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Solving Compound and Absolute Value Inequalities

Solving Compound and Absolute Value Inequalities

Solving Compound and Absolute Value Inequalities Solving Compound and Absolute Value Inequalities

1) compound inequality2) intersection3) union

Solve compound inequalities.

Solve absolute value inequalities.

A compound inequality consists of two inequalities joined by the word and or the word or.



To solve a compound inequality, you must solve each part of the inequality.



To solve a compound inequality, you must solve each part of the inequality.

The graph of a compound inequality containing the word “and” is the intersection of the solution set of the two inequalities.


A compound inequality divides the number line into three separate regions.



x

y z


x


The solution set will be found:

in the blue (middle) region

y z


x




y z

orin the red (outer) regions.


A compound inequality containing the word and is true if and only if (iff), bothinequalities are true.



x5-4 -2 0 2 4-5 -3 1 5-1-5 3

Example:

1x



x5-4 -2 0 2 4-5 -3 1 5-1-5 3

x5-4 -2 0 2 4-5 -3 1 5-1-5 3

Example:

1x

2x



x5-4 -2 0 2 4-5 -3 1 5-1-5 3

x5-4 -2 0 2 4-5 -3 1 5-1-5 3

x5-4 -2 0 2 4-5 -3 1 5-1-5 3

Example:

1x

2x

2

1

x

and

x



x5-4 -2 0 2 4-5 -3 1 5-1-5 3

Example:

2

1

x

and

x


A compound inequality containing the word or is true if one or more, of theinequalities is true.



x5-4 -2 0 2 4-5 -3 1 5-1-5 3

Example:

1x



x5-4 -2 0 2 4-5 -3 1 5-1-5 3

x5-4 -2 0 2 4-5 -3 1 5-1-5 3

Example:

1x

3x



x5-4 -2 0 2 4-5 -3 1 5-1-5 3

x5-4 -2 0 2 4-5 -3 1 5-1-5 3

x5-4 -2 0 2 4-5 -3 1 5-1-5 3

Example:

1x

3x

3

1

x

or

x



x5-4 -2 0 2 4-5 -3 1 5-1-5 3

Example:

3

1

x

or

x


x




y z



Solve an Absolute Value Inequality (<)

You can interpret |a| < 4 to mean that the distance between a and 0 on a numberline is less than 4 units.




To make |a| < 4 true, you must substitute numbers for a that are fewer than 4 unitsfrom 0.


5-4 -2 0 2 4-5 -3 1 5-1-5 3





5-4 -2 0 2 4-5 -3 1 5-1-5 3





5-4 -2 0 2 4-5 -3 1 5-1-5 3




Notice that the graph of |a| < 4 is the sameas the graph a > -4 and a < 4.


5-4 -2 0 2 4-5 -3 1 5-1-5 3





All of the numbers between -4 and 4 are less than 4 units from 0.

The solution set is { a | -4 < a < 4 }


5-4 -2 0 2 4-5 -3 1 5-1-5 3





All of the numbers between -4 and 4 are less than 4 units from 0.

The solution set is { a | -4 < a < 4 }

For all real numbers a and b, b > 0, the following statement is true:

If |a| < b then, -b < a < b


x




y z



Solve an Absolute Value Inequality (>)



You can interpret |a| > 2 to mean that the distance between a and 0 on a numberline is greater than 2 units.




To make |a| > 2 true, you must substitute numbers for a that are more than 2 unitsfrom 0.


5-4 -2 0 2 4-5 -3 1 5-1-5 3





5-4 -2 0 2 4-5 -3 1 5-1-5 3





5-4 -2 0 2 4-5 -3 1 5-1-5 3




Notice that the graph of |a| > 2 is the sameas the graph a < -2 or a > 2.


5-4 -2 0 2 4-5 -3 1 5-1-5 3





All of the numbers not between -2 and 2 are greater than 2 units from 0.

The solution set is { a | a > 2 or a < -2 }


5-4 -2 0 2 4-5 -3 1 5-1-5 3





All of the numbers not between -2 and 2 are greater than 2 units from 0.

The solution set is { a | a > 2 or a < -2 }

For all real numbers a and b, b > 0, the following statement is true:

If |a| > b then, a < -b or a > b



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Using Glencoe’s Algebra 2 text,© 2005

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