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In $15-26,$ solve each inequality and write the solution set if the variable is an element of the set of integers.$$|2 b-7| \geq 9$$

$b \in(-\infty,-1] \cup[8, \infty)$

Algebra

Chapter 1

THE INTEGERS

Section 4

Solving Absolute Value Equations and Inequalities

The Integers

Equations and Inequalities

Polynomials

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In $15-26,$ solve each ine…

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okay were asked to solve the following inequality. And we do that by splitting up are absolute value since it's already isolated. So we have to be minus seven is greater than or equal to nine, and then we're going to have to be minus seven less than minus nine. That's a very and so here, we're gonna solve both sides. So if I add seven here, nine plus seven is 16. So I have to be greater than or equal in 16 which gives me that B is greater than or equal to eight after a divide by two on both sides. And here I add seven to both sides again. And so I'm left to be less than or equal to minus two. And so B should be less than or equal to minus one. And so we can look and draw out our little number line real quick, and I have zero here. We have minus one here and out here we have eight. And so we need to check almost all of our points here just to make sure, because with this less than or equal sign, we need to make sure the core equals case works. So right now, leave open bubbles and minus one in eight, and we'll test both of them. So if you buggin minus one to the original, you get minus two minus seven, which is greater than or equal than nine because it's equal the nine. So that works. If you plug in eight, you get 16 minus seven absolute value, which is also equal to mine. So these work And then if I plug in a number less than minus one, say minus two, you would also get a number greater than or equal nine. So, like minus two, you get minus 11 in the absolute value. So this over here all works. If that bug in zero, you get zero minus seven, which doesn't work. So this in here doesn't work. And then if I plugged in 10 that would work. And so your solutions air drawn out here. And if you wanted a set for the integers, you would get a set of X in the integers Such that Okay, I'm sorry this had to be in the integers said be in the integers such that B is either greater than or equal to eight or B is less than or equal to minus one. It's going to get a good grasp on when to use the word and or in this case it has to be one or the other. Obviously, you can't be less than minus one and greater than eight at the same time. But there are cases where the board and is the correct word to use, but in this case it will be four.

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