Identify the solution of the inequality |x + 3| - 3 >/ 1
I'm assuming you mean: \(|x + 3| - 3 \ge 1\)
Yes.
Hint: Add 3 to both sides
\[x + 3 \ge 4 \]
No, you can't get rid of the absolute value sign yet: \(|x+3| \ge 4\)
How do you get rid of it then?
Hint: \(|x + a| \ge c\) becomes \( -c < x + a < c \)
\[-4 <x+3<4\] Now do you add it all up and set everything equal to X?
Actually, let me correct my formula
\(|x + a| \ge c \) becomes \( -c \ge x + a \ge c\)
Sorry, I'm used to writing these a certain way and also I make typing errors sometimes. But that what you see above is the correct formula.
\[-7 \le x \le1\] Would that be the correct solution?
For one, the inequality symbols are wrong. That's the reason why I corrected myself.
Double check the formula again.
When you do it the way I showed you, you will end up with \(-7 \ge x \ge 1\)
But at this point, you're probably realizing that this cannot be true.
\[x \le -7 or x \ge 1\]
May I ask how you figured that out.
I was going to explain that what you posted is the correct answer.
Isn't it because it would look like \[-7 \le x \ge1\]
No, and you still haven't explained how you figured it out. You would never write it like that.
But that might give me an idea of how to write these in the future.
I'm almost certain you cheated somehow.
I was supposed to explain to you how to arrive at that answer.
There's no way you could have figured that out on your own.
I remember some of what I learned in school, not all of it but certain parts a little. I'm trying to figure out how I remembered it.
Well, anyway, thanks to you, I can re-write my general formula.
\(| x + a | \ge c \) becomes \(-c \le x + a \ge c\)
A little unconventional, but it works
Actually, I just realized that the general formula may not work in every case.
I'll have to test it out.
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