1. Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 2.3 - 1

2. Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 2.3 - 2 Linear Equations and Applications Chapter 2

3. Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 2.3 - 3 2.3 Applications of Linear Equations

4. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 4 2.3 Applications of Linear Equations Objectives 1.Translate from words to mathematical expressions. 2.Write expressions from given information. 3.Distinguish between expressions and equations. 4.Use the six steps in solving an applied problem. 5.Solve percent problems. 6.Solve investment problems. 7.Solve mixture problems.

5. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 5 2.3 Applications of Linear Equations Problem-Solving Hint PROBLEM-SOLVING HINT Usually there are key words and phrases in a verbal problem that translate into mathematical expressions involving addition, subtraction, multiplication, and division. Translations of some commonly used expressions follow.

6. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 6 2.3 Applications of Linear Equations Translating from Words to Mathematical Expressions Verbal Expression The sum of a number and 2 Mathematical Expression (where x and y are numbers) Addition 3 more than a number 7 plus a number 16 added to a number A number increased by 9 The sum of two numbers x + 2 x + 3 7 + x x + 16 x + 9 x + y

7. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 7 2.3 Applications of Linear Equations Translating from Words to Mathematical Expressions Verbal Expression 4 less than a number Mathematical Expression (where x and y are numbers) Subtraction 10 minus a number A number decreased by 5 A number subtracted from 12 The difference between two numbers x – 4 10 – x x – 5 12 – x x – y

8. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 8 2.3 Applications of Linear Equations Translating from Words to Mathematical Expressions Verbal Expression 14 times a number Mathematical Expression (where x and y are numbers) Multiplication A number multiplied by 8 Triple (three times) a number The product of two numbers 14x 8x8x 3x3x xy of a number (used with fractions and percent) 3 4 x 3 4

9. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 9 2.3 Applications of Linear Equations Translating from Words to Mathematical Expressions Verbal Expression The quotient of 6 and a number Mathematical Expression (where x and y are numbers) Division A number divided by 15 The ratio of two numbers or the quotient of two numbers (x ≠ 0) 6 x (y ≠ 0) x y x 15

10. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 10 CAUTION Because subtraction and division are not commutative operations, be careful to correctly translate expressions involving them. For example, “5 less than a number” is translated as x – 5, not 5 – x. “A number subtracted from 12” is expressed as 12 – x, not x – 12. For division, the number by which we are dividing is the denominator, and the number into which we are dividing is the numerator. For example, “a number divided by 15” and “15 divided into x ” both translate as. Similarly, “the quotient of x and y ” is translated as. 2.3 Applications of Linear Equations Caution x 15 x y

11. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 11 2.3 Applications of Linear Equations Indicator Words for Equality Equality The symbol for equality, =, is often indicated by the word is. In fact, any words that indicate the idea of “sameness” translate to =.

12. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 12 2.3 Applications of Linear Equations Translating Words into Equations Verbal SentenceEquation 16x – 25 = 87 If the product of a number and 16 is decreased by 25, the result is 87. = 48 The quotient of a number and the number plus 6 is 48. x + 6 x + x = 54 The quotient of a number and 8, plus the number, is 54. 8 x Twice a number, decreased by 4, is 32. 2x – 4 = 32

13. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 13 2.3 Applications of Linear Equations Distinguishing between Expressions and Equations (a) 4(6 – x) + 2x – 1 (b) 4(6 – x) + 2x – 1 = –15 There is no equals sign, so this is an expression. Because of the equals sign, this is an equation. Decide whether each is an expression or an equation. Note that the expression in part (a) simplifies to the expression –2x + 23 and the equation in part (b) has solution 19.

14. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 14 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 the answer to the problem. Does it seem reasonable? Step 6 Check the answer in the words of the original problem. 2.3 Applications of Linear Equations Six Steps to Solving Application Problems

15. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 15 2.3 Applications of Linear Equations Solving a Geometry Problem Step 1Read the problem. We must find the length and width of the rectangle. The length is 2 ft more than three times the width and the perimeter is 124 ft. The length of a rectangle is 2 ft more than three times the width. The perimeter of the rectangle is 124 ft. Find the length and the width of the rectangle. Step 2Assign a variable. Let W = the width; then 2 + 3W = length. Make a sketch. W 2 + 3W Step 3Write an equation. The perimeter of a rectangle is given by the formula P = 2L + 2W. 124 = 2(2 + 3W) + 2W Let L = 2 + 3W and P = 124.

16. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 16 2.3 Applications of Linear Equations Solving a Geometry Problem Step 4Solve the equation obtained in Step 3. The length of a rectangle is 2 ft more than three times the width. The perimeter of the rectangle is 124 ft. Find the length and the width of the rectangle. 124 = 2(2 + 3W) + 2W 124 = 4 + 6W + 2W 124 – 4 = 4 + 8W – 4 120 8W 8 15 = W Distributive property 124 = 4 + 8W Combine like terms. Subtract 4. Divide by 8. 120 = 8W =

17. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 17 2.3 Applications of Linear Equations Solving a Geometry Problem Step 5State the answer. The width of the rectangle is 15 ft and the length is 2 + 3(15) = 47 ft. The length of a rectangle is 2 ft more than three times the width. The perimeter of the rectangle is 124 ft. Find the length and the width of the rectangle. Step 6Check the answer by substituting these dimensions into the words of the original problem.

18. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 18 2.3 Applications of Linear Equations Finding Unknown Numerical Quantities Step 1Read the problem. We are asked to find the total number of each type of cookie baked. A local grocery store baked a combined total of 912 chocolate chip cookies and sugar cookies. If they baked 336 more chocolate chip cookies than sugar cookies, how many of each did the store bake? Step 2Assign a variable. Let s = the total number of sugar cookies baked. Since there were 336 more chocolate chip cookies baked, s + 336 = the total number of chocolate chip cookies baked. Step 3Write an equation. The sum of the numbers of the cookies baked is 912, so sugarchocolate chip=total+ s(s + 336) =912+

19. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 19 2.3 Applications of Linear Equations Finding Unknown Numerical Quantities Step 4Solve the equation. A local grocery store baked a combined total of 912 chocolate chip cookies and sugar cookies. If they baked 336 more chocolate chip cookies than sugar cookies, how many of each did the store bake? s + (s + 336) = 912 2s + 336 = 912 Combine like terms. Subtract 336. Divide by 2. 2s + 336 – 336 = 912 – 336 2s = 576 2 s = 288 2s 576 =

20. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 20 2.3 Applications of Linear Equations Finding Unknown Numerical Quantities Step 5State the answer. We let s represent the number of sugar cookies, so there were 288 sugar cookies baked. Also, A local grocery store baked a combined total of 912 chocolate chip cookies and sugar cookies. If they baked 336 more chocolate chip cookies than sugar cookies, how many of each did the store bake? (s + 336) = 624 is the number of chocolate chip cookies baked. Step 6Check. 624 is 336 more than 288, and the sum of 624 and 288 is 912. The conditions of the problem are satisfied, and our answer checks.

21. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 21 CAUTION A common error in solving applied problems is forgetting to answer all the questions asked in the problem. In the previous example, we were asked to find the number of each type of cookie baked, so there was an extra step at the end to find the number of chocolate chip cookies baked. 2.3 Applications of Linear Equations Caution

22. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 22 2.3 Applications of Linear Equations Percent Reminder Percent Recall from Section 2.2 that percent means “per one hundred,” so 3% means 0.03, 12% means 0.12, and so on.

23. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 23 2.3 Applications of Linear Equations Solving a Percent Problem Step 1Read the problem. We are given that the number of raffle tickets sold increased by 350% from the first to the second day, and there were 1440 raffle tickets sold altogether. We must find the number of raffle tickets sold the first day. During a 2-day fundraiser, a local school sold 1440 raffle tickets. If they sold 350% more raffle tickets on the second day than the first day, how many raffle tickets did they sell on the first day? Step 2Assign a variable. Let x = the total number raffle tickets sold on the first day of the fundraiser. 350% = 350(.01) = 3.5, so 3.5 x represents the number of raffle tickets sold on the second day of the fundraiser.

24. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 24 2.3 Applications of Linear Equations Solving a Percent Problem During a 2-day fundraiser, a local school sold 1440 raffle tickets. If they sold 350% more raffle tickets on the second day than the first day, how many raffle tickets did they sell on the first day? Step 3Write an equation from the information given. tickets sold the first day =total tickets sold+ x(3.5x) =1440+ Step 4Solve the equation. 1x + 3.5x = 1440 Identity property 4.5x = 1440 Combine like terms. x = 320 Divide by 4.5.

25. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 25 2.3 Applications of Linear Equations Solving a Percent Problem During a 2-day fundraiser, a local school sold 1440 raffle tickets. If they sold 350% more raffle tickets on the second day than the first day, how many raffle tickets did they sell on the first day? Step 5State the answer. There were 320 raffle tickets sold on the first day of the 2-day fundraiser. Step 6Check that the number of raffle tickets sold on the second day, 1440 – 320 = 1120, is 350% of 320.

26. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 26 CAUTION Avoid two common errors that occur in solving problems like the one in the previous example. 1. Do not try to find 350% of 1440 and subtract that amount from 1440. The 350% should be applied to the number of tickets sold on the first day, not the total number of tickets sold. 2. Do not write the equation as The percent must be multiplied by some amount; in this case, the amount is the number of tickets sold on the first day, giving 3.5x. 2.3 Applications of Linear Equations Caution x + 3.5 = 1440.Incorrect

27. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 27 2.3 Applications of Linear Equations Solving an Investment Problem Step 1Read the problem. We must find the amount invested in each account. A local company has $50,000 to invest. It will put part of the money in an account paying 3% interest and the remainder into stocks paying 5%. If the total annual income from these investments will be $2180, how much will be invested in each account? Step 2Assign a variable. Let x = the amount to invest at 3%; 50,000 – x = the amount to invest at 5%. Rate (as a decimal)Interest.030.03x Principle.05 x 50,000 – x The formula for interest is I = p r t. Time 1 1.05(50,000 – x) 50,0002180  Totals 

28. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 28 2.3 Applications of Linear Equations Solving an Investment Problem A local company has $50,000 to invest. It will put part of the money in an account paying 3% interest and the remainder into stocks paying 5%. If the total annual income from these investments will be $2180, how much will be invested in each account? Rate (as a decimal)Interest 0.030.03x Principle 0.05 x 50,000 – x Time 1 10.05(50,000 – x) 50,0002180  Totals  Step 3Write an equation. The last column of the table gives the equation. interest at 3%interest at 5%=total interest+ 0.03x0.05(50,000 – x) =2180+

29. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 29 2.3 Applications of Linear Equations Solving an Investment Problem A local company has $50,000 to invest. It will put part of the money in an account paying 3% interest and the remainder into stocks paying 5%. If the total annual income from these investments will be $2180, how much will be invested in each account? Step 4Solve the equation. We do so without clearing decimals. 0.03x + 0.05(50,000) – 0.05x = 2180 Distributive property 0.03x0.05(50,000 – x) =2180+ 0.03x + 2500 – 0.05x = 2180 Multiply. –0.02x + 2500 = 2180 Combine like terms. –0.02x = –320 Subtract 2500 x = 16,000 Divide by –.02.

30. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 30 2.3 Applications of Linear Equations Solving an Investment Problem A local company has $50,000 to invest. It will put part of the money in an account paying 3% interest and the remainder into stocks paying 5%. If the total annual income from these investments will be $2180, how much will be invested in each account? Step 5State the answer. The company will invest $16,000 at 3%. At 5%, the company will invest $50,000 – $16,000 = $34,000. and Step 6Check by finding the annual interest at each rate; they should total $2180. 0.03($16,000) = $4800.05($34,000) = $1700 $480 + $1700 = $2180, as required.

31. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 31 2.3 Applications of Linear Equations Problem-Solving Hint PROBLEM-SOLVING HINT In the previous example, we chose to let the variable represent the amount invested at 3%. Students often ask, “Can I let the variable represent the other unknown?” The answer is yes. The equation will be different, but in the end the two answers will be the same.

32. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 32 2.3 Applications of Linear Equations Solving a Mixture Problem Step 1Read the problem. The problem asks for the amount of 80% solution to be used. A chemist must mix 12 L of a 30% acid solution with some 80% solution to get a 60% solution. How much of the 80% solution should be used? Step 2Assign a variable. Let x = the number of liters of 80% solution to be used. + = 30%80% 30% 80% 12 L Unknown number of liters, x (12 + x)L

33. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 33 Step 2Assign a variable. Let x = the number of liters of 80% solution 2.3 Applications of Linear Equations Solving a Mixture Problem A chemist must mix 12 L of a 30% acid solution with some 80% solution to get a 60% solution. How much of the 80% solution should be used? Percent (as a decimal)Liters of Pure Acid 0.30 0.30(12) = 3.6 Number of Liters 0.80 12 x 0.08x 12 + x 0.60(12 + x)0.60 Sum must be equal 3.6.08x.60(12 + x) Step 3 Write an equation. += + = 30% 80% 30% 80% 12 Lx L (12 + x)L

34. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 34 Step 4Solve. 2.3 Applications of Linear Equations Solving a Mixture Problem A chemist must mix 12 L of a 30% acid solution with some 80% solution to get a 60% solution. How much of the 80% solution should be used? 3.6 0.08x0.60(12 + x) += 3.6 + 0.80x = 7.2 + 0.60x Distributive property 0.20x = 3.6 Subtract 3.6 and 0.60x. x = 18 Divide by 0.20. Step 5State the answer. The chemist should use 18 L of 80% solution.

35. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 35 Step 6Check. 12 L of 30% solution plus 18 L of 80% solution is 2.3 Applications of Linear Equations Solving a Mixture Problem A chemist must mix 12 L of a 30% acid solution with some 80% solution to get a 60% solution. How much of the 80% solution should be used? 12(0.30) + 18(0.8) = 18 L of acid. Similarly, 12 + 18 or 30 L of 60% solution has 30(0.6) = 18 L Percent (as a decimal)Liters of Pure AcidNumber of Liters 0.30 0.80 12 18 0.60 3.6 14.4 3018 of acid in the mixture. The total amount of pure acid is 18 L both before and after mixing, so the answer checks.

36. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 36 2.3 Applications of Linear Equations Problem-Solving Hint PROBLEM-SOLVING HINT When pure water is added to a solution, remember that water is 0% of the chemical (acid, alcohol, etc.). Similarly, pure chemical is 100% chemical.

37. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 37 2.3 Applications of Linear Equations Solving a Mixture Problem Step 1Read the problem. The problem asks for the amount of pure alcohol to be used. A chemist must mix 8 L of a 10% alcohol solution with pure alcohol to get a a 40% solution. How much of the pure alcohol solution should be used? Step 2Assign a variable. Let x = the number of liters of 100% solution to be used. + = 10%100% 10% 100% 8 L Unknown number of liters, x (8 + x)L

38. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 38 Step 2Assign a variable. Let x = the number of liters of 100% solution 2.3 Applications of Linear Equations Solving a Mixture Problem A chemist must mix 8 L of a 10% alcohol solution with pure alcohol to get a a 40% solution. How much of the pure alcohol solution should be used? Percent (as a decimal) Pure Alcohol (L) 0.10 0.10(8) =.8 Number of Liters 1.00 8 x 1x1x 8 + x0.40(8 + x) 0.40 Sum must be equal.8 1x1x.40(8 + x) Step 3 Write an equation. += + = 10% 100% 10% 100% 8 Lx L (8 + x)L

39. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 39 Step 4Solve. 2.3 Applications of Linear Equations Solving a Mixture Problem A chemist must mix 8 L of a 10% alcohol solution with pure alcohol to get a a 40% solution. How much of the pure alcohol solution should be used? 0.8 1x1x0.40(8 + x) += 0.8 + 1x = 3.2 + 0.40x Distributive property 0.60x = 2.4 Subtract 0.8 and 0.40x. x = 4 Divide by 0.60. Step 5State the answer. The chemist should use 4 L of pure alcohol.

40. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 2.3 - 40 Step 6Check. 8 L of 10% solution plus 4 L pure alcohol is 2.3 Applications of Linear Equations Solving a Mixture Problem A chemist must mix 8 L of a 10% alcohol solution with pure alcohol to get a a 40% solution. How much of the pure alcohol solution should be used? 8(0.10) + 4(1) = 4.8 L of acid. Similarly, 8 + 4 or 12 L of pure alcohol has 12(0.40) = 4.8 L Percent (as a decimal)Pure Alcohol (L)Number of Liters 0.10 1.00 8 4 0.40 0.8 4 124.8 of alcohol in the mixture. The total amount of pure acid is 4.8 L both before and after mixing, so the answer checks.