§ 4.2 Compound Inequalities. Blitzer, Intermediate Algebra, 5e – Slide #2 Section 4.2 Compound Inequalities A compound inequality is formed by joining

§ 4.2

Compound Inequalities

Blitzer, Intermediate Algebra, 5e – Slide #2 Section 4.2

Compound Inequalities

A compound inequality is formed by joining two inequalities with the word and or the word or.

Some Examples of Compound Inequalities

x – 3 < 5 and 2x + 4 < 14

3x – 5 < 13 or 5x + 2 > -3

Compound inequalities illustrate the importance of the words and and or in mathematics, as well as in everyday English.

“And” corresponds to intersection while “Or” corresponds to union.


Intersections & Unions p 253

Intersection & UnionIntersection: The intersection of sets A and B, written , is the set of elements common to both set A and set B. This definition can be expressed in set-builder notation as follows: .

Union: The union of sets A and B, written , is the set of elements that are members of set A or set B or of both sets. This definition can be expressed in set-builder notation as follows: .

NOTE: “Intersection” and “and” can be interpreted as being synonymous. “Union” and “or” can also be interpreted as being synonymous.

BA

BxAxxBA AND |

BA

BxAxxBA OR |


Intersections p 253

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Find the intersection.

Because the two sets have nothing in common, there is no solution. Therefore, we say the solution is the empty set: O.

8,3,27,3,1 (b)

8,3,27,3,1 (b)10,8,6,4,27,5,3,1 (a)

10,8,6,4,27,5,3,1 (a)

Both sets have the element 3. That is the only element they have in common. Therefore, the solution set is {3}.


Linear Inequalities p 253

Check Point Check Point 11

9,8,7,37,6,5,4,3 7,3


Intersections & Compound Inequalities p 254

Solving Compound Inequalities Involving AND1) Solve each inequality separately.

2) Graph the solution set to each inequality on a number line and take the intersection of these solution sets.

Remember, intersection means what the sets have in common.



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Solve the compound inequality. Use graphs to show the solution set to each of the two given inequalities, as well as a third graph that shows the solution set of the compound inequality.

1x

1 and 3 xx

3x

-1[

3)



We see that what the two graphs have in common is from the left-end bracket at x = -1 to the right-end parenthesis at x = 3. You can think of picking up one of the first two graphs and placing it on top of the other. Where they overlap each other is the solution.

1 and 3 xx

-1)

CONTINUECONTINUEDD

[

Therefore the solution is [-1,3).

3



EXAMPLEEXAMPLE

SOLUTIONSOLUTION


813 and 24 xx

24 x

6x

1) Solve each inequality separately. We wish to isolate x in each inequality.

Add 4 to both sides



813 x

3 and 6 xx

93 x

3x

Subtract 1 from both sides

CONTINUECONTINUEDD

Divide both sides by 3

Now we can rewrite the original compound inequality as:



2) Take the intersection of the solution sets of the two inequalities. Now we can solve each half of the compound inequality.

3x

3 and 6 xx

6x

-3 (

6 ]

CONTINUECONTINUEDD

] (

Therefore the solution is (-3,6].

The parenthesis stays in position.

The bracket stays in position.




and 52 x

1

22

x

x

1| xx 1,

3x

242 x

Graph the solutions




and 754 x

1

55

x

x

3

124

x

x

325 x

Graph the solutions



EXAMPLEEXAMPLE

SOLUTIONSOLUTION


1934 and 343 xx

19343 x

224 and 46 xx

1) Solve each inequality separately. We will isolate x in the compound inequality.

Add 3 to both sides

5.5 and 5.1 xx Divide both sides by 4



CONTINUECONTINUEDD2) Take the intersection of the solution sets of the two inequalities. Now we can solve each half of the compound inequality.

5.5x

5.1x

)

[


Intersections & Compound Inequalities

Upon over-laying the preceding two graphs, we get:CONTINUECONTINUE

DD

) [

Therefore the solution is [1.5,5.5).

5.5 and 5.1 xx




11321 x

41

822

x

x

41| xx )4,1[

Graph the solutions


Unions p 256

Solving Compound Inequalities Involving OR1) Solve each inequality separately.

2) Graph the solution set to each inequality on a number line and take the union of these solution sets.

Remember, union means what is in either set – you just “put it all together”


Unions

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Find the union of the sets.

8,3,28,7,3,1

8,7,3,2,1

The solution will be each element of the first set and each element of the second set as well. However, we will not represent any element more than once. Namely the elements 3 and 8 should not be listed twice.


Linear Inequalities p 253


9,8,7,37,6,5,4,3 9,8,7,6,5,4,3


Unions & Compound Inequalities

EXAMPLEEXAMPLE

SOLUTIONSOLUTION


615or 1152 xx

62 x

1152 x

1) Solve each inequality separately.

3x

Add 5 to both sides




Subtract 1 from both sides

615 x

55 x

1x

2) Take the union of the solution sets of the two inequalities.

CONTINUECONTINUEDD


1x

3x

1 [

-3]



Upon over-laying the preceding two graphs, we get:CONTINUECONTINUE

DD

Since the solution set is made up of two distinct intervals (they don’t touch each other), we write the solution as:

[]

1or 3 xx

. ,13,


Union & Compound Inequalities p 258


or 253 x

3

62

x

x

3 1| xorxx ,3]1,(

1

33

x

x

4210 x

Graph the solutions


Union & Compound Inequalities p 258


or 352 x

0

02

x

x

number real a is | xx ,

1

22

x

x

332 x

Graph the solutions

DONE



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Parts for an automobile repair cost $175. The mechanic charges $34 per hour. If you receive an estimate for at least $226 and at most $294 for fixing the car, what is the time interval for hours that the mechanic will be working on the job?

First, we will assign a variable for the unknown quantity.

Let x = the number of hours the mechanic will work on the car.

Next we set up an inequality to represent the situation.



Since the cost of repairing the car is the price of the parts, $175, plus the cost of labor, (x hours) times ($34 per hour), we represent the cost of repairing the car with:

However, the cost of repairing the car has been quoted as being between $226 and $294. This can be represented as:

Now, we just need to solve this inequality.

CONTINUECONTINUEDD

x34175

29434175226 x



Therefore, the time interval for hours that the mechanic will be working on the job is between 1.5 and 3.5 hours.

Subtract 175 from all three parts

CONTINUECONTINUEDD

29434175226 x

1193451 x

5.35.1 x Divide all three parts by 34

DONE

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§ 4.2 Compound Inequalities. Blitzer, Intermediate Algebra, 5e – Slide #2 Section 4.2 Compound Inequalities A compound inequality is formed by joining