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Solving Two-Step Equations

Solving Two-Step Equations

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Solving Two-Step Equations

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  1. Solving Two-Step Equations ALGEBRA 1 LESSON 3-1 (For help, go to Review, Page 118.) Solve each equation and tell which property of equality you used. 1.x – 5 = 14 2.x + 3.8 = 9 3. –7 + x = 7 4.x – 13 = 20 5. = 8 6. 9 = 3x Solve each equation. 7. 10x = 2 8.x = –6 9.x + 2 = 6 x 4 2 3 3 4 1 2 Check Skills You’ll Need 3-1

  2. x 4 5.= 8 • 4 = 8 • 4 Multiplication Property x = 32 of Equality 1. x – 5 = 14 x – 5 + 5 = 14 + 5 Addition Property x = 19 of Equality x 4 2. x + 3.8 = 9 x + 3.8 – 3.8 = 9 – 3.8 Subtraction Property x = 5.2 of Equality 6.9= 3x = Division Property 3 = x of Equality 9 3 3x 3 3.–7 + x = 7 –7 + x + 7 = 7 + 7 Addition Property x = 14 of Equality 7.10x = 2 = x = 4. x – 13 = 20 x – 13 + 13 = 20 + 13 Addition Property x = 33 of Equation 10x 10 2 10 Solving Two-Step Equations ALGEBRA 1 LESSON 3-1 Solutions 1 5 3-1

  3. 1 2 3 4 1 2 3 4 3 4 Solving Two-Step Equations ALGEBRA 1 LESSON 3-1 Solutions (continued) 8. x = –6 • x = • –6 x = –9 2 3 3 2 2 3 3 2 9. x + 2 = 6 x + 2 – 2 = 6 – 2 x= 3 3 4 3 4 3-1

  4. y3 13 – 5 = + 5 – 5 Subtract 5 from each side. y3 8 = Simplify. y3 3 • 8 = 3 • Multiply each side by 3. y 3 Check: 13 = + 5 243 13 + 5 Substitute 24 for y. 13 8 + 5 13 = 13 Solving Two-Step Equations ALGEBRA 1 LESSON 3-1 y3 Solve 13 = + 5. 24 = ySimplify. Quick Check 3-1

  5. 12 12 12 12 12 Relate: cost of used guitar is $25 more than the cost of the new guitar Define: Let = the cost of the new guitar. c 12 Write: $120 = $25 + • c 12 120 = 25 + c 95 = cSimplify. 120 – 25 = 25 – 25 + cSubtract 25 from each side. 2 • 95 = 2•cMultiply each side by 2. Solving Two-Step Equations ALGEBRA 1 LESSON 3-1 A music store sells a used guitar for $120. This is $25 more than the cost of a new guitar of the same brand. What is the cost of a new guitar? 190 = cSimplify. Quick Check The new guitar costs $190. 3-1

  6. Define: Let b = the number of bulbs you order. Write: 0.90 • b + 2.50 = 18.50 3-1 Solving Two-Step Equations ALGEBRA 1 LESSON 3-1 You order iris bulbs from a catalog. Iris bulbs cost $.90 each. The shipping charge is $2.50. If you have $18.50 to spend, how many iris bulbs can you order? Relate: cost per iris times number of iris bulbs plus shipping equals amount to spend. 0.90b + 2.50 = 18.50 0.90b + 2.50 – 2.50 = 18.50 – 2.50 Subtract 2.50 from each side. 0.90b = 16Simplify.

  7. 0.90b 0.90 160.90 = Divide each side by 0.90. b = 17.7 Simplify. Solving Two-Step Equations ALGEBRA 1 LESSON 3-1 (continued) You can order 17 bulbs. Check:  Is the solution reasonable? You can only order whole iris bulbs. Since 18 bulbs would cost 18 • 0.90 = 16.20 plus $2.50 for shipping, which is more than $18.50, you can only order 17 bulbs. Quick Check 3-1

  8. c24 8 – 4 = + 4 – 4 Subtraction Property of Equality c24 4 = Simplify. 24(4) = (24)()Multiplication Property of Equality c24 Solving Two-Step Equations ALGEBRA 1 LESSON 3-1 Solve 8 = + 4. Justify each step. c24 96 = cSimplify. Quick Check 3-1

