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Solving Two-Step Equations

Learn to solve equations involving two steps in this pre-algebra lesson. Practice with various examples and check your skills.

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Solving Two-Step Equations

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  1. 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

  2. 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

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

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

  5. –3b –3 –6 –3 = Divide each side by –3. 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. b = 2 Simplify. Quick Check 7-1

  6. 25w 25 250 25 = Divide each side by 25. 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 – 100Subtract 100 from each side. 25w = 250 Simplify. w = 10 Simplify. It will take you 10 weeks to pay for the bicycle. Quick Check 7-1

  7. q 5 r 3 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 2 –33 –4 –35 16 7-1

  8. 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

  9. Solving Multi-Step Equations PRE-ALGEBRA LESSON 7-2 (For help, go to Lesson 2-3.) Simplify each expression. 1. 2x + 4 + 3x2. 5y + y3. 8a – 5a 4. 2 – 4c + 5c5. 4x + 3 – 2(5 + x) Check Skills You’ll Need 7-2

  10. 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

  11. 4s 4 36 4 = Divide each side by 4. 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 – 5Subtract 5 from each side. 4s = 36 Simplify. s = 9 Simplify. 7-2

  12. 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

  13. sum of three consecutive integers 42 Words is n Let = the least integer. n + 1 Then = the second integer, n + 2 and = the third integer. n n + 1 n + 2 42 Equation + + = Solving Multi-Step Equations PRE-ALGEBRA LESSON 7-2 The sum of three consecutive integers is 42. Find the integers. 7-2

  14. 3n + 3 – 3 = 42 – 3   Subtract 3 from each side. 3n = 39 Simplify. 3n 3 39 3 = Divide each side by 3. n = 13 Simplify. 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. 7-2

  15. 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

  16. 8q – 28 = –4 Use the Distributive Property. 8q – 28 + 28 = –4 + 28 Add 28 to each side. 8q = 24 Simplify. 8q 8 24 8 Divide each side by 8. = q = 3 Simplify. Solving Multi-Step Equations PRE-ALGEBRA LESSON 7-2 Solve each equation. a. 4(2q – 7) = –4 4(2q – 7) = –4 7-2

  17. 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. 24 –6 –6r –6 = Divide each side by –6. –4 = r Simplify. Solving Multi-Step Equations PRE-ALGEBRA LESSON 7-2 (continued) b.    44 = –5(r – 4) – r 44 = –5(r – 4) – r Quick Check 7-2

  18. 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

  19. 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

  20. 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

  21. 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 1 4 9 4 2n 2 –6 2 4m 4 = = 3. –26x – 4 = 100 4. –3a + 15 = 13 –26x – 4 + 4 = 100 + 4 –3a + 15 – 15 = 13 – 15 –26x = 104 x = –4 a = –3a –3 –2 –3 = –26x –26 104 –26 = 2 3 Multi-Step Equations With Fractions and Decimals PRE-ALGEBRA LESSON 7-3 Solutions 7-3

  22. 3 4 3 4 p – 7 + 7 = 11 + 7 Add 7 to each side. 3 4 p = 18 Simplify. 4 3 3 4 p = Multiply each side by , the reciprocal of . 1p = Divide common factors. 4 3 3 4 4 3 3 4 • 18 p – 7 = 11 • 4 • 18 3 6 1 p = 24 Simplify. Multi-Step Equations With Fractions and Decimals PRE-ALGEBRA LESSON 7-3 Solve p – 7 = 11. 7-3

  23. 3 4 Check p – 7 = 11 3 4 (24) – 7 11 Replace p with 24. – 7 11 Divide common factors. 18 – 7 11 Simplify. 3 • 24 4 11 = 11 6 1 Multi-Step Equations With Fractions and Decimals PRE-ALGEBRA LESSON 7-3 (continued) Quick Check 7-3

