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August 29th









August 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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