in-15-26-solve-each-inequality-and-write-the-solution-set-if-the-variable-is-an-element-of-the-set-2

Question

In $15-26,$ solve each inequality and write the solution set if the variable is an element of the set of integers.
$$
|y+2|>7
$$

Joshua Utley · Accepted Answer

all right, We&#x27;re asked to solve the absolute value inequality here, and we can do that by breaking it up into two separate equations. So the first equation makes sense, because if y plus two is greater than seven than the absolute value of Y, plus two is greater than seven. But also, if why plus two happens to be less than negative seven in the absolute value of Y, plus two will then be greater than seven as well. So this is how you&#x27;re gonna flip, do your inequalities. And so on the left side, you can subtract two from both sides. Why is greater than five? And on the right side you can get you could subtract two from both sides and get why less than negative nine. And so the best way to write out the solutions if you want to get a more comprehensive idea, is to draw a number line, and we will include the integers zero and one two. Anything doesn&#x27;t That&#x27;s a one. Then over here you have minus one minus two minus three minus for Carly contained this mother out minus five, minus six minus seven. And basically what they&#x27;re saying is this inequality is true. Whenever we use an open circle to tell you that minus seven isn&#x27;t included This inequality is true. When why is over here? And it&#x27;s also true. One why is over here? So they&#x27;re infinitely many integers that satisfy this. And if I was going to write that in some sort of sanitation, I would write that as the scent of why such that why is greater than five, or why is less than negative tonight, which is a mistake I made over here. This should be minus nine. And so there&#x27;s a pictorial representation of your solution. And so everything in here isn&#x27;t No, it wouldn&#x27;t work. But if you plug in anything bigger than five or less than minus nine, this absolute value inequality holds. And so this would be a set representation. And so they&#x27;re infinitely many managers in this solution.

Aditya Sood · Answer

solve each inequality and write a solution, said It&#x27;s a variable is an element off the set of imagers. The inequality that were given is the absolute value off by plus six is greater than 13 so we can split this up right off the bat into two other equations so we can make this. Y Classics is greater than 13 and we can also make this wipe. Ballistics is less than negative 13. So now we just evaluate. So on the left hand side, we&#x27;ll subtract six from both sides to get why is good at 17 and on the right hand side, we will subtract six and get why is less than negative 19. So now we know the range, so pretty much everything above seven will be in their domain of solutions, and anything less than negative 19 will also be so we can simply rewrite that to say negatives into due to negative 19 and seven to infinity and notice that these are curly rackets. Not really, because these are parentheses, not brackets, and that is important because that just shows that native 19 and seven are both not included within the set of solutions

Joshua Utley · Answer

for us to solve the absolute value inequality, we need to first isolate the absolute value on one side. And so we have seven minus X, less than or equal to 10. And our two equations we get from that is this and this solving both. I&#x27;m going to end up here with minus X, less than or equal to three. Divide by minus one. So I have to flip the sign and that&#x27;s one of my solutions. And then I do the same. Over here I get minus X greater than or equal to minus 17. Divide my negative ones that we put a sign we get 17 here, okay? And so we just need to make sure that our solution makes some sense. And so we have minus three. Here is, you know, here on 17 out here. And so what? We&#x27;re what our solution is saying is that everything in here work. So let&#x27;s just test one to be 100% sure. So we have all of our things less than 17 and we have all of our parts that are greater than negative three. And then if we use zero, which is just one of those points seven minus zero plus two is nine, which is less than 12. So this is going to go here. And then these points out here, though, don&#x27;t work. If you plugged in 20 it wouldn&#x27;t work. And if you plug them minus five, it also wouldn&#x27;t work. And so both of these cannot be true differently. They have to be true at the same time. So your solution is the set of X in the integers such that X is less than or equal to 17. You know, I should include those air included. X is less than or equal to 17 and at the exact same time. X also has to be greater than or equal to minus three. And they&#x27;re all rights of a very large brackets. That&#x27;s what he said.

Aditya Sood · Answer

solve each inequality and read the solution set. If the variable is an element off the set of managers, the inequality that we have been given is absolute value of X is greater than nine. So, uh, frankly, this doesn&#x27;t require too much math. Um, we know that the absolute value function uh, absolutely of X. It will take X and just make it positive. So if X is negative 10 it will simply make that positive benefits already positive. 10. You&#x27;ll keep that as positive 10. And so essentially, we just want to see for what may I use is x greater than, uh, nine. So, uh, did regardless f But regardless of sign. So whether it&#x27;s negative or positive, don&#x27;t care about the sangre schmancy for what values executed than mine. And so, um, we know that it&#x27;s going to be great tonight for value sections 10. 11. Doctor, I&#x27;ve got all the way up to infinity now, since this is absolute, this is this is absolute value. We also know that it will be greater than nine for our values, such as names of 10 you negative 11 all the way to negative infinity. And so, uh, we know that you want to know that this is not greater than on equal to or equal to. It is simply only clear then. So our domain is going to be from negative infinity to negative nine and nine to infinity. Ah, One thing to notice is that right here and right here that it is normal pregnancies that are being used, not square brackets such as these. Um, And the reason is because, uh, nine and negative night are not included within the solution set, so yes, thank you.

Aditya Sood · Answer

solved each inequality and write the solution said. If the variable is an element of the state of managers, So the inequality that were given is absolute value of people. Six is less than or equal to five. So what we can do here is that we can split this up into two parts we can make. This beeps of six is less than or equal to five, and it&#x27;s a B plus. Six is less than or equal to five, and the other part that we could make it is people sticks is greater than or equal to negative five. And so a next step will simply be to evaluate the equation. And so we can subtract six from both sides. To get B is less than or equal to make it a one and Andre inside, we can once against attracts six from both sides. To get B is clearly greater than or equal to negative 11. And so we can. Now we know that the solution set is going to be whatever is in between these two values, while including them too. So we can rewrite this in this notation, which states that the solution set is from negative 11 to negative one for all real imagers

Joshua Utley · Answer

Okay, so we&#x27;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&#x27;ll see. We&#x27;ll have to put some restrictions on it because this alone won&#x27;t work, and we&#x27;ll see why. And this itself also won&#x27;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&#x27;t work. Minus two. Minus three is minus five, which is not less than four in the absolute value. So over here actually doesn&#x27;t work. And obviously, if you plug them number like nine over here, this wouldn&#x27;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&#x27;t familiar, So there&#x27;s a set of your solutions. And then here are the B imposed conditions. By the end quote

Problem

In $15-26,$ solve each inequality and write the s…

Problem 16 Hard Difficulty

In $15-26,$ solve each inequality and write the solution set if the variable is an element of the set of integers.
$$
|y+2|>7
$$

Answer

$x \in(-\infty,-9) \cup(5, \infty)$

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Ann Xavier Gantert

Algebra 2 and Trigonometry

Grace H.

Alayna H.

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Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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$x \in(-\infty,-9) \cup(5, \infty)$

Algebra 2 and Trigonometry

Grace H.

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Kristen K.

Absolute Value - Example 1

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Ann Xavier GantertAlgebra 2 and Trigonometry

Grace H.

Alayna H.

Kayleah T.

Kristen K.

Ann Xavier Gantert

Algebra 2 and Trigonometry