Transcript 1.6 Solving Compound and Absolute Value Inequalities

1.6 Solving Compound and
Absolute Value Inequalities
Lets look at things you can do
with Inequalities
4 < 20 You can add the same number
to both side of the inequality and not
change the sign
4 + 8 < 20 + 8
12 < 28
You also subtract from both side with
changing the sign
12 – 5 < 28 – 5
7 < 23
Multiplication and Division are
different
When you multiply by a positive number
the sign stays the same
4 < 20
4 * 2 < 20 * 2
8 < 40
But when you multiply or divide by a
negative number, the sign changes
direction
4 < 20
4 * - 2 < 20 * - 2
- 8 > - 40
Compound Inequality
Two inequalities joined by
and or the word or
x < - 3 or x > 10
x > -2 and x < 8
And it gives an intersection
-3
10
-2
8
10 < x and x < 30
Can be written as 10 < x < 30
This shows the space between
10 and 30
10
30
14 < x – 8 < 32
We can add to all the part of the
inequality to solve for x
14 < x – 8 < 32
14 + 8 < x – 8 + 8 < 32 + 8
So
22 < x < 40
to graph the answer
Mark 22 and 40 on a line number
and shade between the numbers
22
40
Solve 10  3 y  2  19
Add 2 to all the sides
10  2  3 y  2  2  19  2
Solve 10  3 y  2  19
Add 2 to all the sides
10  2  3 y  2  2  19  2
12  3 y  21
Then divide by 3
10  3 y  2  19
Solve
Add 2 to all the sides
10  2  3 y  2  2  19  2
12  3 y  21
Then divide by 3
4 y7
In Set Building Notation {y| 4  y  7 }
Solve x + 3 < 2 or – x ≤ - 4
Do the problems
By adding -3
x<-1
Graphing the answer
-1
Multiply by - 1
x≥4
4
Filled in point at 4
Written as
x < - 1 or x ≥ 4
Absolute Value Inequalities
If the | x | < a number, then it is an and
statement.
| x | < 5, means x is between – 5 and 5
So | x | < 5 would be written
as – 5 < x < 5
Absolute Value Inequalities
If the | x | > a number, then it is an or
statement.
| x | > 5, means x is less then -5
or greater then 5
So | x | > 5 would be written as
x < - 5 or x > 5
Graphing
| x | < 5 would be graph as
-5
5
| x | > 5 would be graph as
-5
5
Solve | 2x – 2| ≥ 4
2x – 2 ≥ 4
2x – 2 ≤ - 4
add 2 to both sides
2x ≥ 6
2x ≤ - 2
Divide by 2, this will not change the sign
direction
Solve | 2x – 2| ≥ 4
2x – 2 ≥ 4
or
2x – 2 ≤ - 4
add 2 to both sides
2x ≥ 6
or
2x ≤ - 2
Divide by 2, this will not change the sign
direction
x ≥ 3 or x ≤ - 1
Lets work on a
few problem together
Page 43-44
#4 and #5
#10 and #11
Homework
Page 44 – 45
# 15, 19, 21, 24, 27 – 39 odd, 46, 47