Solving Compound Inequalities
After studying this lesson, you will be able to:
- Solve compound inequalities.
Compound inequalities are two inequalities
considered together.
A compound inequality containing the word and is true only if
both inequalities are true. This type of compound inequality is
called a conjunction.
Examples of conjunctions:
x > -5 and x <1
y < 3 and y > -3
A compound inequality containing the word or is true if either
of the inequalities is true. This type of compound inequality is
called a disjunction.
Examples of disjunctions:
x > -5 or x > 1
y < 3 or y > -3
Example 1
| 2y > y - 3 or 3y < y + 6 |
This is a disjunction (it has the word or
). To solve, we work as two separate inequalities |
| 2y > y - 3 |
subtract y from each side |
3y < y + 6 |
subtract y from each side |
| y > -3 |
|
2y <6 |
divide each side by 2 |
| y < 3 |
|
|
|
Therefore, our answer is y> -3 or y <3
(this means that y can be any number since all numbers are
either greater than -3 or less than positive 3)
Example 2
| x - 4 < -1 and x + 4 > 1 |
This is a conjunction (it has the word
and ). To solve, we work as two separate inequalities |
| x - 4 < -1 |
add 4 to each side |
x + 4 > 1 |
subtract 4 from each side |
| x < 3 |
|
x > -3 |
|
Therefore, our answer is x < 3 and x > -3 (this means
that x must be some number between 3 and -3 )
Example 3
| 3m > m + 4 and -2m + m - 6 |
This is a conjunction (it has the word
and ). To solve, we work as two separate inequalities |
| 3m < m + 4 |
subtract m from each side |
2m < 4m - 6 |
subtract 4m from each side |
| 2m < 4 |
divide each side by 2 |
-6m < -6 |
divide each side by -6 (remember to
reverse the symbol) |
| m < 2 |
|
m > 1 |
|
Therefore, our answer is m < 2 and m> 1
(this means that x must be some number between 1 and 2 )
Example 4
| 2 < 3x + 2 < 14 |
This is another way to write a
conjunction. There is no word and there are two
inequality symbols. To solve, we break it down to two
inequalities this way: |
2 < 3x + 2 is the first inequality
3x + 2 < 14 is the second inequality
Now, we solve the way we did in Examples 1 -3: 2 < 3x + 2
and 3x + 2 < 14
| 2 < 3x + 2 |
subtract 2 from each side |
3x + 2 < 14 |
subtract 2 from each side |
| 0 < 3x |
divide each side by 3 |
3x < 12 |
divide each side by 3 |
| 0 < x |
|
x < 4 |
|
Therefore, our answer is x> 0 and x <4 or we can write
it 0 < x < 4
(this means that x must be some number between 0 and 4 ) |