  9. Solving Two-Step Equations ALGEBRA 1 LESSON 3-1 Solve each equation. 1. 3b + 8 = –10 2. –12 = –3x – 9 3. – + 7 = 14 4. –x – 13 = 35 5. What is the justification for the following step? 12 – 2y = 46 12 – 2y – 12 = 46 – 12 –6 1 c 4 –28 –48 Subtraction Property of Equality 3-1

  10. Solving Multi-Step Equations ALGEBRA 1 LESSON 3-2 (For help, go to Lessons 1-2 and 2-4.) Simplify each expression. 1. 2n – 3n2. –4 + 3b + 2 + 5b 3. 9(w – 5) 4. –10(b – 12) 5. 3(–x + 4) 6. 5(6 – w) Evaluate each expression. 7. 28 – a + 4a for a = 5 8. 8 + x – 7x for x = –3 9. (8n + 1)3 for n = –2 10. –(17 + 3y) for y = 6 Check Skills You’ll Need 3-2

  11. Solving Multi-Step Equations ALGEBRA 1 LESSON 3-2 Solutions 1.2n – 3n = (2 – 3)n = –1n = –n 2. –4 + 3b + 2 + 5b = (3 + 5)b + (–4 + 2) = 8b – 2 3. 9(w – 5) = 9w – 9(5) = 9w – 45 4. –10(b – 12) = –10b – (–10)(12) = –10b + 120 5. 3(–x + 4) = 3(–x) + 3(4) = –3x + 12 6. 5(6 – w) = 5(6) – 5w = 30 – 5w 7. 28 – a + 4a for a = 5: 28 – 5 + 4(5) = 28 – 5 + 20 = 23 + 20 = 43 8. 8 + x – 7x for x = –3: 8 + (–3) – 7(–3) = 8 + (–3) + 21 = 5 + 21 = 26 9. (8n + 1)3 for n = –2: (8(–2) + 1)3 = (–16 + 1)3 = (–15)3 = –45 10. –(17 + 3y) for y = 6: –(17 + 3(6)) = –(17 + 18) = –35 3-2

  12. 4a4 844 = Divide each side by 4. 3a+ 6 + a = 90 Check: 3(21)+ 6 + 21 90 Substitute 21 for a. 63+ 6 + 21 90 90 = 90 Solving Multi-Step Equations ALGEBRA 1 LESSON 3-2 Solve 3a + 6 + a = 90 4a+ 6 = 90Combine like terms. 4a + 6– 6 = 90 – 6Subtract 6 from each side. 4a = 84Simplify. a = 21Simplify. Quick Check 3-2

  13. Relate: length plus 25 ft plus length equals amount of side of side of fencing Define: Let x = length of a side adjacent to the house. Write: x + 25 + x = 63 Solving Multi-Step Equations ALGEBRA 1 LESSON 3-2 You need to build a rectangular pen in your back yard for your dog. One side of the pen will be against the house. Two sides of the pen have a length of x ft and the width will be 25 ft. What is the greatest length the pen can be if you have 63 ft of fencing? 3-2

  14. 2x2 382 = Divide each side by 2. Solving Multi-Step Equations ALGEBRA 1 LESSON 3-2 (continued) x+ 25 + x = 63 2x+ 25 = 63 Combine like terms. 2x+ 25 – 25 = 63 – 25Subtract 25 from each side. 2x = 38 Simplify. x = 19 The pen can be 19 ft long. Quick Check 3-2

  15. 2x2 142 = Divide each side by 2. Solving Multi-Step Equations ALGEBRA 1 LESSON 3-2 Solve 2(x – 3) = 8 2x– 6 = 8 Use the Distributive Property. 2x– 6 + 6 = 8 + 6Add 6 to each side. 2x = 14 Simplify. x = 7 Simplify. Quick Check 3-2

  16. 3x2 x5 + = 17 32 15 x+ x = 17 Rewrite the equation. 32 15 1510 210 x+ x = 17 A common denominator of and is 10. 1710 x = 17 Combine like terms. (x) = (17) Multiply each side by the reciprocal of , which is . 1017 1710 1017 1710 1017 Solving Multi-Step Equations ALGEBRA 1 LESSON 3-2 Solve + = 17 3x2 x5 Method 1: Finding common denominators x = 10 Simplify. 3-2

