March 29th
March 29th
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.
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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)
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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 )
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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 )
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