Solving Two-Step Equations PRE-ALGEBRA LESSON 7-1 Solve each equation. a. 15 = 8 + n n = 7 b. p – 19 = 4 p = 23 7-1

Solving Two-Step Equations PRE-ALGEBRA LESSON 7-1 (For help, go to Lesson 2-5.) Solve each equation. 1. 9 + k = 17 2. d – 10 = 1 3. y – 5 = –4 4. x + 16 = 4 5. b + 6 = –4 Check Skills You’ll Need 7-1

Solving Two-Step Equations PRE-ALGEBRA LESSON 7-1 Solutions 1. 9 + k = 17 –9 + 9 + k = –9 + 17 k = 8 2. d – 10 = 1 d – 10 + 10 = 1 + 10 d = 11 3. y – 5 = –4 y – 5 + 5 = –4 + 5 y = 1 4. x + 16 = 4 x + 16 – 16 = 4 – 16 x = –12 5. b + 6 = –4 b + 6 – 6 = –4 – 6 b = –10 7-1

Solving Two-Step Equations PRE-ALGEBRA LESSON 7-1 Solve 5v – 12 = 8. 5v – 12 = 8 5v – 12 + 12 = 8 + 12 Add 12 to each side. 5v = 20 Simplify. = Divide each side by 5. 5v 5 20 v = 4 Simplify. Check  5v – 12 = 8 5(4) – 12 8 Replace v with 4. 20 – 12 8 Multiply. 8 = 8 Simplify. Quick Check 7-1

Solving Two-Step Equations PRE-ALGEBRA LESSON 7-1 Solve 7 – 3b = 1. 7 – 3b = 1 –7 + 7 – 3b = –7 + 1 Add –7 to each side. 0 – 3b = –6 Simplify. –3b = –6 0 – 3b = –3b. = Divide each side by –3. –3b –3 –6 b = 2 Simplify. Quick Check 7-1

Solving Two-Step Equations PRE-ALGEBRA LESSON 7-1 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. 100 + 25w = 350 100 + 25w – 100 = 350 – 100   Subtract 100 from each side. 25w = 250 Simplify. = Divide each side by 25. 25w 25 250 w = 10 Simplify. It will take you 10 weeks to pay for the bicycle. Quick Check 7-1

Solving Two-Step Equations PRE-ALGEBRA LESSON 7-1 Solve each equation. 1. 12x – 14 = 10 2. + 7 = –4 3. 9 – w = 13 4. –22 – = –15 5. 4d – 57 = 7 r 3 2 –33 q 5 –4 –35 16 7-1

Solving Multi-Step Equations PRE-ALGEBRA LESSON 7-2 Simplify each expression. a. 2(18) + 3(21 ÷ 7) 45 b. 21 – 5 + 4x + 2(3 + x) 22 + 6x 7-2

Solving Multi-Step Equations PRE-ALGEBRA LESSON 7-2 (For help, go to Lesson 2-3.) Simplify each expression. 1. 2x + 4 + 3x 2. 5y + y 3. 8a – 5a 4. 2 – 4c + 5c 5. 4x + 3 – 2(5 + x) Check Skills You’ll Need 7-2

Solving Multi-Step Equations PRE-ALGEBRA LESSON 7-2 Solutions 1. 2x + 4 + 3x = (2 + 3)x + 4 = 5x + 4 2. 5y + y = 5y + 1y = (5 + 1)y = 6y 3. 8a – 5a = (8 – 5)a = 3a 4. 2 – 4c + 5c = 2 + (–4c + 5c) = 2 + (–4 + 5)c = 2 + c 5. 4x + 3 – 2(5 + x) = 4x + 3 – 10 – 2x = 4x – 2x – 7 = (4 – 2)x – 7 = 2x – 7 7-2

Solving Multi-Step Equations PRE-ALGEBRA LESSON 7-2 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. s + 3s + 5 = 41 4s + 5 = 41 Combine like terms. 4s + 5 – 5 = 41 – 5   Subtract 5 from each side. 4s = 36 Simplify. = Divide each side by 4. 4s 4 36 s = 9 Simplify. 7-2

Solving Multi-Step Equations PRE-ALGEBRA LESSON 7-2 (continued) Helen has 9 stamps. Jorge has 3(9) + 5 = 32 stamps. Check   Is the solution reasonable? Helen and Jorge have a total of 41 stamps. Since 9 + 32 = 41, the solution is reasonable. Quick Check 7-2

Solving Multi-Step Equations PRE-ALGEBRA LESSON 7-2 The sum of three consecutive integers is 42. Find the integers. 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 + + n n + 1 n + 2 Equation 42 = 7-2

