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§ 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 x3 ) 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. x4 2 x6 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. x6 ] 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 x3 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 x3 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 x3 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 x0 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