Algebra II  To solve equations  To solve problems by writing equations.

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Presentation transcript:

Algebra II

 To solve equations  To solve problems by writing equations

 Substitution ◦ If a = b, then you can replace a with b, and vice versa ◦ If x = y and 9 + x = 15, then 9 + y = 15

 Addition Property ◦ If a = b, then a + c = b + c ◦ If x = 12, then x + 3 =  Subtraction Property ◦ If a = b, then a – c = b – c ◦ If x = 12, then x – 3 = 12 – 3

 Equation – a statement that two expressions are equal  Solution(s) of the equation – All values of the variable that make the equation true  Inverse operations – operations that “undo” each other. + & -

y = 3(y – 3) y = 3y – 9 - 3y -3y y = – y = 18 3 y = 6

4. 3(2x – 1) – 2(3x + 4) = 11x 6x – 3 – 6x – 8 = 11x - 11 = 11x 11 x = - 1

x – 7 = 6x + 5 – 3x 4 + 3x = 3x x -3x 4 = 5 NEVER TRUE!!! No Solution!!

6. 6x + 5 – 2x = 4 + 4x + 1 4x + 5 = 4x x = 4x -4x - 4x 0 = 0 ALWAYS TRUE!!! Infinite Solution!!

7. 7x + 6 – 4x = x - 8 3x + 6 = 3x x 6 = 4 NEVER TRUE!!! No Solution!!

8. 2x + 3(x – 4) = 2(2x – 6) + x 2x + 3x – 12 = 4x – 12 + x 5x – 12 = 5x – x = 5x -5x - 5x 0 = 0 ALWAYS TRUE!!! Infinite Solution!!

 Literal Equation – an equation that uses at least two different letters as variables. ◦ Can solve for any one of its variables. ◦ You solve for a variable “in terms of” the other variables

12. A rectangle has a perimeter of 200 meters. The length is three times the width. What are the dimensions of the rectangle?

x 3x Width Length P = 2(length) + 2(width) 200 = 2(3x) + 2(x) 200 = 6x + 2x 200 = 8x 8 x = 25 Width = 25 m Length = 75 m

 A city park has a pool. You can pay a daily entrance fee of $3 or purchase a membership for the 12-week summer season for $82 and pay only $1 per day to swim. How many days would you have to swim to make the membership worthwhile?