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

Solving Multi-Step Equations PRE-ALGEBRA LESSON 7-2 (continued) 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. Quick Check 7-2

Solving Multi-Step Equations PRE-ALGEBRA LESSON 7-2 Solve each equation. a. 4(2q – 7) = –4 4(2q – 7) = –4 8q – 28 = –4 Use the Distributive Property. 8q – 28 + 28 = –4 + 28 Add 28 to each side. 8q = 24 Simplify. Divide each side by 8. = 8q 8 24 q = 3 Simplify. 7-2

Solving Multi-Step Equations PRE-ALGEBRA LESSON 7-2 (continued) b.    44 = –5(r – 4) – r 44 = –5(r – 4) – r 44 = –5r + 20 – r Use the Distributive Property. 44 = –6r + 20 Combine like terms. 44 – 20 = –6r + 20 – 20 Subtract 20 from each side. 24 = –6r Simplify. Divide each side by –6. = 24 –6 –6r Quick Check –4 = r Simplify. 7-2

Solving Multi-Step Equations PRE-ALGEBRA LESSON 7-2 Solve each equation. 1. b + 2b – 11 = 88 2. 6(2n – 5) = –90 3. 3(x + 6) + x = 86 4. Find four consecutive integers whose sum is –38. 33 –5 17 –11, –10, –9, –8 7-2

Multi-Step Equations With Fractions and Decimals PRE-ALGEBRA LESSON 7-3 Evaluate each algebraic expression. a. c + 5 • 2 for c = 7 17 b. 20 – 8 ÷ p for p = 2 16 7-3

Multi-Step Equations With Fractions and Decimals PRE-ALGEBRA LESSON 7-3 (For help, go to Lesson 7-1.) Solve each equation. 1. 2n + 53 = 47 2. 4m – 37 = –28 3. –26x – 4 = 100 4. –3a + 15 = 13 Check Skills You’ll Need 7-3

Multi-Step Equations With Fractions and Decimals PRE-ALGEBRA LESSON 7-3 1. 2n + 53 = 47 2. 4m – 37 = –28 2n + 53 – 53 = 47 – 53 4m – 37 + 37 = –28 + 37 2n = –6 4m = 9 n = –3 m = 2 3. –26x – 4 = 100 4. –3a + 15 = 13 –26x – 4 + 4 = 100 + 4 –3a + 15 – 15 = 13 – 15 –26x = 104 x = –4 a = = 4m 4 –3a –3 –2 2n 2 –6 –26x –26 104 3 9 1 Solutions 7-3

Multi-Step Equations With Fractions and Decimals PRE-ALGEBRA LESSON 7-3 Solve p – 7 = 11. 3 4 p – 7 = 11 3 4 Add 7 to each side. p – 7 + 7 = 11 + 7 3 4 Simplify. p = 18 3 4 p = 3 4 • • 18 Multiply each side by , the reciprocal of . 1p = 4 • 18 3 1 6 Divide common factors. p = 24 Simplify. 7-3

Multi-Step Equations With Fractions and Decimals PRE-ALGEBRA LESSON 7-3 (continued) Check p – 7 = 11 (24) – 7 11 Replace p with 24. Simplify. 18 – 7 11 11 = 11 3 4 – 7 11 Divide common factors. 3 • 24 1 6 Quick Check 7-3

Multi-Step Equations With Fractions and Decimals PRE-ALGEBRA LESSON 7-3 Solve y + 3 = . 1 2 2 3 y + 3 = 1 2 3 Multiply each side by 6, the LCM of 2 and 3. 6 y + 3 = 1 2 3 Use the Distributive Property. 6 • y + 6 • 3 = 1 2 6 3 3y + 18 = 4 Simplify. 3y + 18 – 18 = 4 – 18 Subtract 18 from each side. 3y = –14 Simplify. Divide each side by 3. 3y 3 –14 = Simplify. y = –4 2 3 Quick Check 7-3

Multi-Step Equations With Fractions and Decimals PRE-ALGEBRA 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. 30 + 0.05x = 36.75 30 – 30 + 0.05x = 36.75 – 30 Subtract 30 from each side. 0.05x = 6.75 Simplify. Divide each side by 0.05. = 6.75 0.05 0.05x x = 135 Simplify. There are 135 minutes on your bill. Quick Check 7-3