  17. 3x2 x5 + = 17 10(+ ) = 10(17) Multiply each side by 10, a common multiple of 2 and 5. 3x2 x5 10() + 10( ) = 10(17) Use the Distributive Property. 3x2 x5 17x17 17017 = Divide each side by 17. Solving Multi-Step Equations ALGEBRA 1 LESSON 3-2 Solve + = 17 (continued) 3x2 x5 Method 2: Multiplying to clear fractions 15x + 2x = 170 Multiply. 17x = 170 Combine like terms. x = 10 Simplify. Quick Check 3-2

  18. 60a60 42060 = Divide each side by 60. Solving Multi-Step Equations ALGEBRA 1 LESSON 3-2 Solve 0.6a + 18.65 = 22.85. 100(0.6a + 18.65) = 100(22.85) The greatest of decimal places is two places. Multiply each side by 100. 100(0.6a) + 100(18.65) = 100(22.85) Use the Distributive Property. 60a + 1865 = 2285 Simplify. 60a + 1865 – 1865 = 2285 – 1865Subtract 1865 from each side. 60a = 420 Simplify. a = 7 Simplify. Quick Check 3-2

  19. Solving Multi-Step Equations ALGEBRA 1 LESSON 3-2 Solve each equation. 1. 4a + 3 – a = 24 2. –3(x – 5) = 66 3. + = 7 4. 0.05x + 24.65 = 27.5 7 –17 n3 n4 12 57 3-2

  20. Equations with Variables on Both Sides ALGEBRA 1 LESSON 3-3 (For help, go to Lessons 2-2 and 3-2.) Simplify. 1. 6x – 2x2. 2x – 6x3. 5x – 5x4. –5x + 5x Solve each equation. 5. 4x + 3 = –5 6. –x + 7 = 12 7. 2t – 8t + 1 = 43 8. 0 = –7n + 4 – 5n Check Skills You’ll Need 3-3

  21. Equations with Variables on Both Sides ALGEBRA 1 LESSON 3-3 Solutions 1. 6x – 2x = (6 – 2)x = 4x2. 2x – 6x = (2 – 6)x = –4x 3. 5x – 5x = (5 – 5)x = 0x = 0 4. –5x + 5x = (–5 + 5)x = 0x = 0 5. 4x + 3 = –5 6. –x + 7 = 12 4x = –8 –x = 5 x = –2 x = –5 7. 2t – 8t + 1 = 43 8. 0 = –7n + 4 – 5n –6t + 1 = 43 0 = –12n + 4 –6t = 42 12n = 4 t = –7 n = 1 3 3-3

  22. = Divide each side by 3. 3x3 153 Equations with Variables on Both Sides ALGEBRA 1 LESSON 3-3 The measure of an angle is (5x – 3)°. Its vertical angle has a measure of (2x + 12)°. Find the value of x. 5x – 3 = 2x + 12 Vertical angles are congruent. 5x – 3 – 2x = 2x + 12 – 2xSubtract 2x from each side. 3x – 3 = 12 Combine like terms. 3x – 3+ 3 = 12 + 3Add 3 to each side. 3x = 15 Simplify. x = 5 Simplify. Quick Check 3-3

  23. Define: let h = the number of hours you must skateboard. Write: 60 + 1.5 h = 5.5 h Equations with Variables on Both Sides ALGEBRA 1 LESSON 3-3 You can buy a skateboard for $60 from a friend and rent the safety equipment for $1.50 per hour. Or you can rent all items you need for $5.50 per hour. How many hours must you use a skateboard to justify buying your friend’s skateboard? Relate: cost of plus equipment equals skateboard and equipment friend’s rental rental skateboard 3-3

  24. 604 4h4 = Divide each side by 4. Equations with Variables on Both Sides ALGEBRA 1 LESSON 3-3 (continued) 60 + 1.5h = 5.5h 60 + 1.5h – 1.5h = 5.5h – 1.5h Subtract 1.5h from each side. 60 = 4hCombine like terms. 15 = hSimplify. You must use your skateboard for more than 15 hours to justify buying the skateboard. Quick Check 3-3

