Questions 19 and 20, give the conditions for the shaded regions.
19.
20.
5
5
4
4
3
3
2
2
1
1
-4 -3 -2 -1 0
-1
1 2
3
4
5
-4 -3 -2 -1 0
-1
-2
-2
-3
-3
-4
-4
1 2
3
4
5
1 2
3
4
5
For Questions 21 and 22, give the conditions for the shaded regions.
21.
22.
5
5
4
4
3
3
2
2
1
1
-4 -3 -2 -1 0
-1
1 2
3
4
5
-4 -3 -2 -1 0
-1
-2
-2
-3
-3
-4
-4
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116 CHAPTER 3
Algebra
See If You
Remember
1. Simplify: 5 x
2
SECTIONS 3.1 – 3.4
− 3[4 − 3( x − 2)]
2. Write an equivalent expression: x ( y + z )
3. Write an equivalent expression: 2 + ( x + y )
4. Name the property: 3xy + 0 = 3xy
(
5. 2 × 10
5
) (7 × 10 3 )
−
6. 0.000125 ÷ 2,500
7. Simplfy:
3 6
3
8. Solve: 2( 3 – y ) + 4 ≤ 5 – y
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SECTION 3.5
Graphing Equations and Inequalities 117
9. Find f ( -2 ) if f (x) = 3x 2 – 2x + 5
10. If A = LW and L = 3 and W = 6.1, find A.
11. Is - 3 a solution to | 2x + 4 | < 7 ?
12. Write in scientific notation: 0.000000001025
13. In a vacuum, light travels at a speed of 299,792,458 meters per second. Write this
number in scientific notation.
14. Find f ( 3 ) if f ( x ) = x 3 – 2x 2 + x – 1
15. Simplify: 5π − 2 + 3π
16. Write in standard form: 8.39 × 10 − 4
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