We remove absolute value bars:
OR , }
Solve the first inequality:
Get 0 on the right:
Critical numbers are when numerator or denominator = 0
2x-4=0 gives critical number 2
x-4=0 gives critical number 4
----------o-----o------
-1 0 1 2 3 4 5 6
We test a value of x to the left of 2. We pick 0 and
substitute it into the original inequality
That is false so we do not shade the number line to the
left of 2.
We test a value of x between 2 and 4. We pick 3 and
substitute it into the original inequality
That is true so we shade the number line between 2 and 4.
----------o=====o------
-1 0 1 2 3 4 5 6
We test a value of x to the right of 4. We pick 5 and
substitute it into the original inequality
That is true so we shade the number line to the
right of 4.
----------o=====o=====>
-1 0 1 2 3 4 5 6
Now we text the critical numbers in the original inequality
This is true so we darken the circle at the value 2
----------@=====o=====>
-1 0 1 2 3 4 5 6
We do not darken the circle at 4 because the denominator in the
original inequality cannot be zero.
Now we consider the other inequality
Get 0 on the right:
The only critical number this time is
x-4=0 gives critical number 4.
We have already considered that critical number
in the first part so nothing is gained from this
second inequality. Therefore the solution is
----------@=====o=====>
-1 0 1 2 3 4 5 6
Which has interval notation
[2, 4) U (4, )
Edwin
