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Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 1 Chapter 2 Equations, Inequalities, and Applications Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 2 2.4 An Introduction to Applications of Linear Equations Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 3 2.4 An Introduction to Applications of Linear Equations Objectives 1. 2. 3. 4. 5. Learn the six steps for solving applied problems. Solve problems involving unknown numbers. Solve problems involving sums of quantities. Solve problems involving supplementary and complementary angles. Solve problems involving consecutive integers. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 4 2.4 An Introduction to Applications of Linear Equations Solving Applied Problems 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. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 5 2.4 An Introduction to Applications of Linear Equations Solving Problems Involving Unknown Numbers Example 1 The product of 3, and a number decreased by 2, is 42. What is the number? Step 1 Read the problem carefully. We are asked to find a number. Step 2 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. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 6 2.4 An Introduction to Applications of Linear Equations Solving Problems Involving Unknown Numbers Example 1 (cont.) The product of 3, and a number decreased by 2, is 42. What is the number? Notice the placement of the commas! Step 3 Write an equation. The product of 3, and 3• a number decreased by 2, ( x – 2 ) is 42. = 42 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. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 7 2.4 An Introduction to Applications of Linear Equations Solving Problems Involving Unknown Numbers Example 1 (cont.) The product of 3, and a number decreased by 2, is 42. What is the number? Step 4 Solve the equation. 3 ( x – 2 ) = 42 3x – 6 = 42 3x – 6 + 6 = 42 + 6 Distribute. Add 6. 3x = 48 Combine terms. 3x 48 = 3 3 Divide by 3. x = 16 Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 8 2.4 An Introduction to Applications of Linear Equations Solving Problems Involving Unknown Numbers Example 1 (cont.) The product of 3, and a number decreased by 2, is 42. What is the number? 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. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 9 2.4 An Introduction to Applications of Linear Equations Solving Problems Involving Sums of Quantities 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? Step 1 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 2 Assign a variable. Let x = the number pencils in the box. Then x + 16 = the number pens in the box. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 10 2.4 An Introduction to Applications of Linear Equations Solving Problems Involving Sums of Quantities 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? Recall: x = # of pencils, x + 16 = # of pens Step 3 Write an equation. The total is the number of pens plus the number of pencils 68 = ( x + 16 ) + x Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 11 2.4 An Introduction to Applications of Linear Equations Solving Problems Involving Sums of Quantities 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? Step 4 Solve the equation. 68 = ( x + 16 ) + x 68 = 2x + 16 68 – 16 = 2x + 16 – 16 Combine terms. Subtract 16. 52 = 2x Combine terms. 52 2x = 2 2 26 = x Divide by 2. or Copyright © 2010 Pearson Education, Inc. All rights reserved. x = 26 2.4 – Slide 12 2.4 An Introduction to Applications of Linear Equations Solving Problems Involving Sums of Quantities 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? Recall: x = # of pencils, x + 16 = # of pens 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. 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. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 13 2.4 An Introduction to Applications of Linear Equations Solving Problems Involving Sums of Quantities 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? Step 1 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 2 Assign a variable. Let x = the number of milliliters of acid required. Then 12x = the number of milliliters of water required. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 14 2.4 An Introduction to Applications of Linear Equations Solving Problems Involving Sums of Quantities 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? Recall: x = ml. of acid, 12x = ml. of water. Step 3 Write an equation. A diagram is sometimes helpful. Beaker Acid Water = 286 x 12x x + Copyright © 2010 Pearson Education, Inc. All rights reserved. 12x = 286 2.4 – Slide 15 2.4 An Introduction to Applications of Linear Equations Solving Problems Involving Sums of Quantities 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? Step 4 Solve. x + 12x = 286 13x = 286 13x = 286 13 13 Combine terms. Divide by 13. x = 22 Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 16 2.4 An Introduction to Applications of Linear Equations Solving Problems Involving Sums of Quantities 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? Recall: x = ml of acid, 12x = ml of water Step 5 State the answer. The beaker requires 22 ml of acid and 12(22) = 264 ml of water. Step 6 Check. Since 22 + 264 = 286, and 264 is 12 times 22, the answer checks. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 17 2.4 An Introduction to Applications of Linear Equations Solving Problems Involving Sums of Quantities 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? Step 1 Read the problem carefully. Three lengths must be found. Step 2 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. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 18 2.4 An Introduction to Applications of Linear Equations Solving Problems Involving Sums of Quantities 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? Step 3 Write an equation. 