1. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 1

2. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 2 Equations, Inequalities, and Applications Chapter 2

3. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 3 2.4 An Introduction to Applications of Linear Equations

4. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 4 Objectives 1.Learn the six steps for solving applied problems. 2.Solve problems involving unknown numbers. 3.Solve problems involving sums of quantities. 4.Solve problems involving supplementary and complementary angles. 5.Solve problems involving consecutive integers. 2.4 An Introduction to Applications of Linear Equations

5. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 5 Solving an Applied Problem Step 1 Read the problem, several times if necessary, until you understand what is given and what is to be found. Step 2 Assign a variable to represent the unknown value, using diagrams or tables as needed. Write down what the variable represents. Express any other unknown values in terms of the variable. Step 3 Write an equation using the variable expression(s). Step 4 Solve the equation. Step 5 State your answer. Does it seem reasonable? Step 6 Check the answer in the words of the original problem. Solving Applied Problems 2.4 An Introduction to Applications of Linear Equations

6. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 6 Example 1 The product of 3, and a number decreased by 2, is 42. What is the number? Solving Problems Involving Unknown Numbers Read the problem carefully. We are asked to find a number. Step 1 Assign a variable to represent the unknown quantity. In this problem, we are asked to find a number, so we write Let x = the number. There are no other unknown quantities to find. Step 2 2.4 An Introduction to Applications of Linear Equations

7. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 7 The equation 3x – 2 = 42 corresponds to the statement “The product of 3 and a number, decreased by 2, is 42.” Be careful when reading these types of problems. The placement of the commas is important. Example 1 (cont.) The product of 3, and a number decreased by 2, is 42. What is the number? Solving Problems Involving Unknown Numbers Write an equation. Step 3 The product of 3,a number decreased by 2,andis42. 3 x–=422)( 2.4 An Introduction to Applications of Linear Equations Notice the placement of the commas!

8. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 8 Example 1 (cont.) The product of 3, and a number decreased by 2, is 42. What is the number? Solving Problems Involving Unknown Numbers Step 4 Solve the equation. 3 ( x – 2 ) = 42 3x – 6 = 42 3x – 6 + 6 = 42 + 6 3x = 48 3x 48 3 = Distribute. Add 6. Combine terms. Divide by 3. x = 16 2.4 An Introduction to Applications of Linear Equations

9. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 9 Example 1 (cont.) The product of 3, and a number decreased by 2, is 42. What is the number? Solving Problems Involving Unknown Numbers Step 5 State the answer. The number is 16. Step 6 Check. When 16 is decreased by 2, we get 16 – 2 = 14. If 3 is multiplied by 14, we get 42, as required. The answer, 16, is correct. 2.4 An Introduction to Applications of Linear Equations

10. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 10 Example 2 A box contains a combined total of 68 pens and pencils. If the box contains 16 more pens than pencils, how many of each are in the box? Solving Problems Involving Sums of Quantities Read the problem. We are given information about the total number of pens and pencils and asked to find the number of each in the box. Step 1 Assign a variable. Let x = the number pencils in the box. Then x + 16 = the number pens in the box. Step 2 2.4 An Introduction to Applications of Linear Equations

11. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 11 Example 2 (cont.) A box contains a combined total of 68 pens and pencils. If the box contains 16 more pens than pencils, how many of each are in the box? Solving Problems Involving Sums of Quantities Write an equation. Step 3 The totalisthe number of pensthe number of pencilsplus 68 = ( x + 16 ) +x Recall: x = # of pencils, x + 16 = # of pens 2.4 An Introduction to Applications of Linear Equations

12. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 12 Example 2 (cont.) A box contains a combined total of 68 pens and pencils. If the box contains 16 more pens than pencils, how many of each are in the box? Solving Problems Involving Sums of Quantities Step 4 Solve the equation. 68 = 2x + 16 68 – 16 = 2x + 16 – 16 52 = 2x 52 2x 2 = 26 = x 68 = ( x + 16 ) + x Combine terms. Subtract 16. Combine terms. Divide by 2. or x = 26 2.4 An Introduction to Applications of Linear Equations

13. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 13 Example 2 (cont.) A box contains a combined total of 68 pens and pencils. If the box contains 16 more pens than pencils, how many of each are in the box? Solving Problems Involving Sums of Quantities Step 5 State the answer. The variable x represents the number of pencils, so there are 26 pencils. Then the number of pens is x + 16 = 26 + 16 = 42. Recall: x = # of pencils, x + 16 = # of pens Step 6 Check. Since there are 26 pencils and 42 pens, the combined total number of pencils and pens is 26 + 42 = 68. Because 42 – 26 = 16, there are 16 more pens than pencils. This information agrees with what is given in the problem, so the answer checks. 2.4 An Introduction to Applications of Linear Equations

14. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 14 Example 3 A mixture of acid and water must be combined to fill a 286 ml beaker. If the mixture must contain 12 ml of water for each milliliter of acid, how many milliliters of water and how many milliliters of acid does it require to fill the beaker? Solving Problems Involving Sums of Quantities Read the problem carefully. We must find how many milliliters of water and how many milliliters of acid are needed to fill the beaker. Step 1 Assign a variable. Let x = the number of milliliters of acid required. Then12x = the number of milliliters of water required. Step 2 2.4 An Introduction to Applications of Linear Equations

15. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 15 Example 3 (cont.) A mixture of acid and water must be combined to fill a 286 ml beaker. If the mixture must contain 12 ml of water for each milliliter of acid, how many milliliters of water and how many milliliters of acid does it require to fill the beaker? Solving Problems Involving Sums of Quantities Write an equation. A diagram is sometimes helpful. Step 3 Acid x Water 12x Beaker = 286 x12x = + 286 Recall: x = ml. of acid, 12x = ml. of water. 2.4 An Introduction to Applications of Linear Equations

16. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 16 Example 3 (cont.) A mixture of acid and water must be combined to fill a 286 ml beaker. If the mixture must contain 12 ml of water for each milliliter of acid, how many milliliters of water and how many milliliters of acid does it require to fill the beaker? Solving Problems Involving Sums of Quantities Step 4 Solve. 13x = 286 13x 286 13 = x = 22 x + 12x = 286 Combine terms. Divide by 13. 2.4 An Introduction to Applications of Linear Equations

17. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 17 Example 3 (cont.) A mixture of acid and water must be combined to fill a 286 ml beaker. If the mixture must contain 12 ml of water for each milliliter of acid, how many milliliters of water and how many milliliters of acid does it require to fill the beaker? Solving Problems Involving Sums of Quantities Step 5 State the answer. The beaker requires 22 ml of acid and 12(22) = 264 ml of water. Recall: x = ml of acid, 12x = ml of water Step 6 Check. Since 22 + 264 = 286, and 264 is 12 times 22, the answer checks. 2.4 An Introduction to Applications of Linear Equations

18. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 18 Example 4 Gerald has a wire 96 inches long that he must cut into three pieces. The longest piece must be three times the length of the middle-sized piece and the shortest piece must be 14 inches shorter than the middle-sized piece. How long must each piece be? Solving Problems Involving Sums of Quantities Read the problem carefully. Three lengths must be found. Step 1 Assign a variable. x = the length of the middle-sized piece, 3x = the length of the longest piece, and x – 14 = the length of the shortest piece. Step 2 2.4 An Introduction to Applications of Linear Equations

19. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 19 Example 4 (cont.) Gerald has a wire 96 inches long that he must cut into three pieces. The longest piece must be three times the length of the middle-sized piece and the shortest piece must be 14 inches shorter than the middle-sized piece. How long must each piece be? Solving Problems Involving Sums of Quantities Step 3 Write an equation. 96 inches 3x3xxx – 14 LongestMiddle-sizedShortestisTotal length 3x3xxx – 14=96++ 2.4 An Introduction to Applications of Linear Equations

20. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 20 Example 4 (cont.) Gerald has a wire 96 inches long that he must cut into three pieces. The longest piece must be three times the length of the middle-sized piece and the shortest piece must be 14 inches shorter than the middle-sized piece. How long must each piece be? Solving Problems Involving Sums of Quantities Step 4 Solve. 96 = 5x – 14 96 + 14 = 5x – 14 + 14 110 = 5x 110 5x 5 = 22 = x 96 = 3x + x + x – 14 Combine terms. Add 14. Combine terms. Divide by 5. or x = 22 2.4 An Introduction to Applications of Linear Equations

21. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 21 Example 4 (cont.) Gerald has a wire 96 inches long that he must cut into three pieces. The longest piece must be three times the length of the middle-sized piece and the shortest piece must be 14 inches shorter than the middle-sized piece. How long must each piece be? Solving Problems Involving Sums of Quantities Step 5 State the answer. The middle-sized piece is 22 in. long, the longest piece is 3(22) = 66 in. long, and the shortest piece is 22 – 14 = 8 in. long. Recall: x = length of middle-sized piece, 3x = length of longest piece, and x – 14 = length of shortest piece. Step 6 Check. The sum of the lengths is 96 in. All conditions of the problem are satisfied. 2.4 An Introduction to Applications of Linear Equations

22. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 22 34 1 2 34Angles and are supplementary. They form a straight angle. Straight angle12Angles and are complementary. They form a right angle, indicated by. Solving with Supplementary and Complementary Angles 2.4 An Introduction to Applications of Linear Equations 180 (degrees)

23. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 23 Problem-Solving Hint If x represents the degree measure of an angle, then 90 – x represents the degree measure of its complement, and 180 – x represents the degree measure of its supplement. x 180 – x x 90 – x Solving with Supplementary and Complementary Angles Complement of x: x Supplement of x: 2.4 An Introduction to Applications of Linear Equations Measure of x: x x

24. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 24 Example 5 Find the measure of an angle whose supplement is 20° more than three times its complement. Solving with Supplementary and Complementary Angles Read the problem. We are to find the measure of an angle, given information about its complement and supplement. Step 1 Assign a variable. Let x = the degree measure of the angle. Then 90 – x = the degree measure of its complement; 180 – x = the degree measure of its supplement. Step 2 Write an equation. Step 3 Supplementis20more thanthree timesits complement. 180 – x=20+3 ( 90 – x ) 2.4 An Introduction to Applications of Linear Equations

25. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 25 Example 5 (cont.) Find the measure of an angle whose supplement is 20° more than three times its complement. Solving with Supplementary and Complementary Angles Solve. Step 4 180 – x = 20 + 3 ( 90 – x ) 180 – x = 20 + 270 – 3x 180 – x = 290 – 3x 180 – x + 3x = 290 – 3x + 3x 2x 110 2 = Distribute. Combine terms. Add 3x. Divide by 2. x = 55 180 + 2x = 290 180 + 2x – 180 = 290 – 180 2x = 110 Combine terms. Subtract 180. Combine terms. 2.4 An Introduction to Applications of Linear Equations

26. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 26 Example 5 (cont.) Find the measure of an angle whose supplement is 20° more than three times its complement. Solving with Supplementary and Complementary Angles State the answer. The measure of the angle is 55°. Step 5 Check. The complement of 55° is 35° and the supplement of 55° is 125°. Also, 125° is equal to 20° more than three times 35° (that is, 125 = 20 + 3(35) is true). Therefore, the answer is correct. Step 6 2.4 An Introduction to Applications of Linear Equations

27. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 27 Problem-Solving Hint When solving consecutive integer problems, if x = the lesser integer, then for any two consecutive integers, use x,x + 1; two consecutive even integers, use x,x + 2; two consecutive odd integers, use x,x + 2. Solving Problems with Consecutive Integers x,x + 2; x,x + 2. 2.4 An Introduction to Applications of Linear Equations

28. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 28 Example 6 The sum of two consecutive checkbook check numbers is 893. Find the numbers. Solving Problems with Consecutive Integers Read the problem. We are to find the check numbers. Step 1 Assign a variable. Let x = the lesser check number. Then x + 1 = the greater check number. Step 2 2.4 An Introduction to Applications of Linear Equations

29. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 29 Example 6 (cont.) The sum of two consecutive checkbook check numbers is 893. Find the numbers. Solving Problems with Consecutive Integers Recall: Let x = the lesser check number. Then x + 1 = the greater check number. Write an equation. The sum of the check numbers is 893, so Step 3 x + ( x + 1 ) = 893 Solve. Step 4 2x + 1 = 893 2x = 892 x = 446 Subtract 1. Combine terms. Divide by 2. 2.4 An Introduction to Applications of Linear Equations

30. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 30 Example 6 (cont.) The sum of two consecutive checkbook check numbers is 893. Find the numbers. Solving Problems with Consecutive Integers State the answer. The lesser check number is 446 and the greater check number is 447. Step 5 Check. The sum of 446 and 447 is 893. The answer is correct. Step 6 2.4 An Introduction to Applications of Linear Equations

31. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 31 4 times the smaller is added to 3 times the larger, the result is 125 Example 7 If four times the smaller of 2 consecutive odd integers is added to three times the larger, the result is 125. Find the integers. Solving Problems with Consecutive Integers Read the problem. We are to find two consecutive odd integers. Step 1 Assign a variable. Let x = the lesser integer. Then x + 2 = the greater integer. Step 2 Write an equation. Step 3 4 x+3 ( x + 2 )=125 2.4 An Introduction to Applications of Linear Equations

32. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 32 Step 4 Example 7 If four times the smaller of 2 consecutive odd integers is added to three times the larger, the result is 125. Find the integers. Solving Problems with Consecutive Integers Solve. State the answer. The lesser integer is 17 and the greater integer is 17 + 2 = 19. Step 5 Check. The sum of 4(17) and 3(19) is 125. The answer is correct. Step 6 4x + 3 ( x + 2 ) = 125 4x + 3x + 6 = 125 7x + 6 = 125 7x + 6 – 6 = 125 – 6 7x = 119 x = 17 Combine terms. Distribute. Subtract 6. Combine terms. Divide by 7. 2.4 An Introduction to Applications of Linear Equations