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

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

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

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

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

Writing and Solving an Equation EXAMPLE 2 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. ANSWER 25 students are needed so that the cost is 20,000 won per student. Cost of School bus & Mr. Rocco tour guide + Cost of Student’s Lunches per student s = KIS’s needed cost per student s s = s s = s = 25

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

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

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

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

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.

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? Jessica is 24 years old. 2 x /3= Jessica’s age x = Frank’s age ½( x – 12) = Jessica’s age 12 yr. ago 2 x /3 = 12 + ½( x – 12) 4 x /6 = x /6 – 6 1 x /6 = 6 Jessica’s age = 12 + Jessica’s age 12yr. ago x = 36

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