96 inches 3x Longest Middle-sized Shortest 3x + x + x – 14 Copyright © 2010 Pearson Education, Inc. All rights reserved. x – 14 x is = Total length 96 2.4 – Slide 19 2.4 An Introduction to Applications of Linear Equations Solving Problems Involving Sums of Quantities 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? Step 4 Solve. 96 = 3x + x + x – 14 96 96 + 14 110 110 5 22 = 5x – 14 Combine terms. Add 14. = 5x – 14 + 14 = 5x Combine terms. 5x Divide by 5. = 5 = x or x = 22 Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 20 2.4 An Introduction to Applications of Linear Equations Solving Problems Involving Sums of Quantities 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? Recall: x = length of middle-sized piece, 3x = length of longest piece, and x – 14 = length of shortest piece. 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. Step 6 Check. The sum of the lengths is 96 in. All conditions of the problem are satisfied. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 21 2.4 An Introduction to Applications of Linear Equations Solving with Supplementary and Complementary Angles 1 2 Angles 1 and 2 are complementary. They form a right angle, indicated by . 3 4 Angles 3 and 4 are supplementary. They form a straight angle. Copyright © 2010 Pearson Education, Inc. All rights reserved. 180 (degrees) Straight angle 2.4 – Slide 22 2.4 An Introduction to Applications of Linear Equations Solving with Supplementary and Complementary Angles Measure of x: Complement of x: Supplement of x: 90 – x 180 – x x x x 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. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 23 2.4 An Introduction to Applications of Linear Equations Solving with Supplementary and Complementary Angles Example 5 Find the measure of an angle whose supplement is 20° more than three times its complement. Step 1 Read the problem. We are to find the measure of an angle, given information about its complement and supplement. Step 2 Assign a variable. Let Then x = the degree measure of the angle. 90 – x = the degree measure of its complement; 180 – x = the degree measure of its supplement. Step 3 Write an equation. Supplement is 20 more than three times its complement. 180 – x = 20 + Copyright © 2010 Pearson Education, Inc. All rights reserved. 3• ( 90 – x ) 2.4 – Slide 24 2.4 An Introduction to Applications of Linear Equations Solving with Supplementary and Complementary Angles Example 5 (cont.) Find the measure of an angle whose supplement is 20° more than three times its complement. Step 4 Solve. 180 – x = 20 + 3 ( 90 – x ) 180 – x = 20 + 270 – 3x Distribute. Combine terms. 180 – x = 290 – 3x 180 – x + 3x = 290 – 3x + 3x Add 3x. Combine terms. 180 + 2x = 290 Subtract 180. 180 + 2x – 180 = 290 – 180 2x = 110 Combine terms. 2x 110 Divide by 2. = 2 2 x = 55 Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 25 2.4 An Introduction to Applications of Linear Equations Solving with Supplementary and Complementary Angles Example 5 (cont.) Find the measure of an angle whose supplement is 20° more than three times its complement. Step 5 State the answer. The measure of the angle is 55°. Step 6 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. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 26 2.4 An Introduction to Applications of Linear Equations Solving Problems with Consecutive Integers 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. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 27 2.4 An Introduction to Applications of Linear Equations Solving Problems with Consecutive Integers Example 6 The sum of two consecutive checkbook check numbers is 893. Find the numbers. Step 1 Read the problem. We are to find the check numbers. Step 2 Assign a variable. Let x = the lesser check number. Then x + 1 = the greater check number. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 28 2.4 An Introduction to Applications of Linear Equations Solving Problems with Consecutive Integers Example 6 (cont.) The sum of two consecutive checkbook check numbers is 893. Find the numbers. Recall: Let Then x = the lesser check number. x + 1 = the greater check number. Step 3 Write an equation. The sum of the check numbers is 893, so Step 4 Solve. x + (x + 1) 2x + 1 2x x Copyright © 2010 Pearson Education, Inc. All rights reserved. = = = = 893 893 892 446 Combine terms. Subtract 1. Divide by 2. 2.4 – Slide 29 2.4 An Introduction to Applications of Linear Equations Solving Problems with Consecutive Integers Example 6 (cont.) The sum of two consecutive checkbook check numbers is 893. Find the numbers. Step 5 State the answer. The lesser check number is 446 and the greater check number is 447. Step 6 Check. The sum of 446 and 447 is 893. The answer is correct. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 30 2.4 An Introduction to Applications of Linear Equations Solving Problems with Consecutive Integers 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. Step 1 Read the problem. We are to find two consecutive odd integers. Step 2 Assign a variable. Let x = the lesser integer. Then x + 2 = the greater integer. Step 3 Write an equation. 4 times the smaller is added to 3 times the larger, the result is 125 4• x + Copyright © 2010 Pearson Education, Inc. All rights reserved. 3• (x+2) = 125 2.4 – Slide 31 2.4 An Introduction to Applications of Linear Equations Solving Problems with Consecutive Integers 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. Step 4 Solve. 4x + 3 ( x + 2 ) = 125 4x + 3x + 6 = 125 Distribute. Combine terms. 7x + 6 = 125 7x + 6 – 6 = 125 – 6 Subtract 6. 7x = 119 Combine terms. x = 17 Divide by 7. Step 5 State the answer. The lesser integer is 17 and the greater integer is 17 + 2 = 19. Step 6 Check. The sum of 4(17) and 3(19) is 125. The answer is correct. Copyright © 2010 Pearson Education, Inc. All rights reserved. 2.4 – Slide 32