California Standards Preview of Algebra 1 5.0 Students solve multistep problems, including word problems, involving linear equations and linear inequalities in one variable and provide justification for each step.

Teacher Example 1A & 1B: Using Inverse Operations to Group Terms with Variables Group the terms with variables on one side of the equal sign, and simplify. A. 60 – 4y = 8y 60 – 4y = 8y 60 – 4y + 4y = 8y + 4y Add 4y to both sides. 60 = 12y Simplify. B. –5b + 72 = –2b –5b + 72 = –2b –5b + 5b + 72 = –2b + 5b Add 5b to both sides. 72 = 3b Simplify.

Student Practice 1A & 1B: Group the terms with variables on one side of the equal sign, and simplify. A. 40 – 2y = 6y 40 – 2y = 6y 40 – 2y + 2y = 6y + 2y Add 2y to both sides. 40 = 8y Simplify. B. –8b + 24 = –5b –8b + 24 = –5b –8b + 8b + 24 = –5b + 8b Add 8b to both sides. 24 = 3b Simplify.

Teacher Example 2A: Solving Equations with Variables on Both Sides Solve. 7c = 2c + 55 7c = 2c + 55 – 2c Subtract 2c from both sides. 5c = 55 Simplify. 5c = 55 5 Divide both sides by 5. c = 11 Check 7c = 2c + 55 Substitute 11 for c. 7(11) = 2(11) + 55 ? 77 = 77 11 is the solution. 

Additional Example 2B: Solving Equations with Variables on Both Sides Solve. 49 – 3m = 4m + 14 49 – 3m = 4m + 14 Add 3m to both sides. + 3m + 3m 49 = 7m + 14 Simplify. 49 = 7m + 14 Subtract 14 from both sides. – 14 – 14 35 = 7m 35 = 7m 7 Divide both sides by 7. 5 = m

Additional Example 2C: Solving Equations with Variables on Both Sides 5 x = 1 x – 12 2 5 x = 1 x – 12 Subtract 1 5 x from both sides. 15 – x 1 5 x –12 = Simplify. 1 5 x (5)(–12) = (5) Multiply both sides by 5. x = –60

Student Practice 2A: Solve. 8f = 3f + 65 8f = 3f + 65 –3f –3f Subtract 3f from both sides. 5f = 65 Simplify. 5f = 65 5 Divide both sides by 5. f = 13

Student Practice 2B: Solve. 54 – 3q = 6q + 9 54 – 3q = 6q + 9 Add 3q to both sides. + 3q + 3q 54 = 9q + 9 Simplify. 54 = 9q + 9 Subtract 9 from both sides. – 9 45 = 9q 45 = 9q 9 Divide both sides by 9. 5 = q

Student Practice 2C: Solve. 2 3 w = 1 w – 9 2 3 w = 1 w – 9 Subtract 1 3 w from both sides. 13 – w 1 3 w –9 = Simplify. 1 3 w (3)(–9) = (3) Multiply both sides by 3. w = –27

Teacher Example 3: Consumer Math Application Christine can buy a new snowboard for $136.50. 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?

Teacher Example 3 Continued Let d represent the number of days. 18.25d = 136.5 + 8.5d Subtract 8.5d from both sides. – 8.5d – 8.5d 9.75d = 136.5 Simplify. 9.75d = 136.5 Divide both sides by 9.75. 9.75 9.75 d = 14 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.

Student Practice 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?

Student Practice 3: Let m represent the number of minutes. 75 = 40 + 0.10m Subtract 40 from both sides. –40 –40 35 = 0.10m Simplify. 35 0.10m 0.10 = Divide both sides by 0.10. 350 = m If you are going to use more than 350 minutes, it will be cheaper to subscribe to the unlimited plan.

Lesson Quiz: Part I Group the terms with variables on one side of the equal sign, and simplify. 1. 14n = 11n + 81 2. –14k + 12 = –18k Solve. 3. 58 + 3y = –4y – 19 4. –6x = x – 112 3n = 81 4k = –12 y = –11 x = 16

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