Chapter 1
Section 5
ALGEBRA 2 LESSON 1-5
ALGEBRA 2 LESSON 1-5
Solve each equation.
1. 5(x – 6) = 40 2. 5b = 2(3b – 8) 3. 2y + 6y = 15 – 2y + 8
4. 4x + 8 > 20
(For help, go to Lessons 1-3 and 1-4.)
Absolute Value Equations and Inequalities
Absolute Value Equations and Inequalities
Solve each inequality.
6. 4(t – 1) < 3t + 5
1. 5(x – 6) = 40 2. 5b = 2(3b – 8)
= 5b = 6b – 16
x – 6 = 8 –b = –16
x = 14 b = 16
Solutions
ALGEBRA 2 LESSON 1-5
ALGEBRA 2 LESSON 1-5
Absolute Value Equations and Inequalities
Absolute Value Equations and Inequalities
5(x – 6) 5
40 5
5. 3a – 2 a + 6 6. 4(t – 1) < 3t + 5 3a – a 6 + 2 4t – 4 < 3t + 5 2a 8 4t – 3t < 5 + 4
a 4 t < 9
3. 2y + 6y = 15 – 2y + 8 4. 4x + 8 > 20 2y + 6y + 2y = 15 + 8 4x > 12
10y = 23 x > 3
Absolute Value
Absolute Value – the distance from zero a number is
on the number line – it is always positive
Symbol :
│x│
Definition:
If x is positive ( x > 0) then
│x│ = x
If x is negative ( x < 0) then
│x │ = -x
Absolute Value Equations have a possibility of two solutions
This is because the value inside the │ │ can equal either the
negative or the positive of the value on the other side of the
equal sign
Absolute Value Equations and Inequalities
Absolute Value Equations and Inequalities
Solve |15 – 3
x
| = 6.
|15 – 3x| = 6
15 – 3x = 6 or15 – 3x = –6
The value of 15 – 3x can be 6 or –6 since |6| and |–6| both equal 6.
x = 3 or x = 7
Divide each side of both equations by –3.
–3x = –9 –3x = –21
Subtract 15 from each side of both equations.
Check: |15 – 3x| = 6 |15 – 3x| = 6 |15 – 3(3)| 6 |15 – 3(7)| 6
|6| 6 |–6| 6
Try This Problem
│3x + 2 │ = 7
3x + 2 = 7
or
3x + 2 = -7
3x = 5 or
3x = -9
x =
5/
3
or
x = -3
Check your answer by plugging it back into the equation.
Absolute Value Equations and Inequalities
Absolute Value Equations and Inequalities
Solve 4 – 2|
x
+ 9| = –5.
4 – 2|x + 9| = –5
–2|x + 9| = –9 Add –4 to each side.
x = –4.5 or x
= –13.5Subtract 9 from each side of both
equations.
Check: 4 – 2 |x + 9| = –5 4 – 2|x + 9| = –5 4 – 2 |–4.5 + 9| –5 4 – 2 |–13.5 + 9| –5 4 – 2 |4.5| –5 4 – 2 |–4.5| –5 4 – 2 (4.5) –5 4 – 2 (4.5) –5
–5 = –5 –5 = –5
|x + 9| = 92 Divide each side by –2.
Try This Problem
Solve 2
│3x - 1 │ + 5 = 33.
2 │3x - 1 │ = 28
│3x – 1│ = 14
3x – 1 = 14
or
3x – 1 = -14
3x = 15
or
3x = -14
x = 5
or
x = -8/3
Check the solutions by plugging them into the
Extraneous Solutions
Extraneous solution – a solution of an
equation derived from an original
equation that is not a solution of the
original equation
This is why we MUST check all
Solve
│2x + 5 │ = 3x + 4
2x + 5 = 3x + 4
5 = x + 4
1 = x
Check:
2(1) + 5 = 3(1) + 4
2 + 5 = 3 + 4
7 = 7
2x + 5 = -(3x + 4)
2x + 5 = -3x – 4
5x + 5 = -4
5x = -9
x = -9/5
Check:
│2(-
9/
5
) + 5 │ = 3(-
9/
5) + 4
│-
18/
5
+
25/
5│ = -
27/
5+
20/
5│
7/
5
│ = -
7/
5│
7/
5
│ ≠ -
7/
5Try These Problems
Solve and check for extraneous solutions.
a)
│2x + 3 │ = 3x + 2
2x + 3 = 3x + 2 or 2x + 3 = -3x – 2 3 = x + 2 or 5x + 3 = -2
1 = x or 5x = -5 1 = x or x = -1 Check:
The solution is x = 1
b)
│x │ = x – 1
x = x – 1 or x = -x + 1
0 = -1 or 2x = 1
x = ½
Check:
Homework
Absolute Value Inequalities
Let
k
represent a positive real number
│x │ ≥ k is equivalent to x ≤ -k or x ≥ k
│x │ ≤ k is equivalent to -k ≤ x ≤ k
Remember to isolate the absolute value
Solve |2
x
– 5| > 3. Graph the solution.
|2
x
– 5| > 3
2
x
– 5 < –3 or 2
x
– 5 > 3
Rewrite as a compound inequality.
x
< 1 or
x
> 4
Try This Problem
Solve
│2x - 3 │ > 7
2x – 3 > 7 or 2x – 3 < -7
Solve –2|
x
+ 1| + 5 –3. Graph the solution.
>–|x + 1| 4 Divide each side by – 2 and reverse the inequality.
<– –2|x + 1| + 5 –3>–
–2|x + 1| –8
Isolate the absolute value expression. Subtract 5 from each side.
> –
–4 x + 1 4
Rewrite as a compound inequality.
<– <–
–5 x 3
Solve for x.
Try This Problem
Solve |5z + 3| - 7 < 34. Graph the solution.
|5z + 3| -7 < 34
|5z + 3| < 41
-41 < 5z + 3 < 41
-44 < 5z < 38
-44
/
5
< z <
38/
5-8
4/
Ranges in Measurement
Absolute value inequalities and
compound inequalities can be used to
specify an allowable range in
measurement.
Tolerance – the difference between a
measurement and its maximum and
minimum allowable values
Tolerance
For example, if a manufacturing specification calls
for a dimension
d
of 10 cm with a tolerance of
0.1 cm, then the allowable difference between
d
and 10 is less than or equal to 0.1
|
d
- 10 | ≤ 0.1
absolute value inequality
d
– 10 ≤ 0.1 and
d
– 10
≥-0.1
equivalent compound
inequality
-0.1 ≤ d
– 10
≤ 0.1 equivalent compound
inequality
The area
A
in square inches of a square photo is required to
satisfy 8.5 ≤
A
≤ 8.9. Write this requirement as an absolute
value inequality.
Write an inequality.
–0.2 <– A – 8.7 0.2<–
Rewrite as an absolute value inequality.
|A – 8.7| 0.2<–
Find the tolerance.
8.9 – 8.5
2 = 0.42 = 0.2
Find the average of the maximum and minimum values.
8.9 + 8.5
Try This Problem
The specification for the circumference C in inches of a
basketball for junior high school is 27.75 ≤ C ≤ 30.
Write the specification as an absolute value inequality.
375
.
2
75
.
2
75
.
27
5
.
28
125
.
28
2
25
.
56
2
75
.
27
5
.
28
375
.
125
.
28
C
Find the tolerance.
Find the average from min and max values.
