2. Solving Inequalities, Compound Inequalities and Absolute Value Inequalities Sec 1.5 &1.6 pg. 33 - 43

3. Hallford © 2007Glencoe © 2003 Objectives The learner will be able to (TLWBAT):  solve inequalities  solve real-world problems with inequalities  solve compound inequalities  solve absolute value inequalities

4. Hallford © 2007Glencoe © 2003 Properties Trichotomy Property  For any two real numbers a and b, a b

5. Hallford © 2007Glencoe © 2003 Properties of Inequalities For any real number a, b, and c  Addition Prop. If a>b, then a + c > b + c If a<b, then a + c < b + c  Subtraction Prop. If a>b, then a – c > b – c If a<b, then a – c < b – c See pg. 33

6. Hallford © 2007Glencoe © 2003 Work this problem x – 7 > 2x + 2 -x -x -7 > x + 2 -2 -2 -9 > x Now we need to graph our answer on a number line.

7. Hallford © 2007Glencoe © 2003 Graphing on a Number Line Let’s look at our previous answer, -9 > x -9 0 What does this answer mean? Are my “x’s” here? Or here? -9 is GREATER than x, so my “x’s” that make sense are -10, -10.1, -11, etc. Anything smaller than -9 For you use an opening dot or point - For you use a closed dot or point -

8. Hallford © 2007Glencoe © 2003 Properties of Inequalities Multiplication Prop. FFor any real numbers a, b, and c IIf c is positive IIf a>b, then ac>bc IIf a<b, then ac<bc IIf c is negative IIf a>b, then ac<bc IIf a<b, then ac>bc Division Prop. FFor any real numbers a, b, and c IIf c is positive IIf a>b, then a / c > b / c IIf a<b, then a / c < b / c IIf c is negative IIf a>b, then a / c < b / c IIf a<b, then a / c > b / c

9. Hallford © 2007Glencoe © 2003 Work this problem 3x – 7 < 7x + 13 -4x – 7 < 13 7 7 -4x < 20 -4 x > -5 Now let’s graph -7x 0-5 You can also use set builder notation to express your answer. {x | x > -5} This is read as the set of all x such that x is greater than or equal to -5

10. Hallford © 2007Glencoe © 2003 Work this problem 3(a +4) – 2(3a +4) ≤ 4a - 1 3a + 12 – 6a – 8 ≤ 4a – 1 -3a + 4 ≤ 4a – 1 4 ≤ 7a - 1 5 ≤ 7a 5 / 7 ≤ a Now graph and express in set builder notation. 0 {a | a ≥ 5 / 7 } You can also use interval notation. Interval notation uses ( & ) for and [ & ] for ≤ & ≥. We also use -∞ (negative infinity) & +∞ (positive infinity) Interval notation for this problem would be [5/7, +∞)

11. Hallford © 2007Glencoe © 2003 Compound Inequalities A compound inequality consists of two inequalities joined by the word “and” or the word “or”. You must solve both inequalities and then graph.  The final graph of the “and” inequality is the intersection of both individual solution graphs.  The final graph of the “or” inequality in the union of both individual solution graphs

12. Hallford © 2007Glencoe © 2003 Let’s solve an “and” compound inequality 7 < 2x - 1 < 15 Method 1 – Divide into two problems 7 < 2x – 1 8 < 2x 4 < x Method 2 – Work together 2x – 1 < 15 2x < 16 x < 8 7 < 2x – 1 < 15 8 < 2x < 16 4 < x < 8 Whatever you do to one side do to the other! Now Graph 4 < x x < 8 0 0 0 4 4 4 8 8 8 { x | 4 ≤ x < 8} “Means 7 < 2x -1 and 2x – 1< 15” 4 < x < 8

13. Hallford © 2007Glencoe © 2003 Let’s work about the same problem as an “or” inequality 7 < 2x -1 or 2x – 1 < 15 We know the answer is 4 < x and x < 8, but this time the answer graph is different. It is the UNION of the two graphs. 4 < x x < 8 0 0 0 4 4 8 8 The solution set is all real numbers. 4 < x or x < 8

14. Hallford © 2007Glencoe © 2003 Work this problem 2x + 7 10 2x + 7 < -1 3x + 7 > 10 2x < -8 x < -4 3x > 3 x > 1 0 -4 0 or 1 0 -4 1

15. Hallford © 2007Glencoe © 2003 Absolute Value Inequalities |a| 0  work as an “and” problem  -b < a < b |a| > b, where b > 0  work as an “or” problem  a > b or a < -b

16. Hallford © 2007Glencoe © 2003 Work these problems |x – 1| < 3 -3 < x -1 < 3 -2 < x < 4 |x -1 | > 3 x -1 > 3 or x -1 < -3 x > 4 or x < -2 0 0

17. Hallford © 2007Glencoe © 2003 Absolute Value Inequalities |a| < b, where b < 0  if b is less than zero, it is negative, so there is no solution |a| > b, where b < 0  if b is less than zero, it is negative, so every real number is a solution or all reals.

18. Hallford © 2007Glencoe © 2003 Work these problems |2x – 7| < -5 There is no solution for the above since the absolute value cannot be less than zero.  |3x – 1| + 9 > 2 |3x – 1| > -7 Any value of “x” will make this statement true, since the absolute value is always greater than a negative number All Real numbers