  24. 1 2 2 3 1 2 2 3 y + 3 = Multiply each side by 6, the LCM of 2 and 3. 1 2 2 3 6 y + 3 = 6 1 2 6 •y + 6 • 3 = Use the Distributive Property. 2 3 6 3y + 18 = 4 Simplify. 3y + 18 – 18 = 4 – 18 Subtract 18 from each side. 3y = –14 Simplify. 3y 3 –14 3 = Divide each side by 3. 2 3 –4 y = Simplify. Multi-Step Equations With Fractions and Decimals PRE-ALGEBRA LESSON 7-3 Solve y + 3 = . Quick Check 7-3

  25. 30 – 30 + 0.05x = 36.75 – 30 Subtract 30 from each side. 0.05x = 6.75 Simplify. Divide each side by 0.05. 0.05x 0.05 6.75 0.05 = x = 135 Simplify. 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 Quick Check There are 135 minutes on your bill. 7-3

  26. 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

  27. 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

  28. 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

  29. 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

  30. number of days 150 mi $137.80 Words • $29.95/d + $.12/mi • = d Let = number of days Mr. Reynolds had the van. d 150 137.80 Equation • 29.95 + 0.12 • = 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? 7-4

  31. 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. 29.95d 29.95 119.80 29.95 Divide each side by 29.95. = d = 4 Simplify. Problem Solving Strategy: Write an Equation PRE-ALGEBRA LESSON 7-4 (continued) d • 29.95 + 0.12 • 150 = 137.80 Quick Check Mr. Reynolds had the van for 4 days. 7-4

  32. 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

  33. 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

  34. 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

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

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

  37. Jenny’s time Steve’s time Words 20 words/min • = 15 words/min • x Let = Jenny’s time. x + 5 Then = Steve’s time. (x + 5) = 15 • x Equation 20 • 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 7-5

  38. 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 5 x = 15 Simplify. = Solving Equations With Variables on Both Sides PRE-ALGEBRA LESSON 7-5 (continued) 20x = 15(x + 5) Jenny will catch up with Steve in 15 min. 7-5

  39. 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

  40. 1 9 – 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 + 5q4. 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? –9 –3 9 4 deliveries 7-5

  41. 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

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

  43. 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 Solving Two-Step Inequalities PRE-ALGEBRA LESSON 7-6 7-6

  44. 7g + 11 – 11 > 67 – 11 Subtract 11 from each side. 7g > 56 Simplify. 7g 7 56 7 > Divide each side by 7. g > 8 Simplify. Solving Two-Step Inequalities PRE-ALGEBRA LESSON 7-6 Solve and graph 7g + 11 > 67. 7g + 11 > 67 Quick Check 7-6

  45. < < < < < 2 3 6 – r – 6 > 2 3 6 + 6 – r – 6 + 6 Add 6 to each side. 2 3 12 – r Simplify. 3 2 3 2 3 2 2 3 – Multiply each side by . Reverse the direction of the inequality symbol. – (12) – – r > –18 r, or r –18 Simplify. Solving Two-Step Inequalities PRE-ALGEBRA LESSON 7-6 2 3 Solve 6 – r – 6. Quick Check 7-6

  46. < number of rides is less thanor equal to Words $1.50/ride $25 $4 admission + • r Let = number of rides Dale goes on. 1.5 r 25 Inequality 4 + • 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? 7-6

  47. < < < < < 4 + 1.5r 25 Subtract 4 from each side. 4 + 1.5r– 4 25 – 4 Simplify. 1.5r 21 1.5r 1.5 21 1.5 Divide each side by 1.5. r 14 Simplify. Solving Two-Step Inequalities PRE-ALGEBRA LESSON 7-6 (continued) The greatest number of rides Dale can go on is 14. Quick Check 7-6

  48. < k 3 – k 3 > > > a 3 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 d < 6 g > –12 7-6

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

  50. 1 2 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. Check Skills You’ll Need 7-7

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