  25. Step 1: For Y1= enter 2.5x + 1. For Y2= enter 4x – 2.6. Step 2: Use the GRAPH feature to display the graph. You can adjust the window by using the ZOOM or WINDOW features. Equations with Variables on Both Sides ALGEBRA 1 LESSON 3-3 Solve 2.5x ± 1 ≠ 4x – 2.6 using a graphing calculator. Step 3: Use the CALC feature. Select intersect to find the point where the lines intersect. The lines intersect at (2.4, 7). The x-value 2.4 is the solution of the equation. Quick Check 3-3

  26. Equations with Variables on Both Sides ALGEBRA 1 LESSON 3-3 Solve each equation. a. 4 – 4y = –2(2y – 2) 4 – 4y = –2(2y – 2) 4 – 4y = –4y + 4Use the Distributive Property. 4 – 4y + 4y = –4y + 4 + 4yAdd 4y to each side. 4 = 4 Always true! The equation is true for every value of y, so the equation is an identity. b. –6z + 8 = z + 10 – 7z –6z + 8 = z + 10 – 7z –6z + 8 = –6z + 10 Combine like terms. –6z + 8 + 6z = –6z + 10 + 6zAdd 6z to each side. 8 = 10 Not true for any value of z! This equation has no solution Quick Check 3-3

  27. 1 9 – Equations with Variables on Both Sides ALGEBRA 1 LESSON 3-3 Solve each equation. 1. 3 – 2t = 7t + 4 2. 4n = 2(n + 1) + 3(n – 1) 3. 3(1 – 2x) = 4 – 6x 4. 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 with Plan A to earn as much as with Plan B? 1 no solution 4 deliveries 3-3

  28. Ratio and Proportion ALGEBRA 1 LESSON 3-4 (For help, go to the Skills Handbook pages 758 and 761.) Write each fraction in simplest form. 1. 2. 3. Simplify each product. 4. 5. 6. 49 84 24 42 135 180 35 25 40 14 99 144 96 88 21 81 108 56    Check Skills You’ll Need 3-4

  29. Ratio and Proportion ALGEBRA 1 LESSON 3-4 1. 2. 3. 4. 5. 6. Solutions 49 7 • 7 7 84 7 • 12 12 = = 246 • 4 4 42 6 • 7 7 = = 13545 • 33 180 45 • 4 4 = = 3540 5 • 7 5 • 8 5 • 7 • 5 • 8 8 25 14 5 • 5 7 • 2 5 • 5 • 7 • 2 2   = = = = 4 99 96 9 • 11 8 • 12 9 • 11 • 8 • 12 9 3 144 88 12 • 12 8 • 11 12 • 12 • 8 • 11 12 4   = = = = 21 108 3 • 7 3 • 3 • 3 • 4 3 • 7 • 4 4 1 81 56 3 • 3 • 3 • 3 7 • 8 3 • 7 • 8 8 2   4 = = = = 4 3-4

  30. cost $1.56 ounces 48 oz = $.0325/oz Ratio and Proportion ALGEBRA 1 LESSON 3-4 Another brand of apple juice costs $1.56 for 48 oz. Find the unit rate. The unit rate is 3.25¢/oz. Quick Check 3-4

  31. Define: Let t = time needed to ride 295 km. Relate: Tour de France average speed equals 295-km trip average speed kilometers hours 3630 92.5 295 t Write: = 3630 92.5 295 t = Quick Check 3630t = 92.5(295) Write cross products. 92.5(295) 3630 t = Divide each side by 3630. t 7.5 Simplify. Round to the nearest tenth. Ratio and Proportion ALGEBRA 1 LESSON 3-4 In 2000, Lance Armstrong completed the 3630-km Tour de France course in 92.5 hours. Traveling at his average speed, how long would it take Lance Armstrong to ride 295 km? Traveling at his average speed, it would take Lance approximately 7.5 hours to cycle 295 km. 3-4

  32. 40 mi 1 h 5280 ft 1 mi 1 h 60 min 1 min 60 s • • • Use appropriate conversion factors. 40 mi 1 h 5280 ft 1 mi 1 h 60 min 1 min 60 s • • • Divide the common units. = 58.6 ft/s Simplify. Ratio and Proportion ALGEBRA 1 LESSON 3-4 The fastest recorded speed for an eastern gray kangaroo is 40 mi per hour. What is the kangaroo’s speed in feet per second? The kangaroo’s speed is about 58.7 ft/s. Quick Check 3-4

