1. Pre-Algebra Solve 5v – 12 = 8. Solving Two-Step Equations Lesson 7-1 5v – 12 = 8 5v – 12 + 12 = 8 + 12Add 12 to each side. 5v = 20Simplify. v = 4Simplify. = Divide each side by 5. 5v55v5 20 5 Check: 5v – 12 = 8 5(4) – 12 8Replace v with 4. 20 – 12 8Multiply. 8 = 8Simplify. Additional Examples

2. Pre-Algebra Solve 7 – 3b = 1. Solving Two-Step Equations Lesson 7-1 7 – 3b = 1 –7 + 7 – 3b = –7 + 1 Add –7 to each side. 0 – 3b = –6Simplify. –3b = –60 – 3b = –3b. b = 2Simplify. = Divide each side by –3. –3b –3 –6 –3 Additional Examples

3. Pre-Algebra You borrow $350 to buy a bicycle. You agree to pay $100 the first week, and then $25 each week until the balance is paid off. To find how many weeks w it will take you to pay for the bicycle, solve 100 + 25w = 350. Solving Two-Step Equations Lesson 7-1 It will take you 10 weeks to pay for the bicycle. 100 + 25w = 350 100 + 25w – 100 = 350 – 100 Subtract 100 from each side. 25w = 250 Simplify. w = 10Simplify. = Divide each side by 25. 25w 25 250 25 Additional Examples

4. Pre-Algebra Solving Multi-Step Equations In his stamp collection, Jorge has five more than three times as many stamps as Helen. Together they have 41 stamps. Solve the equation s + 3s + 5 = 41. Find the number of stamps each one has. Lesson 7-2 s + 3s + 5 = 41 4s + 5 = 41 Combine like terms. 4s + 5 – 5 = 41 – 5 Subtract 5 from each side. 4s = 36 Simplify. s = 9 Simplify. = Divide each side by 4. 4s44s4 36 4 Additional Examples

5. Pre-Algebra Solving Multi-Step Equations (continued) Lesson 7-2 Check: Is the solution reasonable? Helen and Jorge have a total of 41 stamps. Since 9 + 32 = 41, the solution is reasonable. Helen has 9 stamps. Jorge has 3(9) + 5 = 32 stamps. Additional Examples

6. Pre-Algebra Solving Multi-Step Equations The sum of three consecutive integers is 42. Find the integers. Lesson 7-2 sum of three consecutive integers 42 is Words Let = the least integer. n Then = the second integer, n + 1 and = the third integer. n + 2 + nn + 1n + 2 Equation 42 = Additional Examples

7. Pre-Algebra Solving Multi-Step Equations (continued) Lesson 7-2 n + (n + 1) + (n + 2) = 42 (n + n + n) + (1 + 2) = 42Use the Commutative and Associative Properties of Addition to group like terms together. 3n + 3 = 42Combine like terms. 3n + 3 – 3 = 42 – 3 Subtract 3 from each side. 3n = 39Simplify. n = 13Simplify. = Divide each side by 3. 3n33n3 39 3 Additional Examples

8. Pre-Algebra Solving Multi-Step Equations (continued) Lesson 7-2 If n = 13, then n + 1 = 14, and n + 2 = 15. The three integers are 13, 14, and 15. Check:Is the solution reasonable? Yes, because 13 + 14 + 15 = 42. Additional Examples

9. Pre-Algebra Solving Multi-Step Equations Solve each equation. Lesson 7-2 a. 4(2q – 7) = –4 4(2q – 7) = –4 8q – 28 = –4Use the Distributive Property. 8q – 28 + 28 = –4 + 28Add 28 to each side. 8q = 24Simplify. q = 3Simplify. Divide each side by 8.= 8q88q8 24 8 Additional Examples

10. Pre-Algebra Solving Multi-Step Equations (continued) Lesson 7-2 b. 44 = –5(r – 4) – r 44 = –5(r – 4) – r 44 = –5r + 20 – rUse the Distributive Property. 44 = –6r + 20Combine like terms. 44 – 20 = –6r + 20 – 20Subtract 20 from each side. 24 = –6rSimplify. –4 = rSimplify. Divide each side by –6.= 24 –6 –6r –6 Additional Examples

11. Pre-Algebra Multi-Step Equations With Fractions and Decimals Solve p – 7 = 11. Lesson 7-3 p – 7 = 11 3434 Add 7 to each side.p – 7 + 7 = 11 + 7 3434 Simplify. p = 18 3434 p = 3434 4343 4343 18Multiply each side by, the reciprocal of. 3434 4343 1p = 4 18 3 1 6 Divide common factors. p = 24Simplify. 3434 Additional Examples

12. Pre-Algebra Multi-Step Equations With Fractions and Decimals (continued) Lesson 7-3 Check: p – 7 = 11 (24) – 7 11Replace p with 24. Simplify. 18 – 7 11 11 = 11 3434 3434 – 7 11 Divide common factors. 3 24 4 1 6 Additional Examples

