1 block
1 block
You walk directly east from your house one block. How far
from your house are you?
You walk directly west from your house one block. How far
from your house are you?
It didn't matter which direction you walked, you were still 1 block
from your house.
This is like absolute value. It is the distance from zero. It doesn't
matter whether we are in the positive direction or the negative
direction, we just care about how far away we are.
4  4
4 units away from 0
4 units away from 0
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
4 4
x 6
What we are after here are values of x such
that they are 6 away from 0.
x  6 or x  6
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
6 and -6 are both 6 units away from 0
x  4  10
The "stuff" inside the absolute value signs could = 10 (the
positive direction) or the "stuff" inside the absolute value signs
could = 10 (the negative direction)
x  4  10
Let's check it:
x  4  10
x  6 or x  14
6  4  10
 14  4  10
10  10
10  10
Let's look at absolute value with an inequality.
x 5
This is asking, "For what numbers is the distance from 0 less than 5
units?"
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
5  x  5
Everything
inbetween these
lines is less than 5
units away from 0
Inequality notation
So if we have x  a it is equivalent to
(5, 5)
Interval notation
a  x  a
This means x is greater than -a and x is less than a
(or x is inbetween -a and a)
3x  1  5
Based on what we just observed, the "stuff"
inside the absolute value signs is inbetween
-5 and 5 or equal to either end since the
inequality sign has "or equal to".
To solve this we get x isolated in the middle.
Whatever steps we do to get it alone, we do
to each end. We keep in our minds the fact
that if we multiply or divide by a negative, we
must turn the signs the other way.
 5  3x 1  5
+1
-4
+1 +1
6
 4  3x  6
3
3
3
4
 x2
3
So x is inbetween or equal to
- 4/3 and 2
 4 
 3 , 2
Let's graph the solution:
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
What if the inequality is greater than?
x 5
This is asking, "When is our distance from 0 more than 5 units away?"
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
Everything outside these
Everything outside these
lines is more than 5
lines is more than 5
units away from 0
We'll have to express
units away from 0
this with two
difference pieces
x  5
In interval notation:
OR
x5
(,5) or (5, )
So if we have x  a it is equivalent to
x  a or
xa
1  2x  3  3
+3 +3
6
1  2x  6
We must first isolate the absolute value.
This means if there are other things on the
left hand side of the inequality that are
outside of the absolute value signs, we
must get rid of them first.
From what we saw previously, the "stuff"
inside the absolute value is either less than
or equal to -6 or greater than or equal to 6
1  2x  6 or 1  2x  6
 2x  7
-2
-2
7
or
x
2
 2x  5
-2
-2
5
x
2
Isolate x, remembering that if
you multiply or divide by a
negative you must turn the sign.
We are dividing by a negative
so turn the signs!
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au