 # 1. 6(2x – 3) – 2(6x + 1) = 10x 12x – 18 – 12x – 2 = 10x - 20 = 10x 10 x = –2.

## Presentation on theme: "1. 6(2x – 3) – 2(6x + 1) = 10x 12x – 18 – 12x – 2 = 10x - 20 = 10x 10 x = –2."— Presentation transcript:

1. 6(2x – 3) – 2(6x + 1) = 10x 12x – 18 – 12x – 2 = 10x - 20 = 10x 10 x = –2

2. 15 + 2x – 4 = 9x + 11 – 7x 11 + 2x = 11 + 2x -11 2x = 2x -2x - 2x 0 = 0 ALWAYS TRUE!!! Infinite Solution!!

Algebra II

 To solve & graph inequalities  To write & solve compound inequalities

Less Than <  Is fewer than  Is less than  Is at most  Is no more than  Is at maximum

Greater Than >  Is more than  Is larger than  Is at least  Is no less than  Is at minimum

ALWAYS FLIP THE INEQUALITY SIGN WHEN YOU DIVIDE BY – #

FLIP!!!

2. -2(x + 9) + 5 < 3 -2x – 18 + 5 < 3 -2x – 13 < 3 +13 +13 -2x < 16 -2 - 2 x >-8 FLIP!!!

3. -2(3x + 1) > -6x + 7 -6x – 2 > - 6x + 7 +2 +2 -6x > -6x + 9 +6x +6x 0 > 9 NEVER TRUE!No Solution

4. 5(2x – 3) – 7x < 3x + 8 10x – 15 – 7x < 3x + 8 3x – 15 < 3x + 8 -3x -3x - 15 < 8 ALWAYS TRUE!Infinite Solutions

5. 4(2x – 3) > 8(x + 1) 8x – 12 > 8x + 8 -8x -8x - 12 > 8 NEVER TRUE!No Solution

 Compound Inequality – join two inequalities with the word and or or  “And” compound inequality – Find all values of the variable that make BOTH inequalities true. Need overlap  “Or” compound inequality – Find all values of the variables that make at least one of the inequalities true

8. 5< x + 1 < 13 -1 -1 - 1 4 < x < 12 Read: x + 1 is greater than 5 AND less than 13

00

00

11. 16 < 5x + 1 or 3x + 9 < 6 -1 -1 -9 -9 15 < 5xor3x < -3 55 3 3 3 < x or x < -1 x > 3 or x < -1 00

 Your friend solved the problem as shown.  What was his error?

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