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Solve each inequality or compound inequality. Wri…

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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52 Problem 53 Problem 54 Problem 55 Problem 56 Problem 57 Problem 58 Problem 59 Problem 60 Problem 61 Problem 62 Problem 63 Problem 64 Problem 65 Problem 66 Problem 67 Problem 68 Problem 69 Problem 70 Problem 71 Problem 72 Problem 73 Problem 74 Problem 75 Problem 76 Problem 77 Problem 78 Problem 79 Problem 80 Problem 81 Problem 82 Problem 83 Problem 84 Problem 85 Problem 86 Problem 87 Problem 88 Problem 89 Problem 90 Problem 91 Problem 92 Problem 93 Problem 94 Problem 95 Problem 96 Problem 97 Problem 98 Problem 99 Problem 100 Problem 101 Problem 102 Problem 103 Problem 104 Problem 105 Problem 106 Problem 107 Problem 108 Problem 109 Problem 110 Problem 111 Problem 112 Problem 113 Problem 114 Problem 115 Problem 116 Problem 117 Problem 118 Problem 119 Problem 120 Problem 121 Problem 122 Problem 123 Problem 124

Problem 67 Hard Difficulty

Solve each inequality or compound inequality. Write the solution set in interval notation and graph it.
$$
0<5(x+2) \leq 15
$$

Answer

$$(-2,1]$$

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Chapter 2

Equations, Inequalities, and Problem Solving

Section 7

Solving Inequalities

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Video Transcript

soft the compound inequality, right? The solution set an interval notation, then Graff it. We have the compound inequality zero is less than the product of five in the binomial X plus two is less than or equal to 15. In order to solve the compound inequality, we need to use properties of inequality to isolate the variable as the middle part of the inequality. The first part thing we want to dio is to rewrite the inequality without the parentheses. In order to do that, we need to distribute the five into the binomial X Plus two. The other parts of the inequality are unchanged. We're not doing anything to them yet, so we're gonna rewrite the inequality as zero is less than five times X. Divac's plus five times two, which is 10 is less than or equal to 15. From here, we can work on isolating the variable. Right now we have five X plus 10 and we need it to be ex all by itself. So we're going to use inverse operations to get X all by itself. Let's start by subtracting 10 from all parts of the inequality. Now we have negative. 10 is less than five x is less than or equal to five. We still have a product of five index at their middle part of the inequality. So to undo the product, we're going to use its inverse operation of division and divide all parts of the inequality by five. This leaves us the compound. Inequality of negative, too, is less than X, this less than or equal to one. This indicates that the values of X are greater than negative, too, but also less than and or equal to one. We can represent this solution set on a number line by starting at negative, too, with an open circle. We're using an open circle because negative, too, is not included in the solution set. And then we're going to closed the interval at one with a closed circle. It's a closed circle because one is part of the solutions that X can be equal to one. We're going to shade the number line between negative too and positive one. To represent our solution. Set an interval. Notation will open with a parentheses at negative, too, and we will close with a bracket at positive one

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