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Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides 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. California Standards

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Holt CA Course 1 11-4 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 = 8y 60 – 4y + 4y = 8y + 4y 60 = 12y B. –5b + 72 = –2b –5b + 72 = –2b –5b + 5b + 72 = –2b + 5b 72 = 3b Add 4y to both sides. Simplify. 60 – 4y = 8y Add 5b to both sides. Simplify.

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Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides Group the terms with variables on one side of the equal sign, and simplify. A. 40 – 2y = 6y 40 – 2y + 2y = 6y + 2y 40 = 8y B. –8b + 24 = –5b –8b + 24 = –5b –8b + 8b + 24 = –5b + 8b 24 = 3b Add 2y to both sides. Simplify. 40 – 2y = 6y Add 8b to both sides. Simplify. Check It Out! Example 1

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Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides Solve. Additional Example 2A: Solving Equations with Variables on Both Sides 7c = 2c + 55 5c = 55 5 5 c = 11 Subtract 2c from both sides. Simplify. Divide both sides by 5. Check 7c = 2c + 55 7(11) = 2(11) + 55 ? 77 = 77 – 2c Substitute 11 for c. 11 is the solution.

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Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides Additional Example 2B: Solving Equations with Variables on Both Sides Solve. 49 – 3m = 4m + 14 49 = 7m + 14 35 = 7m 7 7 5 = m Add 3m to both sides. Simplify. Subtract 14 from both sides. Divide both sides by 7. + 3m – 14

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Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides Additional Example 2C: Solving Equations with Variables on Both Sides 2525 x = 1515 x– 12 2525 x =x = 1515 x 1515 x = 1515 x(5)(–12) = (5) x = –60 Subtract 1515 x from both sides. Simplify. Multiply both sides by 5. 1515 –x 1515 –x

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Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides Solve. 8f = 3f + 65 5f = 65 5 5 f = 13 Subtract 3f from both sides. Simplify. Divide both sides by 5. Check It Out! Example 2A –3f

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Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides Solve. 54 – 3q = 6q + 9 54 = 9q + 9 45 = 9q 9 9 5 = q Add 3q to both sides. Simplify. Subtract 9 from both sides. Divide both sides by 9. Check It Out! Example 2B + 3q – 9

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Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides 2323 w = 1313 w– 9– 9 2323 w =w = 1313 w– 9 1313 w = 1313 w(3)(–9) = (3) w = –27 Subtract 1313 w from both sides. Simplify. Multiply both sides by 3. Check It Out! Example 2C Solve. 13 13 –w –w 13 13

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Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides 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? Additional Example 3: Consumer Math Application

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Holt CA Course 1 11-4 Solving Equations with Variables on Both Sides Additional Example 3 Continued 18.25d = 136.5 + 8.5d 9.75d = 136.5 9.75 d = 14 Let d represent the number of days. Subtract 8.5d from both sides. Simplify. Divide both sides by 9.75. 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

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Holt CA Course 1 11-4 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?

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