#### jUnIoRjUnCtIoNmAtHmAtH

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So I have the problem |y-2| + 11 > 0, and I got, for my two solutions, |y| > -9 (which would mean all real numbers are true) and |y| < 13. How would I display my answer?

Thanks!

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So I have the problem |y-2| + 11 > 0, and I got, for my two solutions, |y| > -9 (which would mean all real numbers are true) and |y| < 13. How would I display my answer?

Thanks!

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Also, can you show how you got them?

The correct answer is that the inequality is true for all real numbers, but the inequality |y-2| + 11 > 0 is not inherently equivalent to |y| > -9. I think you took an illegal step. The important part of answering your question will be to correct your work.

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Thanks! I realized I had a computation error, thus seeming like I took an "illegal step".

Also, can you show how you got them?

The correct answer is that the inequality is true for all real numbers, but the inequality |y-2| + 11 > 0 is not inherently equivalent to |y| > -9. I think you took an illegal step. The important part of answering your question will be to correct your work.

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Please think about the following:So I have the problem |y-2| + 11 > 0, and I got, for my two solutions, |y| > -9 (which would mean all real numbers are true) and |y| < 13. How would I display my answer?

If \(\displaystyle c>0\) then the solution to \(\displaystyle |x+5|+c>0\) has a solution set of \(\displaystyle (-\infty,\infty)\), all real numbers.

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It will still be good to show your work, because I think there's more than just a wrong addition. Your result should not have lookedThanks! I realized I had a computation error, thus seeming like I took an "illegal step".