Holt CA Course Solving Equations with Variables on Both Sides Warm Up Warm Up California Standards California Standards Lesson Presentation Lesson PresentationPreview
Holt CA Course Solving Equations with Variables on Both Sides Warm Up Solve. 1. 6n + 8 – 4n = –4w + 16 – 4w = – t – 17 – 13t = = 2(x + 7) + 4 n = 6 w = 6 t = 7 x = –3
Holt CA Course Solving Equations with Variables on Both Sides Objective: You will learn how to solve equations that have variables on both sides.
Holt CA Course Solving Equations with Variables on Both Sides Preview of Algebra Students solve multistep problems, including word problems, involving linear equations and linear inequalities in one variable and provide justification for each step. California Standards
Holt CA Course Solving Equations with Variables on Both Sides Group the terms with variables on one side of the equal sign, and simplify. Additional Example 1: Using Inverse Operations to Group Terms with Variables A. 60 – 4y = 8yB. –5b + 72 = –2b
Holt CA Course Solving Equations with Variables on Both Sides Group the terms with variables on one side of the equal sign, and simplify. A. 40 – 2y = 6yB. –8b + 24 = –5b Check It Out! Example 1
Holt CA Course Solving Equations with Variables on Both Sides Solve. Additional Example 2A: Solving Equations with Variables on Both Sides 7c = 2c + 55
Holt CA Course Solving Equations with Variables on Both Sides Additional Example 2B: Solving Equations with Variables on Both Sides Solve. 49 – 3m = 4m + 14
Holt CA Course Solving Equations with Variables on Both Sides Additional Example 2C: Solving Equations with Variables on Both Sides 2525 x = 1515 x– 12
Holt CA Course Solving Equations with Variables on Both Sides Solve. 8f = 3f + 65 Check It Out! Example 2A
Holt CA Course Solving Equations with Variables on Both Sides Solve. 54 – 3q = 6q + 9 Check It Out! Example 2B
Holt CA Course Solving Equations with Variables on Both Sides 2323 w = 1313 w– 9– 9 Check It Out! Example 2C Solve.
Holt CA Course Solving Equations with Variables on Both Sides
Holt CA Course Solving Equations with Variables on Both Sides Christine can buy a new snowboard for $ She will still need to rent boots for $8.50 a day. She can rent a snowboard and boots for $18.25 a day. How many days would Christine need to rent both the snowboard and the boots to pay as much as she would if she buys the snowboard and rents only the boots for the season? Additional Example 3: Consumer Math Application
Holt CA Course Solving Equations with Variables on Both Sides Additional Example 3 Continued 18.25d = d 9.75d = d = 14 Let d represent the number of days. Subtract 8.5d from both sides. Simplify. Divide both sides by Christine would need to rent both the snowboard and the boots for 14 days to pay as much as she would have if she had bought the snowboard and rented only the boots. – 8.5d
Holt CA Course Solving Equations with Variables on Both Sides Check It Out! Example 3 A local telephone company charges $40 per month for services plus a fee of $0.10 a minute for long distance calls. Another company charges $75.00 a month for unlimited service. How many minutes does it take for a person who subscribes to the first plan to pay as much as a person who subscribes to the unlimited plan?
Holt CA Course Solving Equations with Variables on Both Sides Check It Out! Example 3 Continued Let m represent the number of minutes. 75 = m 350 = m Subtract 40 from both sides. Simplify. If you are going to use more than 350 minutes, it will be cheaper to subscribe to the unlimited plan. Divide both sides by = 0.10m m 0.10 = –40
Holt CA Course Solving Equations with Variables on Both Sides Home Learning Click to view Lesson Videos On-Line Support: Equation Manipulatives
Holt CA Course Solving Equations with Variables on Both Sides Lesson Quiz: Part I Group the terms with variables on one side of the equal sign, and simplify n = 11n –14k + 12 = –18k Solve y = –4y – –6x = x – 112 4k = –12 3n = 81 y = –11 x = 16
Holt CA Course Solving Equations with Variables on Both Sides Lesson Quiz Part II 5. Mary can purchase ice skates for $57 and then pay a $6 entry fee at the ice skating rink. She can also rent skates there for $3 and pay the entry fee. How many times must Mary skate to pay the same amount whether she purchases or rents the skates? 19 times