§ 4.2 compound inequalities. blitzer, intermediate algebra, 4e – slide #22 intersections &...

§ 4.2

Compound Inequalities

Blitzer, Intermediate Algebra, 4e – Slide #22

Intersections & Unions

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

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}.


Intersections & Compound Inequalities

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 they 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

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6[

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6)



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 ever they overlap each other is the solution.

1 and 3 xx

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6)

CONTINUECONTINUEDD

[

Therefore the solution is [-1,3).



EXAMPLEEXAMPLE

SOLUTIONSOLUTION


813 and 24 xx

24 x

6x

1) Solve each inequality separately. We will to isolate x in both inequalities.

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

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6(

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6]

CONTINUECONTINUEDD

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6](

Therefore the solution is (-3,6].

The parenthesis stays in position.

The bracket stays in position.



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

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6)

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6[



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

DD

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6)[

Therefore the solution is [1.5,5.5).

5.5 and 5.1 xx


Unions

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 everything from both graphs.


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}.


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

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6[

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6]



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

DD

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

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,



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 that the mechanic will be working on the job?

First, I will assign a variable for the unknown quantity.Let x = the number of hours the mechanic will work on the car.

Next I 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 during which the mechanic will be working on the job is between 1.5 and 3.5 hours.

Subtract 175 from all three sides

CONTINUECONTINUEDD

29434175226 x

1193451 x

5.35.1 x Divide all three sides by 34

§ 4.2 compound inequalities. blitzer, intermediate algebra, 4e – slide #22 intersections &...

Documents