anonymous
  • anonymous
Find a and b. If |x-3|<7, then a
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
-7
anonymous
  • anonymous
If you were to substitute in -4 and10 in to a
SnuggieLad
  • SnuggieLad
Are we doing piece wise functions here?

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anonymous
  • anonymous
I do not understand? Piece wise functions?
SnuggieLad
  • SnuggieLad
You haven't learned that? Then I got my answer.
anonymous
  • anonymous
LOL! Probably not, still in the beginning of College Algebra
SnuggieLad
  • SnuggieLad
This is not college algebra.
SnuggieLad
  • SnuggieLad
This is Algebra 2.
anonymous
  • anonymous
That makes senses, but it is called College Algebra at my school, and towards end of semester it gets in to Trig
SnuggieLad
  • SnuggieLad
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.
anonymous
  • anonymous
Thank you
jdoe0001
  • jdoe0001
\(\bf |x-3|<7 \implies \begin{cases} +(x-3)<7\\ \quad \\ \bf -(x-3)<7 \end{cases}\) if you were to solve those 2 cases, or scenarios, what would be the "values" of "x"?
anonymous
  • anonymous
I am so lost
jdoe0001
  • jdoe0001
hmm ok, what part?
anonymous
  • anonymous
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
anonymous
  • anonymous
So would I have multiplied both sides by -1 and then added three to both sides?
jdoe0001
  • jdoe0001
those answers are correct
jdoe0001
  • jdoe0001
now to make each of the inequalities look like " x + 4 " just add 4 to each side :)
jdoe0001
  • jdoe0001
hmm... the 2nd one would be -4l
jdoe0001
  • jdoe0001
other than that, is ok... so \(\large {|x-3|<7 \implies \begin{cases} +(x-3)<7 \implies x < 10\\ \quad \\ -(x-3)<7 \implies x > -4 \end{cases}\\ \quad \\ x < 10 \implies x +4 < 14\\ \quad \\ x > -4 \implies x+4 > 0}\)
jdoe0001
  • jdoe0001
so you can see what "a" and "b" are
anonymous
  • anonymous
So I did end up with -4
jdoe0001
  • jdoe0001
\(\large {\begin{cases} x < 10 \implies x +4 < 14\\ \quad \\ x > -4 \implies x+4 > 0 \end{cases}\implies \bf 0 < x +4 < 14}\)
anonymous
  • anonymous
So I would add 4 to both sides and not subtract?
jdoe0001
  • jdoe0001
well, they want \(\large x \color{red}{+}4\) so you'd need to add, yes
anonymous
  • anonymous
-4
jdoe0001
  • jdoe0001
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
anonymous
  • anonymous
-4
anonymous
  • anonymous
\[or we can say a \le 0,b \ge 14 .\]

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