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Transcript
Solving Absolute Value
INEQUALITIES
Section 2.7
Advanced Algebra 1
x 3
An absolute value equation is an equation that contains
a variable inside the absolute value sign.
This absolute value equation represents the
numbers on the number line whose distance from 0
is equal to 3.
Two numbers satisfy this equation. Both 3 and -3 are
3 units from 0.
Look at the number line and notice the distance from 0
of -3 and 3.
3 units 3 units
-3
0
3
Do you
remember how to
solve this:
3|x + 2| -7 = 14
• Isolate the absolute value expression by adding 7 and dividing by 3.
3|x + 2| -7 = 14
3|x + 2| = 21
|x + 2| = 7
• Set up two equations to solve.
x+2=7
x=5
or
x + 2 = -7
x = -9
Solving Absolute Value Inequalities
Definition:
x > a means x > a or x < -a
-a
0
a
Definition:
x  a means x  a and x  a
-a
0
a
Keys to Solving Absolute Value Inequalities
GreatOR
Less ThAND
x a
x a
xa
OR
x  a
xa
AND
x  a
2
x3  2
3
2
x3 2
3
OR
GreatOR
2
x  3  2
3
2

3 x  3  3 2
3

2

3  x  3   3  2 
3

2x  9  6
2x  9  6
2x  3
3
x
2
2x  15
OR
15
x
2
5  2x  4
5  2x  4
2x  1
1
x
2
1
9
<x<
2
2
Less ThAND
5  2x  4
AND
2 x  9
9
x
2
AND
-1
0
1
2
3
4
5
Your Turn…
Make two cases.
Solve the two
cases
independently.
Solve | 2x + 3 | < 6.
2 x  3  6 AND 2 x  3  6
2 x  3 AND
3
x
AND
2
2 x  9
9
x
2
Your Turn…
Make two cases.
Solve the two
cases
independently.
Solve | 2x – 3 | > 5.
2 x  3  5 OR 2 x  3  5
2 x  8 OR
x  4 OR
2 x  2
x  1
Example 1:
|2x + 1| > 7
2x + 1 > 7 or 2x + 1 >7
2x + 1 >7 or 2x + 1 <-7
This is an ‘or’ statement.
(Greator). Rewrite.
In the 2nd inequality, reverse the
inequality sign and negate the
right side value.
Solve each inequality.
x > 3 or
x < -4
-4
Graph the solution.
3
Example 2:
This is an ‘and’ statement.
(Less thand).
|x -5|< 3
x -5< 3 and x -5< 3
x -5< 3 and x -5> -3
Rewrite.
In the 2nd inequality, reverse the
inequality sign and negate the
right side value.
x < 8 and x > 2
2<x<8
Solve each inequality.
Graph the solution.
2
8
Pretend that you are allowed to go within 9 of
the speed limit of 65mph without getting a ticket.
Write an absolute value inequality that models
this situation.
|x – 65| < 9
Desired amount
Acceptable Range
Check Answer: x-65< 9 AND x-65> -9
x<74 AND x >56 
56<x<74
If a bag of chips is within .4 oz of 6 oz then it
is allowed to go on the market. Write an
inequality that models this situation.
|x – 6| < .4
Desired amount
Acceptable Range
Check Answer: x – 6 < .4 AND x – 6 > -.4
x < 6.4 AND x > 5.6
5.6< x < 6.4
In a poll of 100 people, Misty’s approval
rating as a dog is 78% with a 3% of error.
ticket. Write an absolute value inequality
that models this situation.
|x – 78| < 3
Desired amount
Acceptable Range
Check answer: x-78 < 3 AND x-78>-3
x<81 AND x>75 
75<x<81
Challenges…
HINT:
Think
midpoint
and
distance.
Find the absolute-value
inequality statement
that corresponds to the
inequality
–2 < x < 4.
Find the absolute-value
inequality statement
that corresponds to the
inequalities
x < 19 or x > 24.