  33. Multiply each side by the least common multiple of 3 and 4, which is 12. y 3 3 4 • 12 = • 12 4y = 9 Simplify. 4y 4 9 4 = Divide each side by 4. y = 2.25 Simplify. Ratio and Proportion ALGEBRA 1 LESSON 3-4 y 3 3 4 Solve = . Quick Check 3-4

  34. w 4.5 6 5 = – w(5) = (4.5)(–6) Write cross products. 5w = –27 Simplify. 5w 5 –27 5 = Divide each side by 5. w = –5.4 Simplify. Ratio and Proportion ALGEBRA 1 LESSON 3-4 w 4.5 6 5 Use cross products to solve the proportion = – . Quick Check 3-4

  35. z + 3 4 z – 4 6 = (z + 3)(6) = 4(z – 4) Write cross products. 6z + 18 = 4z – 16 Use the Distributive Property. 2z + 18 = –16 Subtract 4z from each side. 2z = –34 Subtract 18 from each side. 2z 2 –34 2 = Divide each side by 2. z = –17 Simplify. Ratio and Proportion ALGEBRA 1 LESSON 3-4 z + 3 4 z – 4 6 Solve the proportion = . Quick Check 3-4

  36. Ratio and Proportion ALGEBRA 1 LESSON 3-4 Solve. 1. Find the unit rate of a 12-oz bottle of orange juice that sells for $1.29. 2. If you are driving 65 mi/h, how many feet per second are you driving? Solve each proportion. 3. 4. 5. 6. 10.75¢/oz. about 95.3 ft/s c 6 12 15 21 12 7 y 4.8 4 = = 1 2 3 + x 7 4 8 2 + x x – 4 25 35 –17 = = 3-4

  37. Proportions and Similar Figures ALGEBRA 1 LESSON 3-5 (For help, go to the Skills Handbook and Lesson 3-4.) Simplify 1.2.3. Solve each proportion. 4. 5. 6. 7. 8. 9. 36 42 81 108 26 52 x 12 7 30 y 12 8 45 w 15 12 27 = = = n + 1 24 9 a 81 10 25 75 z 30 n 9 = = = Check Skills You’ll Need 3-5

  38. x 12 7 30 = 90 81 96 45 84 30 y = x = a = 180 27 750 75 4 5 z = w = x = 2 Proportions and Similar Figures ALGEBRA 1 LESSON 3-5 1.5. 7. 9. 2. 3. 4. 6. 8. Solutions y 12 8 45 9 a 81 10 n 9 n + 1 24 36 42 6 • 66 6 • 7 7 = = = = = 45y = 12(8) 81a = 9(10) 24n = 9(n + 1) 81 108 27 • 33 27 • 4 4 = = 45y = 96 81a = 90 24n = 9n + 9 26 52 26 • 11 26 • 2 2 15n = 9 = = 9 15 2 15 1 9 n = y = 2 a = 1 3 5 n = 30x = 12(7) w 15 12 27 25 75 z 30 = = 30x = 84 27w = 15(12) 75z = 25(30) 27w = 180 75z = 750 2 3 w = 6 z = 10 3-5

  39. EF BC DE AB Relate: = Write a proportion comparing the lengths of the corresponding sides. Define: Let x = AB. 6 9 8 x Write: = Substitute 6 for EF, 9 for BC, 8 for DE, and x for AB. 6x = 9(8) Write cross products. 6x 6 72 6 = Divide each side by 6. x = 12 Simplify. Proportions and Similar Figures ALGEBRA 1 LESSON 3-5 In the figure below, ABC ~ DEF. Find AB. Quick Check AB is 12 mm. 3-5

  40. Proportions and Similar Figures ALGEBRA 1 LESSON 3-5 ∆ DEF has vertices D(–2, 2), E(2, 1), and F(–2, –1). It is dilated by a scale factor of 3, and the origin is at the center of the dilation. Graph the figure and its dilation. Multiply the x- and y- coordinates of each point by 3. Then graph the figure. D (–2, 2) → D´(–6, 6) E (2, 1) → E´(6, 3) F (–2, –1) → F´(–6, –3) Quick Check 3-5