13. Pre-Algebra Multi-Step Equations With Fractions and Decimals Solve y + 3 =. Lesson 7-3 y + 3 = 1212 2323 Multiply each side by 6, the LCM of 2 and 3. 6y + 3 = 1212 6 2323 3y + 18 – 18 = 4 – 18Subtract 18 from each side. 3y = –14Simplify. 3y + 18 = 4Simplify. Use the Distributive Property. 6 y + 6 3 = 1212 6 2323 Divide each side by 3. 3y33y3 –14 3 = Simplify.y = –4 2323 2323 1212 Additional Examples

14. Pre-Algebra Multi-Step Equations With Fractions and Decimals Lesson 7-3 Suppose your cell phone plan is $30 per month plus $.05 per minute. Your bill is $36.75. Use the equation 30 + 0.05x = 36.75 to find the number of minutes on your bill. There are 135 minutes on your bill. 30 + 0.05x = 36.75 x = 135Simplify. 30 – 30 + 0.05x = 36.75 – 30Subtract 30 from each side. 0.05x = 6.75Simplify. Divide each side by 0.05.= 6.75 0.05 0.05x 0.05 Additional Examples

15. Pre-Algebra Problem Solving Strategy: Write an Equation A moving van rents for $29.95 a day plus $.12 a mile. Mr. Reynolds’s bill was $137.80 and he drove the van 150 mi. For how many days did he have the van? Lesson 7-4 Words number of days $29.95/d + $.12/mi 150 mi$137.80 = Let d = number of days Mr. Reynolds had the van. Equation 137.80d 29.95 + 0.12 150 = Additional Examples

16. Pre-Algebra Problem Solving Strategy: Write an Equation (continued) Lesson 7-4 Mr. Reynolds had the van for 4 days. d 29.95 + 0.12 150 = 137.80 29.95d + 18 = 137.80Multiply 0.12 and 150. 29.95d + 18 – 18 = 137.80 – 18Subtract 18 from each side. 29.95d = 119.80Simplify. d = 4Simplify. Divide each side by 29.95. = 29.95d 29.95 119.80 29.95 Additional Examples

17. Pre-Algebra Solving Equations With Variables on Both Sides Solve 4c + 3 = 15 – 2c. Lesson 7-5 4c + 3 = 15 – 2c 4c + 2c + 3 = 15 – 2c + 2cAdd 2c to each side. 6c + 3 = 15Combine like terms. 6c + 3 – 3 = 15 – 3Subtract 3 from each side. 6c = 12Simplify. c = 2Simplify. Divide each side by 6.= 6c66c6 12 6 Check: 4c + 3 = 15 – 2c 4(2) + 3 15 – 2(2) 8 + 3 15 – 4 11 = 11 Substitute 2 for c. Multiply. Additional Examples

18. Pre-Algebra Solving Equations With Variables on Both Sides Lesson 7-5 Steve types at a rate of 15 words/min and Jenny types at a rate of 20 words/min. Steve and Jenny are both typing the same document, and Steve starts 5 min before Jenny. How long will it take Jenny to catch up with Steve? words Jenny types = words Steve types Words 20 words/min Jenny’s time = 15 words/min Steve’s time Let = Jenny’s time. x Then x + 5 = Steve’s time. Equation20 x = 15 (x + 5) Additional Examples

19. Pre-Algebra Solving Equations With Variables on Both Sides Lesson 7-5 (continued) 20x = 15(x + 5) Jenny will catch up with Steve in 15 min. 20x = 15x + 75Use the Distributive Property. 20x – 15x = 15x – 15x + 75Subtract 15x from each side. 5x = 75Combine like terms. x = 15Simplify. Divide each side by 5. = 5x55x5 75 5 Additional Examples

20. Pre-Algebra Solving Equations With Variables on Both Sides Lesson 7-5 (continued) Check: Test the result. At 20 words/min for 15 min, Jenny types 300 words. Steve’s time is five min longer. He types for 20 min. At 15 words/min for 20 min, Steve types 300 words. Since Jenny and Steve each type 300 words, the answer checks. Additional Examples

21. Pre-Algebra Solving Two-Step Inequalities Solve and graph 7g + 11 > 67. Lesson 7-6 7g + 11 > 67 7g + 11 – 11 > 67 – 11Subtract 11 from each side. 7g > 56Simplify. g > 8Simplify. Divide each side by 7. > 7g77g7 56 7 Additional Examples

22. Pre-Algebra Solving Two-Step Inequalities Solve 6 – r – 6. Lesson 7-6 < 2323 6 – r – 6 < 2323 6 + 6 – r – 6 + 6 < 2323 Add 6 to each side. Simplify. 12 – r < 2323 Simplify. > –18 r, or r –18 < 3232 –(12) 3232 – 2323 –r Multiply each side by. Reverse the direction of the inequality symbol. 3232 – > Additional Examples

