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Summer Assignment Review Graphing Compound Inequalities And Absolute Value Inequalities Compound Inequalities AND compound inequalities should create a range of values. Donβt assume that the given inequality creates this range. β2 β€ π₯ β€ 3 β2 β₯ π₯ β₯ 3 π₯ β₯ β2 π΄ππ· π₯ β€ 3 π₯ β€ β2 π΄ππ· π₯ β₯ 3 π₯ falls between -2 and 3 on the number line. π₯ can not be both to the left of -2 and to the right of 3 on the number line . There is no solution. Compound Inequalities OR compound inequalities should create two ranges of values. π₯ β€ β2 ππ π₯β₯3 π₯ is either to the left of -2 or to the right of 3 on the number line. AND compound inequalities should create a range of values. π₯ β₯ β2 ππ π₯ β€ 3 π₯ can be anywhere on the number line. The solution is All Real Numbers. Practice Graph each compound inequality. 1) β3 < π₯ < 4 2) π₯ β₯ β3 ππ π₯ < 3 3) π₯ < β4 ππ π₯ β₯ 2 4) π₯ β€ 0 πππ π₯ β₯ 3 5) π₯ > β1 πππ π₯ β€ 5 All Real Numbers No Solutions Absolute Value Inequalities When we say that π₯ = 3, we are saying that the distance between x and 0 is 3. When we say that π₯ < 3, we are saying that the distance between x and 0 is less than 3. The inequalities < and β€ lead to βandβ relationships β3 < π₯ < 3 When we say that π₯ > 3, we are saying that the distance between x and 0 is greater than 3. The inequalities > and β₯ lead to βorβ relationships π₯ < β3 ππ π₯ > 3 Solving Absolute Value Inequalities Case 2 - modified method: Drop the absolute value bars, flip the inequality symbol and change the sign of the term on the right. 2π₯ β 1 < β7 2π₯ < β6 π₯ < β3 Solve and Graph an Abs. Val. Ineq. Example : Solve π₯ β 5 β₯ 7 . Graph your solution. Step 1: Is it and or or? This is β₯ so it is βorβ. Set up two inequalities. π₯β5 β₯7 For the second, flip the inequality and change the sign of the 7 π₯β5β₯7 π₯ β 5 β€ β7 Step 2: Solve both π₯ β€ β2 π₯ β₯ 12 Step 3: Graph π₯ β₯ 12 -3 -2 -1 or π₯ β€ β2 · · · 11 12 13 Solve and Graph an Abs. Val. Ineq. Example : Solve β4π₯ β 5 + 3 < 9 . Graph your solution. Step 1: Is it and or or? We donβt know yet, get the abs. val. alone. β4π₯ β 5 + 3 < 9 β4π₯ β 5 < 6 Step 2: Solve Now : It is < , so it is an βandβ inequality. Drop the abs. val. Bars and put -6 < on the left. β6 < β4π₯ β 5 < 6 +5 +5 +5 β1 < β4π₯ < 11 -4 -4 -4 0.25 > π₯ > β2.75 β2.75 < π₯ < 0.25 Step 3: Graph Practice π₯ > 5 ππ π₯ < β11 -12 -11 -10 ··· 4 5 6 β5 < π€ < 6 -6 -5 -4 · · · 5 6 7 β0.2 β€ π β€ 2.6