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ALGEBRA II HONORS/GIFTED - SECTION 1-5 (Solving Inequalities)

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Presentation on theme: "ALGEBRA II HONORS/GIFTED - SECTION 1-5 (Solving Inequalities)"— Presentation transcript:

1 ALGEBRA II HONORS/GIFTED - SECTION 1-5 (Solving Inequalities)
@ SECTION 1-5 : SOLVING INEQUALITIES

2 Human Number Line activity.
INEQUALITY : A mathematical sentence such that the sides are not equal. Inequalities use one or more of the symbols : Compound inequalities are composed of two or more inequalities. 1) How many solutions can an inequality have? None, one, infinite! 2) Given 6 < 8, let’s do some… a) adding c) multiplying b) subtracting d) dividing

3 When you multiply or divide an inequality by a negative number, remember to switch the direction of the inequality.

4 Solve and graph on a number line.
3) 3x – 5 < 16 {real numbers} will always be the solution set. 3x < 21 7 x < 7 < >

5 4) -4 • • -4 j > 16 16

6 5) x – 3 < or x + 3 > 6 How many parts of an “or” statement must be true for the entire statement to be true? x < 1 or x > 3 Just This is a compound inequality because it involves two or more inequalities. As long as either part is true, GRAPH IT! So, what is graphed on the number line? 6) 2x – 7 < -11 or -3x + 5 < 3 Both parts!

7 7) 3 < 2x + 5 < 11 Means 3 < 2x + 5 and 2x + 5 < 11. Solving this compound inequality in this format gets kind of tricky. Let’s do something different – something us middle children will love! Solve for x in the middle. -2 < 2x < 6 -1 < x < 3 So, where is x located? Yep, in the middle! So, graph, yep, in the middle.

8 8) 5 > -3x + 2 > -4 smaller # < x < larger # larger # > x > smaller #


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