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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.$$|x-3|<4$$

$x \in(-1,7)$

Algebra

Chapter 1

THE INTEGERS

Section 4

Solving Absolute Value Equations and Inequalities

The Integers

Equations and Inequalities

Polynomials

Missouri State University

McMaster University

Idaho State University

Lectures

01:32

In mathematics, the absolu…

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In $15-26,$ solve each ine…

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In $15-26,$ find and graph…

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Okay, so we're asked to solve the following absolute value inequality. And we can do this by splitting it up into two separate equations. Namely, if X minus three is less than four. And then if X minus three is greater, then minus four. And so if we do this, yeah, if we do that, if we do this, we have a situation that this will satisfy the inequality. But we'll see. We'll have to put some restrictions on it because this alone won't work, and we'll see why. And this itself also won't work completely. So if we solve both, if we add three to both sides here we get AKs is less than seven. Okay. And for this one, if we had three to be the side, we get X greater than minus one. And on their own, both of these solutions will work. If you plug them back into your original equation, both of them work out. But if you draw a number line for your solutions. And here you have 012345 67 Here. Then you have minus one here. This is zero. Well, if X is less than seven, that covers all of this. However, we have imposed another restriction here. Namely, the X is greater than minus one. And so here we have. This is well, so if you were to test negative to, for instance in your inequality, it wouldn't work. Minus two. Minus three is minus five, which is not less than four in the absolute value. So over here actually doesn't work. And obviously, if you plug them number like nine over here, this wouldn't work either. Now, if you plug anything here in between and it works out and so actually you have to kind of combine your answers to get one hole answer. And so this is a full picture of your solutions here. But if you wanted to write that out and set, you would just have to say that you would have the set of war of exes such that X is less than seven and accidents greater on minus one. And if you wanted to make sure there were intruders, you could just add some piece about, you know, x in ze, which is a symbol for the integers. If you weren't familiar, So there's a set of your solutions. And then here are the B imposed conditions. By the end quote

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