Multi-Step Equations With Fractions and Decimals PRE-ALGEBRA LESSON 7-3 Solve each equation. 1. d + 13 = 20 2. (s – 8) = 9 3. 5 + = 14 4. 0.07x + 0.03 = 0.38 5. 0.015w – 1.85 = 1.615 1 6 3 5 2c 4 42 23 18 5 231 7-3

Problem Solving Strategy: Write an Equation PRE-ALGEBRA LESSON 7-4 Write each phrase as an algebraic expression. a. 12 times a number 12n b. 8 less than a number n – 8 c. twice the sum of 5 and a number 2(5 + n) 7-4

Problem Solving Strategy: Write an Equation PRE-ALGEBRA LESSON 7-4 (For help, go to Lesson 2-4.) Write an equation to represent each situation. 1. Pierre bought a puppy for $48. This is $21 less than the original price. What was the original price of the puppy? 2. A tent weighs 6 lb. Together, your backpack and the tent weigh 33 lb. How much does your backpack weigh? 3. A veterinarian weighs 140 lb. She steps on a scale while holding a large dog. The scale shows 192 lb. What is the weight of the dog? Check Skills You’ll Need 7-4

Problem Solving Strategy: Write an Equation PRE-ALGEBRA LESSON 7-4 Solutions 1. Price of puppy minus $21 is $48. Let p = the original price of the puppy. p – 21 = 48 2. Weight of backpack plus weight of tent is 33 lb. Let b = the weight of the backpack. b + 6 = 33 3. The weight of the veterinarian plus the weight of the dog is 192 lb. Let d = the weight of the dog. 140 + d = 192 7-4

Problem Solving Strategy: Write an Equation PRE-ALGEBRA LESSON 7-4 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? Words number of days • $29.95/d + $.12/mi • 150 mi $137.80 = Let d = number of days Mr. Reynolds had the van. Equation 137.80 d • 29.95 + 0.12 • 150 = 7-4

Problem Solving Strategy: Write an Equation PRE-ALGEBRA LESSON 7-4 (continued) d • 29.95 + 0.12 • 150 = 137.80 29.95d + 18 = 137.80 Multiply 0.12 and 150. 29.95d + 18 – 18 = 137.80 – 18 Subtract 18 from each side. 29.95d = 119.80 Simplify. Divide each side by 29.95. = 29.95d 29.95 119.80 d = 4 Simplify. Mr. Reynolds had the van for 4 days. Quick Check 7-4

Problem Solving Strategy: Write an Equation PRE-ALGEBRA LESSON 7-4 Write an equation. Then solve. 1. You buy 2 pounds of sliced roast beef for $3 per pound and some smoked turkey for $5 per pound. You spend $13.50. How much turkey did you buy? 2. A phone call costs $.35 for the first minute and $.15 for each additional minute. The phone call lasted 14 minutes. How much did the call cost? 1.5 lb $2.30 7-4

Solving Equations With Variables on Both Sides PRE-ALGEBRA LESSON 7-5 On Jim’s vacation, he collected the same number of shells each day for 7 days. When he came home, he gave away 14 shells and had 28 left over. How many shells did he collect each day of his vacation? 6 shells 7-5

Solving Equations With Variables on Both Sides PRE-ALGEBRA LESSON 7-5 (For help, go to Lesson 7-2.) Solve each equation. 1. k + 3k = 20 2. 8x – 3x = 35 3. 3b + 2 – b = –18 4. –8 – y + 7y = 40 Check Skills You’ll Need 7-5

Solving Equations With Variables on Both Sides PRE-ALGEBRA LESSON 7-5 Solutions 1. k + 3k = 20 2. 8x – 3x = 35 4k = 20 5x = 35 k = 5 x = 7 3. 3b + 2 – b = –18 4. –8 – y + 7y = 40 (3b – b) + 2 = –18 –8 + 6y = 40 2b + 2 = –18 –8 + 8 + 6y = 40 + 8 2b + 2 – 2 = –18 – 2 6y = 48 2b = –20 y = 8 b = –10 = 20 4 4k 35 5 5x 48 6 6y –20 2 2b 7-5

Solving Equations With Variables on Both Sides PRE-ALGEBRA LESSON 7-5 Solve 4c + 3 = 15 – 2c. 4c + 3 = 15 – 2c 4c + 2c + 3 = 15 – 2c + 2c Add 2c to each side. 6c + 3 = 15 Combine like terms. 6c + 3 – 3 = 15 – 3 Subtract 3 from each side. 6c = 12 Simplify. c = 2 Simplify. Divide each side by 6. = 6c 6 12 Check 4c + 3 = 15 – 2c 4(2) + 3 15 – 2(2)   8 + 3 15 – 4 11 = 11 Substitute 2 for c. Multiply. Quick Check 7-5

