< 2
Subtract 2 from both sides to get 0 on the right:
- 2 < 0
Get a common denominator of 2x. Multiply the 2 by - 2· < 0
- < 0
< 0
< 0
< 0
Factor the numerator
< 0
Divide both sides by 3
< 0
Multiply both sides by 2
< 0
Find all critical values by setting numerator and denominator = 0
1-x=0 x=0
-x=-1
x=1
So the critical values are 0 and 1
Put these on a number line:
----------o--o---------
-3 -2 -1 0 1 2 3 4
test a value to the left of 0, say x=-1
Substitute it in
< 0
< 0
< 0
< 0
-2 < 0
This is true so we shade the portion of the
number line left of 0
< ==========o--o---------
-3 -2 -1 0 1 2 3 4
test a value between 0 and 1 say x=0.5
Substitute it in
< 0
< 0
1 < 0
That is false so we do not shade the portion of the
number line between 0 and 1. So we still have
< ==========o--o---------
-3 -2 -1 0 1 2 3 4
We test a value to the right of 1, say x=2
Substitute it in
< 0
< 0
< 0
< 0
This is true so we shade the portion of the
number line right of 1
< =========o--o======== >
-3 -2 -1 0 1 2 3 4
That's the graph of the solution set, in interval notation
that is
Edwin