  41. 102 17 x 6 = Write a proportion. 17x = 102 • 6 Write cross products. 17x = 612 Simplify. 17x 17 612 17 = Divide each side by 17. x = 36 Simplify. Proportions and Similar Figures ALGEBRA 1 LESSON 3-5 A flagpole casts a shadow 102 feet long. A 6 ft tall man casts a shadow 17 feet long. How tall is the flagpole? The flagpole is 36 ft tall. Quick Check 3-5

  42. 1 • x = 10 • 2.25 Write cross products. x = 22.5 Simplify. Proportions and Similar Figures ALGEBRA 1 LESSON 3-5 The scale of a map is 1 inch : 10 miles. The map distance from Valkaria to Gifford is 2.25 inches. Approximately how far is the actual distance? map actual map actual Write a proportion. 1 10 2.25 x = The actual distance from Valkaria to Gifford is approximately 22.5 mi. Quick Check 3-5

  43. Proportions and Similar Figures ALGEBRA 1 LESSON 3-5 1. In the figure below, ABC ~ DEF. Find DF. 2. A boy who is 5.5 feet tall casts a shadow that is 8.25 feet long. The tree next to him casts a shadow that is 18 feet long. How tall is the tree? 3. The scale on a map is 1 in.: 20 mi. What is the actual distance between two towns that are 3.5 inches apart on the map? About 19.7 cm 12 ft 70 mi 3-5

  44. Equations and Problem Solving ALGEBRA 1 LESSON 3-6 (For help, go to Lesson 1-1.) Write a variable expression for each situation. 1. value in cents of q quarters 2. twice the length 3. number of miles traveled at 34 mi/h in h hours 4. weight of 5 crates if each crate weighs x kilograms 5. cost of n items at $3.99 per item Check Skills You’ll Need 3-6

  45. Equations and Problem Solving ALGEBRA 1 LESSON 3-6 Solutions 1. value in cents of q quarters: 25q 2. twice the length : 2 3. number of miles traveled at 34 mi/h in h hours: 34h 4. weight of 5 crates if each crate weighs x kilograms: 5x 5. cost of n items at $3.99 per item: 3.99n 3-6

  46. Define: Let x = the length. The width is described in terms of the length. So define a variable for the length first. Then x – 3 = the width. Write: P = 2 + 2w Use the perimeter formula. 26 = 2 x + 2( x – 3)Substitute 26 for P, x for , and x – 3 for w. Equations and Problem Solving ALGEBRA 1 LESSON 3-6 The width of a rectangle is 3 in. less than its length. The perimeter of the rectangle is 26 in. What is the width of the rectangle? Relate: The width is 3 in. less than the length. 3-6

  47. 26 = 2x + 2x – 6Use the Distributive Property. 26 = 4x – 6Combine like terms. 26 + 6 = 4x – 6 + 6Add 6 to each side. 32 = 4x Simplify. = Divide each side by 4. 324 4x4 8 = x Simplify. Equations and Problem Solving ALGEBRA 1 LESSON 3-6 (continued) The width of the rectangle is 3 in. less than the length, which is 8 in. So the width of the rectangle is 5 in. Quick Check 3-6

  48. Define: Let x = the first integer. Then x + 1 = the second integer, and x + 2 = the third integer. Write: x + x + 1 + x + 2 = 72 Equations and Problem Solving ALGEBRA 1 LESSON 3-6 The sum of three consecutive integers is 72. Find the integers. Relate: first plus second plus third is 72 integer integer integer 3-6

  49. 3x + 3 – 3 = 72 – 3Subtract 3 from each side. 3x = 69 Simplify. 3x3 693 = Divide each side by 3. x = 23 Simplify. Equations and Problem Solving ALGEBRA 1 LESSON 3-6 (continued) x + x + 1 + x + 2 = 72 3x+ 3 = 72 Combine like terms. If x = 23, then x + 1 = 24, and x + 2 = 25. The three integers are 23, 24, and 25. Quick Check 3-6

  50. Define: Let t = the time the airplane travels. Then t – 1 = the time the jet travels. Aircraft Rate Time Distance Traveled Airplane 180 t 180t Jet 330 t – 1 330(t – 1) Equations and Problem Solving ALGEBRA 1 LESSON 3-6 An airplane left an airport flying at 180 mi/h. A jet that flies at 330 mi/h left 1 hour later. The jet follows the same route as the airplane at a different altitude. How many hours will it take the jet to catch up with the airplane? 3-6