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Find a and b. If |x-3|<7, then a

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If you were to substitute in -4 and10 in to a
Are we doing piece wise functions here?

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Other answers:

I do not understand? Piece wise functions?
You haven't learned that? Then I got my answer.
LOL! Probably not, still in the beginning of College Algebra
This is not college algebra.
This is Algebra 2.
That makes senses, but it is called College Algebra at my school, and towards end of semester it gets in to Trig
That is algebra two. And this is piece wise functions. I am going to go offline for a while be back in about ten or fifteen minutes.
Thank you
\(\bf |x-3|<7 \implies \begin{cases} +(x-3)<7\\ \quad \\ \bf -(x-3)<7 \end{cases}\) if you were to solve those 2 cases, or scenarios, what would be the "values" of "x"?
I am so lost
hmm ok, what part?
When I work those problems, which I am probably not doing correctly I get:\[+(x−3)<7 = x<10\] \[−(x−3)<7 = x>4\] but the answer is -4, so I believe I am doing this wrong
So would I have multiplied both sides by -1 and then added three to both sides?
those answers are correct
now to make each of the inequalities look like " x + 4 " just add 4 to each side :)
hmm... the 2nd one would be -4l
other than that, is ok... so \(\large {|x-3|<7 \implies \begin{cases} +(x-3)<7 \implies x < 10\\ \quad \\ -(x-3)<7 \implies x > -4 \end{cases}\\ \quad \\ x < 10 \implies x +4 < 14\\ \quad \\ x > -4 \implies x+4 > 0}\)
so you can see what "a" and "b" are
So I did end up with -4
\(\large {\begin{cases} x < 10 \implies x +4 < 14\\ \quad \\ x > -4 \implies x+4 > 0 \end{cases}\implies \bf 0 < x +4 < 14}\)
So I would add 4 to both sides and not subtract?
well, they want \(\large x \color{red}{+}4\) so you'd need to add, yes
yeap when yo have only left and right sides, you add/divide/multiply to BOTH sides when you have left, middle, and right sides, you add/divide/multiply to ALL sides
\[or we can say a \le 0,b \ge 14 .\]

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