Solving Equations With Variables on Both Sides PRE-ALGEBRA 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. Equation 20 • x = 15 • (x + 5) 7-5

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

Solving Equations With Variables on Both Sides PRE-ALGEBRA 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. Quick Check 7-5

Solving Equations With Variables on Both Sides PRE-ALGEBRA LESSON 7-5 Solve each equation. 1. 3 – 2t = 7t + 4 2. 18 + 6z = 4z 3. 2q – 4 = 5 + 5q 4. 7(v – 4) = 3(3 + v) – 1 5. You work for a delivery service. With Plan A, you can earn $5 per hour plus $.75 per delivery. With Plan B, you can earn $7 per hour plus $.25 per delivery. How many deliveries must you make per hour to earn the same amount from either plan? 1 9 – –9 –3 9 4 deliveries 7-5

Solving Two-Step Inequalities PRE-ALGEBRA LESSON 7-6 What number is 42,625 less than the sum of 62,345 and 51,284? 71,004 7-6

Solving Two-Step Inequalities PRE-ALGEBRA LESSON 7-6 (For help, go to Lesson 2-9.) Solve each inequality. Graph the solutions. 1. w + 4 –5 2. 7 < z – 3 3. 4 > a + 6 4. x – 5 –6 > < Check Skills You’ll Need 7-6

Solving Two-Step Inequalities PRE-ALGEBRA LESSON 7-6 Solutions 1. w + 4 –5 2. 7 < z – 3 w + 4 – 4 –5 – 4 7 + 3 < z – 3 + 3 w –9 10 < z z > 10 > 3. 4 > a + 6 4. x – 5 –6 4 – 6 > a + 6 – 6 x – 5 + 5 –6 + 5 –2 > a x –1 a < –2 < 7-6

Solving Two-Step Inequalities PRE-ALGEBRA LESSON 7-6 Solve and graph 7g + 11 > 67. 7g + 11 > 67 7g + 11 – 11 > 67 – 11 Subtract 11 from each side. 7g > 56 Simplify. Divide each side by 7. > 7g 7 56 g > 8 Simplify. Quick Check 7-6

Solving Two-Step Inequalities PRE-ALGEBRA LESSON 7-6 Solve 6 – r – 6. 2 3 < 6 – r – 6 < 2 3 6 + 6 – r – 6 + 6 < 2 3 Add 6 to each side. Simplify. 12 – r < 2 3 3 2 – (12) r Multiply each side by . Reverse the direction of the inequality symbol. > Simplify. > –18 r, or r –18 < Quick Check 7-6

Solving Two-Step Inequalities PRE-ALGEBRA LESSON 7-6 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? Let = number of rides Dale goes on. Words $4 admission + $1.50/ride number of rides is less than or equal to $25 • r 25 Inequality 4 + • 1.5 r < 7-6

Solving Two-Step Inequalities PRE-ALGEBRA LESSON 7-6 (continued) 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 < Simplify. r 14 < The greatest number of rides Dale can go on is 14. Quick Check 7-6

Solving Two-Step Inequalities PRE-ALGEBRA LESSON 7-6 Solve each inequality. 1. 14 > 4d – 10 2. + 8 7 3. 32 – 12g < 176 4. 8 + 5a 23 – k 3 < d < 6 k 3 > > g > –12 a 3 > 7-6

Transforming Formulas PRE-ALGEBRA LESSON 7-7 How many miles are in 21,120 yd? (Hint: 1 mi = 5,280 ft) 12 7-7

Transforming Formulas PRE-ALGEBRA LESSON 7-7 (For help, go to Lesson 3-4.) Use each formula for the values given. 1. Use the formula d = rt to find d when r = 80 km/h and t = 4 h. 2. Use the formula P = 2 + 2w to find P when = 9 m and w = 7 m. 3. Use the formula A = bh to find A when b = 12 ft and h = 8 ft. 1 2 Check Skills You’ll Need 7-7

Transforming Formulas PRE-ALGEBRA LESSON 7-7 Solutions 1. d = rt d = (80)(4) d = 320 km 2. P = 2 + 2w P = 2(9) + 2(7) P = 18 + 14 P = 32 m 3. A = bh A = (12)(8) A = (96) A = 48 ft2 1 2 7-7

Transforming Formulas PRE-ALGEBRA LESSON 7-7 Solve the circumference formula C = 2 r for r. C = 2 r Use the Division Property of Equality. C 2 2 r = Simplify. C 2 = r, or r = Quick Check 7-7