23. Pre-Algebra Solving Two-Step Inequalities Dale has $25 to spend at a carnival. If the admission to the carnival is $4 and the rides cost $1.50 each, what is the greatest number of rides Dale can go on? Lesson 7-6 25 Inequality4 + 1.5r < Let = number of rides Dale goes on. Words $4 admission + $1.50/ride number of rides is less than or equal to $25 r Additional Examples

24. Pre-Algebra Solving Two-Step Inequalities (continued) Lesson 7-6 The greatest number of rides Dale can go on is 14. 4 + 1.5r 25 < Subtract 4 from each side.4 + 1.5r – 4 25 – 4 < Simplify.1.5r 21 < Divide each side by 1.5. 1.5r 1.5 21 1.5 < Simplify.r 14 < Additional Examples

25. Pre-Algebra Solve the circumference formula C = 2 r for r. Transforming Formulas Lesson 7-7 C = 2 r Use the Division Property of Equality. C2C2 2 r 2 = Simplify. C2C2 = r, or r = C2C2 Additional Examples

26. Pre-Algebra Transforming Formulas Solve the perimeter formula P = 2 + 2w for w. Lesson 7-7 P = 2 + 2w Simplify.P – 2 = 2w Subtract 2 from each side.P – 2 = 2 + 2w – 2 Multiply each side by. 1212 1212 (P – 2 ) = (2w) 1212 1212 P – = wUse the Distributive Property and simplify. Additional Examples

27. Pre-Algebra Transforming Formulas You plan a 600-mi trip to New York City. You estimate your trip will take about 10 hours. To estimate your average speed, solve the distance formula d = rt for r. Then substitute to find the average speed. Lesson 7-7 d = rt Your average speed will be about 60 mi/h. = 60Simplify. Divide each side by t.= dtdt rt t Simplify.= r, or r = dtdt dtdt Replace d with 600 and t with 10.r = 600 10 Additional Examples

28. Pre-Algebra Transforming Formulas The high temperature one day in San Diego was 32°C. Solve C = (F – 32) for F. Then substitute to find the temperature in degrees Fahrenheit. Lesson 7-7 5959 C = (F – 32) 5959 (C) = (F – 32) 9595 9595 5959 Multiply each side by. 9595 Simplify.C = F – 32 9595 Additional Examples

29. Pre-Algebra Transforming Formulas (continued) Lesson 7-7 32°C is 89.6°F. Add 32 to each side. C + 32 = F – 32 + 32 9595 Simplify and rewrite.C + 32 = F, or F = C + 32 9595 9595 Replace C with 32. Simplify. F = (32) + 32 = 89.6 9595 Additional Examples

30. Pre-Algebra Simple and Compound Interest Suppose you deposit $1,000 in a savings account that earns 6% in interest per year. Lesson 7-8 a. Find the interest earned in two years. Find the total of principal plus interest. The account will earn $120 in two years. The total of principal plus interest will be $1,120. I = prtUse the simple interest formula. I = 1,000 0.06 2Replace p with 1,000, r with 0.06, and t with 2. I = 120Simplify. total = 1,000 + 120 = 1,120Find the total. Additional Examples

31. Pre-Algebra Simple and Compound Interest (continued) Lesson 7-8 b. Find the interest earned in six months. Find the total of principal plus interest. The account will earn $30 in six months. The total of principal plus interest will be $1,030. I = prtUse the simple-interest formula. I = 1,000 0.06 0.5Replace p with 1,000, r with 0.06, and t with 0.5. I = 30Simplify. Total = 1,000 + 30 = 1,030Find the total. Write the months as part of a year. t = = = 0.5 1212 6 12 Additional Examples

32. Pre-Algebra Year 5 : $486.20486.20 0.05 = 24.31486.20 + 24.31 = 510.51 Year 6 : $510.51510.51 0.05 25.53510.51 + 25.53 = 536.04 Year 7 : $536.04536.04 0.05 26.80536.04 + 26.80 = 562.84 Year 8 : $562.84562.84 0.05 28.14562.84 + 28.14 = 590.98 Simple and Compound Interest You deposit $400 in an account that earns 5% interest compounded annually (once per year). The balance after the first four years is $486.20. What is the balance in your account after another 4 years, a total of 8 years? Round to the nearest cent. Lesson 7-8 After the next four years, for a total of 8 years, the balance is $590.98. Interest Balance Principal at Beginning of Year Additional Examples

33. Pre-Algebra Simple and Compound Interest Find the balance on a deposit of $2,500 that earns 3% interest compounded semiannually for 4 years. Lesson 7-8 The interest rate r for compounding semiannually is 0.03 ÷ 2, or 0.015. The number of payment periods n is 4 years  2 interest periods per year, or 8. The balance is $2,816.23. B = p(1 + r) n Use the compound interest formula. B = 2,500(1 + 0.015) 8 Replace p with 2,500, r with 0.015, and n with 8. Use a calculator. Round to the nearest cent.B 2,816.23 Additional Examples