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Solving Compound Inequalities

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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factoring and expanding expressions 
finding LCM and GCF 
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solving a system of two and three linear equations (including Cramer's rule) 
graphing curves (lines, parabolas, hyperbolas, circles, ellipses, equation and inequality solutions) 
graphing general functions 
operations with functions (composition, inverse, range, domain...) 
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