Chapter Four 4.2

advertisement
§ 4.2
Compound Inequalities
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
3x – 5 < 13
and
2x + 4 < 14
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.
Blitzer, Intermediate Algebra, 5e – Slide #2 Section 4.2
Intersections & Unions
p 253
Intersection & Union
Intersection: The intersection of sets A and B, written A  B , is the set of
elements common to both set A and set B. This definition can be expressed in
set-builder notation as follows: A  B  x | x  A AND x  B .
Union: The union of sets A and B, written A  B , 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: A  B  x | x  A OR x  B.
NOTE: “Intersection” and “and” can be interpreted as being synonymous.
“Union” and “or” can also be interpreted as being synonymous.
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 4.2
Intersections
p 253
EXAMPLE
Find the intersection.
(a) 1,3,5,7 2,4,6,8,10 (b) 1,3,7 2,3,8
SOLUTION
(a) 1,3,5,7 2,4,6,8,10
Because the two sets have nothing in common, there is no
solution. Therefore, we say the solution is the empty set: O.
(b) 1,3,7 2,3,8
Both sets have the element 3. That is the only element
they have in common. Therefore, the solution set is {3}.
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 4.2
Linear Inequalities
p 253
Check Point 1
3,4,5,6,7 3,7,8,9
3,7
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 4.1
Intersections & Compound Inequalities
p 254
Solving Compound Inequalities Involving AND
1) 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.
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 4.2
Intersections & Compound Inequalities
p 254
EXAMPLE
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.
x  3 and x  1
SOLUTION
x3
)
3
x  1
[
-1
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 4.2
Intersections & Compound Inequalities
p 254
CONTINUED
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.
x  3 and x  1
[
-1
)
3
Therefore the solution is [-1,3).
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 4.2
Intersections & Compound Inequalities
p 254
EXAMPLE
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.
x  4  2 and 3x  1  8
SOLUTION
1) Solve each inequality separately. We wish to isolate x
in each inequality.
x4 2
x6
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 4.2
Add 4 to both sides
Intersections & Compound Inequalities
p 254
CONTINUED
3 x  1  8
3x  9
x  3
Subtract 1 from both sides
Divide both sides by 3
Now we can rewrite the original compound inequality as:
x  6 and x  3
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 4.2
Intersections & Compound Inequalities
p 254
CONTINUED
2) Take the intersection of the solution sets of the two
inequalities. Now we can solve each half of the
compound inequality.
x6
]
6
x  3
The bracket
stays in position.
(
The parenthesis
stays in position.
-3
x  6 and x  3
(
Therefore the solution is (-3,6].
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 4.2
]
Intersections & Compound Inequalities
Check Point 2
x  2  5 and
2 x  4  2
x3
2x  2
x 1
Graph the solutions
x | x  1
 ,1
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 4.1
p 255
Intersections & Compound Inequalities
Check Point 3
4 x  5  7 and
5x  2  3
4 x  12
5x  5
x 1
x3
Graph the solutions

Blitzer, Intermediate Algebra, 5e – Slide #13 Section 4.1
p 255
Intersections & Compound Inequalities
p 256
EXAMPLE
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.
3  4 x  3  19
SOLUTION
1) Solve each inequality separately. We will isolate x in the
compound inequality.
3  4 x  3 and 4 x  3  19
6  4 x and 4 x  22
Add 3 to both sides
1.5  x and x  5.5
Divide both sides by 4
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 4.2
Intersections & Compound Inequalities
CONTINUED
2) Take the intersection of the solution sets of the two
inequalities. Now we can solve each half of the compound
inequality.
x  1 .5
[
x  5 .5
)
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 4.2
p 256
Intersections & Compound Inequalities
CONTINUED
Upon over-laying the preceding two graphs, we get:
x  1.5 and x  5.5
[
Therefore the solution is [1.5,5.5).
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 4.2
)
Intersections & Compound Inequalities
Check Point 4
1  2 x  3  11
 2  2x  8
1  x  4
Graph the solutions
x | 1  x  4
[1,4)
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 4.1
p 256
Unions
p 256
Solving Compound Inequalities Involving OR
1) 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”
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 4.2
Unions
EXAMPLE
Find the union of the sets.
1,3,7,8 2,3,8
SOLUTION
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.
1,2,3,7,8
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 4.2
Linear Inequalities
p 253
Check Point 5
3,4,5,6,7 3,7,8,9
3,4,5,6,7,8,9
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 4.1
Unions & Compound Inequalities
EXAMPLE
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.
2 x  5  11 or 5 x  1  6
SOLUTION
1) Solve each inequality separately.
2 x  5  11
2 x  6
x  3
Add 5 to both sides
Divide both sides by 2
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 4.2
Unions & Compound Inequalities
CONTINUED
5x  1  6
5x  5
Subtract 1 from both sides
x 1
Divide both sides by 5
2) Take the union of the solution sets of the two inequalities.
x  3
]
-3
x 1
[
1
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 4.2
Unions & Compound Inequalities
CONTINUED
Upon over-laying the preceding two graphs, we get:
x  3 or x  1
]
[
Since the solution set is made up of two distinct intervals
(they don’t touch each other), we write the solution as:
 ,3 1, .
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 4.2
Union & Compound Inequalities
Check Point 6
3 x  5  2 or 10  2 x  4
3x  3
x 1
 2 x  6
x3
Graph the solutions
x | x  1 or x  3
(,1]  3, 
Blitzer, Intermediate Algebra, 5e – Slide #24 Section 4.1
p 258
Union & Compound Inequalities
Check Point 7
2 x  5  3 or
2x  3  3
2 x  2
x  1
2x  0
x0
Graph the solutions
x | x is a real number
 , 
Blitzer, Intermediate Algebra, 5e – Slide #25 Section 4.1
p 258
DONE
Unions & Compound Inequalities
EXAMPLE
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?
SOLUTION
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.
Blitzer, Intermediate Algebra, 5e – Slide #27 Section 4.2
Unions & Compound Inequalities
CONTINUED
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:
175  34 x
However, the cost of repairing the car has been quoted as
being between $226 and $294. This can be represented as:
226  175  34 x  294
Now, we just need to solve this inequality.
Blitzer, Intermediate Algebra, 5e – Slide #28 Section 4.2
Unions & Compound Inequalities
CONTINUED
226  175  34 x  294
51  34 x  119
Subtract 175 from all three parts
1 .5  x  3 .5
Divide all three parts by 34
Therefore, the time interval for hours that the mechanic will
be working on the job is between 1.5 and 3.5 hours.
Blitzer, Intermediate Algebra, 5e – Slide #29 Section 4.2
DONE
Download
Related flashcards

Algebra

20 cards

Polynomials

21 cards

Abstract algebra

19 cards

Group theory

35 cards

Ring theory

15 cards

Create Flashcards