Download Chapter 1 Section 7- Inequalities Involving Quotients and Absolute

Chapter 1 Section 7- Inequalities Involving Quotients and
Absolute Value
For these problems we will be first making sure that all terms are on
one side of the equation, while on the other side will be zero. Then find
values of x that make the denominator and the numerator equal to zero.
For example in the equation:
__x-2__  0
x+3
We find that when x = 2 the numerator equals
zero and when x = -3 the equation is undefined.
We are trying to find the solution set for this equation so we plug in
three different values like -4 (less than -3), 1 (in between 2 and -3), and 3
(more than 2) to find the solution set.
When we plug in -4 the numerator is negative (-4-2 = -6) and so is the
denominator (-4+3= -1) therefore it is positive and since we are looking for a
positive value, part of the solution set will include numbers less than -3; (-, 3). Then when we plug in 1, the numerator is negative (1-2=-1) and the
denominator is positive (1+3= 4) therefore it is negative and since we are
looking for a positive value we will not include this in our solution set. Finally
when we plug in 3 the numerator is positive (3-2=1) and so is the
denominator (3+3= 0) therefore it is positive and since we are looking for a
positive value, the final part of the solution set will include numbers greater
than 2; (2, ).
So the final solution set would be (-,-3)  (2, ).
In the equation:
__x + 2__  3
x+4
x+2
x+4
-3 < 0
We are trying to get all the variables on one
side. So we subtract 3 from each side
x + 2 -3x -12 < 0 Then we need them to have a common
x+4
denominator so we can add them.
-2x – 10 < 0 Then we proceed with the steps in the last example
x+4
.
(make the numerator and denominator zero and
take those numbers to find number values
greater, less than, or in between )
So the final solution would be (-, -5]  (-4, ).
Absolute Value Properties
For any real number k > 0, x = k
if x = k or x = -k
For any real number k > 0, x< k
if –k < x < k
For any real number k > 0, x> k
if x < -k or x > k
For example in the equation:
4x - 3 > 9
4x – 3 > 9
4x – 3 < - 9
4x > 12
4x < -6
x>3
x < -3/4
so the solution set is (-, -3/2)  (3, )
Additionally in the equation:
3x - 1 = x + 4
3x – 1 = x + 4
2x = 5
x = 5/2
So the solution set is (-3/4, 5/2)
3x – 1 = -x -4
4x = -3
x = -3/4