October 28, 2013

Solving Compound and Absolute Value Inequalities

Warm Up

1)Three –fourths of the difference between a number and six is no more than the quotient of that number and four

2. 4x + 1 > x - 2

2 3

4. Glynn has to drive 450 miles. His car has an 18 gallon gas tank and he would like to make the trip on one tank of gas. What is the minimum miles per gallon his car would have to get to make the trip on one tank? Write an inequality to show .

5. Solve for x if ax + by = c

Warm Up

6. -3(2x - 3) + 7x = -4(x - 5) + 6x – 2

7. Simplify: -6 - (-3) + (-2) * 4 =

8.

10. -2|x - 7|- 4 = -22

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

To solve a compound inequality, you must solve each part of the inequality separately. Conjunctions are solved when both parts of the inequality are true

A. Conjunctions: Two inequalities joined by the word ‘and’.

For example: -1 < x and x < 4; This can also be written -1 < x < 4

x

-1 4

-1 < x < 4; x > -1 < 4

The graph of a compound inequality containing the word ‘and’ is the intersection of the solution set of the two inequalities. TheIntersection is the solution for the compound inequality.


All compound inequalities divide the number line into three separate regions.


x

y z

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

x

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

Example:

1x



x

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

x

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

Example:

1x

2x


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

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

x

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

x

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

x

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

Example:

1x

2x

2

1

x

and

x



x

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

Example:

2

1

x

and

x


x

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

Example:

1x


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

x

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

x

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

Example:

1x

3x



x

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

x

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

x

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

Example:

1x

3x

3

1

x

or

x



x

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

Example:

3

1

x

or

x



x

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

The solution set will be found:

in the blue (middle) region

zy

or

in the red (outer) regions.


Solve an Absolute Value Inequality (<)

You can interpret |a| < 4 to mean that the distance between a and 0 on a number

line is less than 4 units.





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

from 0.


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





from 0.


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


You can interpret |x| < 4 to mean that the distance between a and 0 on a number


To make |a| < 4 true, you must substitute numbers for x that are fewer than 4 units

from 0.


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


You can interpret |x| < 4 to mean that the distance betweenax and 0 on a number line is less than 4 units.

To make |x| < 4 true, you must substitute numbers for x that are fewer than 4 units

from 0.

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


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


You can interpret |x| < 4 to mean that the distance between xa and 0 on a number


To make |x| < 4 true, you must substitute numbers for x that are fewer than 4 units

from 0.

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

All of the numbers between -4 and 4 are less than 4 units from 0.The solution set is { x | -4 < x < 4 }


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





from 0.

Notice that the graph of |a| < 4 is the same

as the graph a > -4 and a < 4.

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

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

The solution set will be found:

in the blue (middle) region

y z

or

in the red (outer) regions.


Solve an Absolute Value Inequality (>)



You can interpret |a| > 2 to mean that the distance between a and 0 on a number

line is greater than 2 units.





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

from 0.


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





from 0.


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





from 0.

Notice that the graph of |a| > 2 is the same

as the graph a < -2 or a > 2.


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





from 0.



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





from 0.



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


October 28, 2013

Technology

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