Transforming Formulas PRE-ALGEBRA LESSON 7-7 Solve the perimeter formula P = 2 + 2w for w. P = 2 + 2w Subtract 2 from each side. P – 2 = 2 + 2w – 2 Simplify. P – 2 = 2w Multiply each side by . 1 2 (P – 2 ) = (2w) 1 2 P – = w Use the Distributive Property and simplify. Quick Check 7-7

Transforming Formulas PRE-ALGEBRA LESSON 7-7 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. d = rt Divide each side by t. = d t rt Simplify. = r, or r = d t Replace d with 600 and t with 10. r = 600 10 = 60 Simplify. Your average speed will be about 60 mi/h. Quick Check 7-7

Transforming Formulas PRE-ALGEBRA LESSON 7-7 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. 5 9 C = (F – 32) 5 9 (C) = (F – 32) 9 5 Multiply each side by . Simplify. C = F – 32 9 5 7-7

Transforming Formulas PRE-ALGEBRA LESSON 7-7 (continued) Add 32 to each side. C + 32 = F – 32 + 32 9 5 Simplify and rewrite. C + 32 = F, or F = C + 32 9 5 Replace C with 32. Simplify. F = (32) + 32 = 89.6 9 5 32°C is 89.6°F. Quick Check 7-7

Transforming Formulas PRE-ALGEBRA LESSON 7-7 Solve for the given variable. 1. Solve the area formula A = bh for b. 2. Solve the averaging formula a = for c. 3. The speed v of a satellite as it orbits Earth may be found using the formula v2 = . Solve this formula for m, the mass of Earth. 1 2 b = 2A h b + c 2 c = 2a – b Gm r v2r G m = 7-7

Simple and Compound Interest PRE-ALGEBRA LESSON 7-8 Use your classmates as subjects and estimate these percents. a. About what percent are girls? b. About what percent are twins? Check students’ answers. 7-8

Simple and Compound Interest PRE-ALGEBRA LESSON 7-8 (For help, go to Lesson 6-6.) Find each amount. 1. 6% of $400 2. 55% of $2,000 3. 4.5% of $700 4. 5 % of $325 1 2 Check Skills You’ll Need 7-8

Simple and Compound Interest PRE-ALGEBRA LESSON 7-8 1. 2. 6(400) = 100x 55(2000) = 100x $24 = x $1,100 = x 3. 4. 4.5(700) = 100x 5.5(325) = 100x $31.50 = x $17.88 = x Solutions 6 100 x 400 = 55 100 x 2000 = x 100 = 6(400) = 100x 100 55(2000) 4.5 100 x 700 = 5.5 100 x 325 = 100x 100 = 4.5(700) 100x 100 = 5.5(325) 7-8

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

Simple and Compound Interest PRE-ALGEBRA LESSON 7-8 (continued) b.  Find the interest earned in six months. Find the total of principal plus interest. Write the months as part of a year. t = = = 0.5 1 2 6 12 I = prt Use the simple interest formula. I = 1,000 • 0.06 • 0.5 Replace p with 1,000, r with 0.06, and t with 0.5. I = 30 Simplify. Total = 1,000 + 30 = 1,030 Find the total. The account will earn $30 in six months. The total of principal plus interest will be $1,030. Quick Check 7-8

Simple and Compound Interest PRE-ALGEBRA LESSON 7-8 Quick Check 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. Principal at Beginning of Year Interest Balance Year 5 : $486.20 486.20 • 0.05 = 24.31 486.20 + 24.31 = 510.51 Year 6 : $510.51 510.51 • 0.05 25.53 510.51 + 25.53 = 536.04 Year 7 : $536.04 536.04 • 0.05 26.80 536.04 + 26.80 = 562.84 Year 8 : $562.84 562.84 • 0.05 28.14 562.84 + 28.14 = 590.98 After the next four years, for a total of 8 years, the balance is $590.98. 7-8

Simple and Compound Interest PRE-ALGEBRA LESSON 7-8 Find the balance on a deposit of $2,500 that earns 3% interest compounded semiannually for 4 years. 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. 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 The balance is $2,816.23. Quick Check 7-8

Simple and Compound Interest PRE-ALGEBRA LESSON 7-8 Find the simple interest and the balance. 1. $1,200 at 5.5% for 2 years 2. $2,500 at 8% for 6 months 3. Find the balance on a deposit of $1,200, earning 9.5% interest compounded semiannually for 10 years. $132; $1,332 $100; $2,600 $3,035.72 7-8