§ 4.2 Compound Inequalities

Blitzer, Intermediate Algebra, 4e – Slide #22 Intersections & Unions Intersection & Union Intersection: 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.

Blitzer, Intermediate Algebra, 4e – Slide #23 IntersectionsEXAMPLE SOLUTION 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. Both sets have the element {3}. That is the only element they have in common. Therefore, the solution set is {3}.

Blitzer, Intermediate Algebra, 4e – Slide #24 Intersections & Compound Inequalities 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 they have in common.

Blitzer, Intermediate Algebra, 4e – Slide #25 Intersections & Compound InequalitiesEXAMPLE SOLUTION 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 [ )

Blitzer, Intermediate Algebra, 4e – Slide #26 Intersections & Compound Inequalities 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 ) CONTINUED [ Therefore the solution is [-1,3).

Blitzer, Intermediate Algebra, 4e – Slide #27 Intersections & Compound InequalitiesEXAMPLE SOLUTION 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. 1) Solve each inequality separately. We will to isolate x in both inequalities. Add 4 to both sides

Blitzer, Intermediate Algebra, 4e – Slide #28 Intersections & Compound Inequalities Subtract 1 from both sides CONTINUED Divide both sides by 3 Now we can rewrite the original compound inequality as:

Blitzer, Intermediate Algebra, 4e – Slide #29 Intersections & Compound Inequalities 2) Take the intersection of the solution sets of the two inequalities. Now we can solve each half of the compound inequality ( ] CONTINUED ]( Therefore the solution is (-3,6]. The parenthesis stays in position. The bracket stays in position.

Blitzer, Intermediate Algebra, 4e – Slide #30 Intersections & Compound InequalitiesEXAMPLE SOLUTION 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. 1) Solve each inequality separately. We will isolate x in the compound inequality. Add 3 to both sides Divide both sides by 4

Blitzer, Intermediate Algebra, 4e – Slide #31 Intersections & Compound InequalitiesCONTINUED 2) Take the intersection of the solution sets of the two inequalities. Now we can solve each half of the compound inequality ) [

Blitzer, Intermediate Algebra, 4e – Slide #32 Intersections & Compound Inequalities Upon over-laying the preceding two graphs, we get: CONTINUED )[ Therefore the solution is [1.5,5.5).

Blitzer, Intermediate Algebra, 4e – Slide #33 Unions 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 everything from both graphs.

Blitzer, Intermediate Algebra, 4e – Slide #34 UnionsEXAMPLE SOLUTION Find the union of the sets. 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}.

Blitzer, Intermediate Algebra, 4e – Slide #35 Unions & Compound InequalitiesEXAMPLE SOLUTION 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. 1) Solve each inequality separately. Add 5 to both sides Divide both sides by 2

Blitzer, Intermediate Algebra, 4e – Slide #36 Unions & Compound Inequalities Subtract 1 from both sides 2) Take the union of the solution sets of the two inequalities. CONTINUED Divide both sides by [ ]

Blitzer, Intermediate Algebra, 4e – Slide #37 Unions & Compound Inequalities Upon over-laying the preceding two graphs, we get: CONTINUED Since the solution set is made up of two distinct intervals (they don’t touch each other), we write the solution as: [ ]

Blitzer, Intermediate Algebra, 4e – Slide #38 Unions & Compound InequalitiesEXAMPLE SOLUTION 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.

Blitzer, Intermediate Algebra, 4e – Slide #39 Unions & Compound Inequalities 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. CONTINUED

Blitzer, Intermediate Algebra, 4e – Slide #40 Unions & Compound Inequalities 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 CONTINUED Divide all three sides by 34