Download 2-1 Absolute Value Equations

2-1
Absolute Value Equations
TEKS FOCUS
VOCABULARY
ĚAbsolute value – The absolute
TEKS (6)(E) Solve absolute value linear
equations.
value of a real number x, written
|x|, is its distance from zero on a
number line.
TEKS (1)(B) Use a problem–solving
model that incorporates analyzing
given information, formulating a plan
or strategy, determining a solution,
justifying the solution, and evaluating
the problem–solving process and the
reasonableness of the solution.
formulate a plan or strategy to
solve a problem.
ĚStrategy – a plan or method for
solving a problem
ĚExtraneous solution – An extraneous ĚReasonableness – the quality of
solution is a solution derived from
an original equation that is not a
solution to the original equation.
being within the realm of common
sense or sound reasoning. The
reasonableness of a solution is
whether or not the solution
makes sense.
ĚFormulate – create with careful
Additional TEKS (1)(D), (1)(E), (6)(D)
effort and purpose. You can
ESSENTIAL UNDERSTANDING
An absolute value quantity is nonnegative. Since opposites have the same absolute
value, an absolute value equation can have two solutions.
Key Concept
Absolute Value
Definition
Numbers
The absolute value of a real number x, written 0 x 0 , is
its distance from zero on the number line.
Absolute Value Equation
0x0 = a
040 = 4
0 -4 0 = 4
Meaning
the distance between x and 0
is a units
Symbols
0 x 0 = x, if x Ú 0
0 x 0 = -x, if x 6 0
Graph and Solution
Problem 1
P
a
0
a
x = -a or x = a
TEKS Process Standard (1)(E)
Solving an Absolute Value Equation
How is solving this
equation different
from solving a linear
equation?
In the absolute value
equation, 2x - 1 can
represent two opposite
quantities.
What is the solution of ∣ 2x − 1 ∣ = 5? Graph the solution.
W
0 2x - 1 0 = 5
2x - 1 = 5
Lesson 2-1
2x - 1 = -5
2x = 6
x=3
3 2 1
36
or
0
2x = -4
x = -2
or
1
Absolute Value Equations
2
3
Rewrite as two equations.
2x 1 could be 5 or 5.
Add 1 to each side of both equations.
Divide each side of both equations by 2.
continued on next page ▶
Problem 1
continued
Check 0 2(3) - 1 0 ≟ 5
0 2( -2) - 1 0 ≟ 5
06 - 10 ≟5
0 -4 - 1 0 ≟ 5
050 = 5 ✔
0 -5 0 = 5 ✔
Problem
bl
2
Solving a Multi-Step Absolute Value Equation
Is there a simpler
way to think of this
problem?
Solving
30x + 20 - 1 = 8
is similar to solving
3y - 1 = 8.
What is the solution of 3 ∣ x + 2 ∣ − 1 = 8? Graph the solution.
W
30x + 20 - 1 = 8
30x + 20 = 9
Add 1 to each side.
0x + 20 = 3
Divide each side by 3.
x+2=3
or
x + 2 = -3
x=1
or
x = -5
5 4 3 2 1
0
1
2
Rewrite as two equations.
Subtract 2 from each side of both equations.
3
Check 3 0 (1) + 2 0 - 1 ≟ 8
3 0 ( -5) + 2 0 - 1 ≟ 8
3030 - 1≟8
3 0 -3 0 - 1 ≟ 8
8=8 ✔
8=8 ✔
Problem
P
bl
3
Can you solve this
the same way as you
solved Problem 1?
Yes, let 3x + 2 equal
4x + 5 and -(4x + 5).
Checking for Extraneous Solutions
C
What is the solution of ∣ 3x + 2 ∣ = 4x + 5? Check for extraneous solutions.
W
0 3x + 2 0 = 4x + 5
3x + 2 = 4x + 5
or
-x = 3
3x + 2 = -(4x + 5)
Rewrite as two equations.
3x + 2 = -4x - 5
Solve each equation.
7x = -7
x = -3
or
Check 0 3( -3) + 2 0 ≟ 4( -3) + 5
0 -9 + 2 0 ≟ -12 + 5
0 -7 0 ≠ -7 ✘
x = -1
0 3( -1) + 2 0 ≟ 4( -1) + 5
0 -3 + 2 0 ≟ -4 + 5
0 -1 0 = 1 ✔
Since x = -3 does not satisfy the original equation, -3 is an extraneous solution.
The only solution to the equation is x = -1.
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Problem 4
P
TEKS Process Standard (1)(B)
Formulating an Absolute Value Equation
You drive 8 miles from home to a gym. After your workout, you drive to a farm that
is m miles from your home on the same road as your home and the gym. After
buying some fruits and vegetables, you drive back to the gym where you have left
your cell phone. So far, you have driven a total of 15 miles. Formulate an absolute
value equation to represent this situation.
You know the distance
from home to the gym.
The distance from home
to the farm is m.
The farm is m miles
from home, in either
direction from the gym.
So, use absolute value to
represent the distance
from the gym to the farm.
