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Lesson 3.3 Solving Equations with Variables on Both Sides

Lesson 3.3 Solving Equations with Variables on Both Sides. Skill Check. Lesson Presentation. Lesson Quiz. Skill Check. Simplify. 1 . 5 x + 6 + 2 x. 7 x + 6. 2 . 1 – 3 y – 5 + 4 y. y – 4. 3 . -2( k + 94). -2 k – 188. 4. -3(117 – 22 y ). 66 y – 351. Vocabulary.

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Lesson 3.3 Solving Equations with Variables on Both Sides

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  1. Lesson 3.3Solving Equations with Variables on Both Sides Skill Check Lesson Presentation Lesson Quiz

  2. Skill Check Simplify. 1. 5x + 6 + 2x 7x + 6 2. 1 – 3y – 5 + 4y y – 4 3. -2(k + 94) -2k – 188 4. -3(117 – 22y) 66y – 351

  3. Vocabulary Equation: Is a mathematical sentence showing the relationship between two expressions. Solution of an Equation: Is a number that produces a true statement when it is substituted for a variable. Inverse Operations: Are two operations that undo each other, such as addition and subtraction.

  4. Study Strategy Be Careful when combining steps, moving variable terms and constant terms in a single step, until you become proficient at the solution process. Check your answer, always 

  5. 1 EXAMPLE Solving Equations with the Variable on Both Sides Solve. a. 5k – 8 = 7k + 18 b. 8y + 4 = 11y – 17 c. m – 1 = 9m + 15 SOLUTION b. 8y + 4 = 11y – 17 c. m – 1 = 9m + 15 a. 5k – 8 = 7k + 18 -2k = 26 -3y = -21 -8m = 16 k = -13 y = 7 m = -2

  6. 2 EXAMPLE Writing and Solving an Equation School Bus KIS’s school bus tour of Jeju costs 300,000 won for the bus and tour guide Mr. Rocco, plus 8,000 won per student for lunch. How many students are needed so that the cost is 20,000 won per student? Let s represent the number of students. Write a verbal model. 300000 + 8000s = 20000s -12000s = -300000 s = 25 ANSWER 25 students are needed so that the cost is 20,000 won per student .

  7. 3 EXAMPLE An Equation with No Solution Solve. a. 4(-2 + 3x) = 12x SOLUTION a. 4(-2 + 3x) = 12x -8 + 12x = 12x -8 ≠ 0, No Solution

  8. 4 EXAMPLE Solving an Equation with All Numbers as Solutions Solve. a. 7(2x + 3) = 21 + 14x SOLUTION a. 7(2x + 3) = 21 + 14x 14x + 21 = 21 + 14x x = x, Notice that all values of x, this statement will always be true.

  9. 5 EXAMPLE Solving an Equation to Find a Perimeter Find the perimeter of the square. 6 + x 5x + 14 SOLUTION 5x + 14 = 6 + x Perimeter of Square = 4s 4x = -8 5(-2) + 14 or 6 + (-2) x = -2 4 or 4 side = 4 Therefore, the perimeter of the square is 16units. 

  10. Lesson Quiz Solve the equation. 1. 13x + 9 = 11x + 13 x = 2 2.-3k – 25 = 5k – 1 k = -3 3. 6(1 + 5y) = 30y – 2 6 ≠ -2, No Solution 4. -14 + 7m = 7(m – 2) m = m, All Numbers 5. ChallengeFor what value(s) of a does the equation 5y + 10 = 5(2 + ay) have all numbers as a solution? a = 0,1

  11. Closure • 1.) How do you solve an equation with variables on both sides? Use inverse property to get all the variable terms on one side and all the constant terms on the other side. Then use the division property of equality to find the solution. 

  12. Closure • 2.) Jessica’s age is two thirds of Frank’s age. Twelve years ago, Jessica’s age was half of Frank’s age. How old is Jessica now? 2x/3= Jessica’s age x = Frank’s age ½(x – 12) = Jessica’s age 12 yr. ago Jessica’s age = 12 + Jessica’s age 12yr. ago 2x/3 = 12 + ½(x – 12) x = 36 4x/6 = 12 + 3x/6– 6 Jessica is 24 years old.  1x/6 = 6

  13. Lesson 3.3 Homework: • pp.133-135 Exs. (20, 24, 26, 28, 30, 31, 32, 34, 36, 38, 50, 51) • Challenge/Bonus 43

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