Write an equation. You
traveled the distance
between the gym and the
farm twice.
distance to gym = 8
distance to farm = m
distance from gym to
farm = |8 – m|
total distance traveled
15 = 8 + |8 – m| + |8 – m|
NLINE
HO
ME
RK
O
An absolute value equation representing this situation is 15 = 8 + 2 0 8 - m 0 .
WO
PRACTICE and APPLICATION EXERCISES
Scan page for a Virtual Nerd™ tutorial video.
Solve each equation. Check your answers.
For additional support when
completing your homework,
go to PearsonTEXAS.com.
1. 0 3x 0 = 18
2. 0 -4x 0 = 32
3. 0 x - 3 0 = 9
4. 2 0 3x - 2 0 = 14
5. 0 3x + 4 0 = -3
6. 0 2x - 3 0 = -1
7. 0 x + 4 0 + 3 = 17
8. 0 4 - z 0 - 10 = 1
Solve each equation. Check for extraneous solutions.
9. 0 x - 1 0 = 5x + 10
38
Lesson 2-1
10. 0 2z - 3 0 = 4z - 1
11. 0 3x + 5 0 = 5x + 2
12. 0 2y - 4 0 = 12
13. 3 0 4w - 1 0 - 5 = 10
14. 0 2x + 5 0 = 3x + 4
Absolute Value Equations
15. Write an absolute value equation to describe the graph.
4 2
0
2
4
Solve each equation.
16. - 0 4 - 8b 0 = 12
17. 4 0 3x + 4 0 = 4x + 8
18. 0 3x - 1 0 + 10 = 25
1
19. 2 0 3c + 5 0 = 6c + 4
20. 5 0 6 - 5x 0 = 15x - 35
21. 7 0 8 - 3h 0 = 21h - 49
22. 6 0 2x + 5 0 = 6x + 24
23. 14 0 4x + 7 0 = 8x + 16
24. 23 0 3x - 6 0 = 4(x - 2)
Create Representations to Communicate Mathematical Ideas (1)(E) Write an
absolute value equation that represents each set of numbers.
25. all real numbers exactly 6 units from 0
26. all real numbers exactly 5 units from 3
Apply Mathematics (1)(A) Write an absolute value equation to represent
each situation.
27. A fence extends from the side of a house. A dog is tied to a fence
post ten feet from the side of the house. The dog’s leash is six
feet long. Let d be the greatest and least distance from the house
that the dog can reach along the fence.
10 ft
28. A factory produces widgets whose length must be within 1.5 mm
of an ideal length of 47 mm. A factory supervisor wants to mark
rulers with the least and greatest acceptable length for each
widget for workers to use in quality control inspections. Let ℓ
be the length of the markings along the ruler.
6 ft
29. You and your friends Jose and Sarah all live on the same street.
You know that Jose lives five blocks away from you. Jose says that
Sarah lives two blocks from him, but he didn’t say whether she lives closer to you
than he does or further. Let b be the number of blocks Sarah might live from you.
Is the absolute value equation always, sometimes, or never true? Explain.
30. 0 x 0 = -6
31. 0 x 0 = x
32. 0 x 0 + 0 x 0 = 2x
33. 0 x + 2 0 = x + 2
Solve each equation for x.
34. 0 ax 0 - b = c
35. 0 cx - d 0 = ab
36. a 0 bx - c 0 = d
Write an absolute value equation that has the given solution.
37. - 3 and 3
38. 2 and 4
39. - 1 and 5
40. 3 and 4
41. 0
42. 7
43. - 11
44. no solution
45. infinitely many solutions
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Use Multiple Representations to Communicate Mathematical Ideas (1)(D) In the
diagram, your office is shown as 12 miles from home. The question marks represent
possible locations of a pizza parlor, which is x miles from your home. Use the
diagram for Exercises 46–49.
Office
Home
?
12 miles
46. Formulate an absolute value equation to represent d, where d represents the
distance from the office to the pizza parlor.
47. Formulate an absolute value equation to represent e, where e represents the total
distance to drive from the office to the pizza parlor, and then back to the office.
48. If f = 12 + 0 12 - x 0 , describe in words what the distance f might represent.
49. Is it possible to formulate an absolute value equation to represent g, if g is the
total distance driven from home to the office, to the pizza parlor, and then home?
Explain.
TEXAS Test Practice
T
50. What is the positive solution of 0 3x + 8 0 = 19?
51. What is the solution of 0 2x - 4 0 = 16?
A. 10
C. 10, - 6
B. 10, - 10
D. 6, - 6
52. How many solutions does the equation 0 4x + 7 0 = -3 have?
F. 0
G. 1
H. 2
J. infinitely many
53. The packing material for a particular computer needs to be within 0.5 mm of the
desired thickness, which is 27.5 mm. Which equation represents the limits of the
width of the packing material?
40
Lesson 2-1
A. 0 w + 27.5 0 = 0.5
C. 0 w + 0.5 0 = 27.5
B. 0 w - 0.5 0 = 27.5
D. 0 w - 27.5 0 = 0.5
Absolute Value Equations
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