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surjithayer Group TitleBest ResponseYou've already chosen the best response.0
7<x3<7 7+3<x<7+3 4<x<10 add4 in each inequality and get the value of a and b.
 one year ago

jdjrferg4 Group TitleBest ResponseYou've already chosen the best response.0
If you were to substitute in 4 and10 in to a<x+4<b, wouldn't you subtract 4 from each side to get 8<x<6?
 one year ago

SnuggieLad Group TitleBest ResponseYou've already chosen the best response.0
Are we doing piece wise functions here?
 one year ago

jdjrferg4 Group TitleBest ResponseYou've already chosen the best response.0
I do not understand? Piece wise functions?
 one year ago

SnuggieLad Group TitleBest ResponseYou've already chosen the best response.0
You haven't learned that? Then I got my answer.
 one year ago

jdjrferg4 Group TitleBest ResponseYou've already chosen the best response.0
LOL! Probably not, still in the beginning of College Algebra
 one year ago

SnuggieLad Group TitleBest ResponseYou've already chosen the best response.0
This is not college algebra.
 one year ago

SnuggieLad Group TitleBest ResponseYou've already chosen the best response.0
This is Algebra 2.
 one year ago

jdjrferg4 Group TitleBest ResponseYou've already chosen the best response.0
That makes senses, but it is called College Algebra at my school, and towards end of semester it gets in to Trig
 one year ago

SnuggieLad Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

jdjrferg4 Group TitleBest ResponseYou've already chosen the best response.0
Thank you
 one year ago

jdoe0001 Group TitleBest ResponseYou've already chosen the best response.0
\(\bf x3<7 \implies \begin{cases} +(x3)<7\\ \quad \\ \bf (x3)<7 \end{cases}\) if you were to solve those 2 cases, or scenarios, what would be the "values" of "x"?
 one year ago

jdjrferg4 Group TitleBest ResponseYou've already chosen the best response.0
I am so lost
 one year ago

jdoe0001 Group TitleBest ResponseYou've already chosen the best response.0
hmm ok, what part?
 one year ago

jdjrferg4 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

jdjrferg4 Group TitleBest ResponseYou've already chosen the best response.0
So would I have multiplied both sides by 1 and then added three to both sides?
 one year ago

jdoe0001 Group TitleBest ResponseYou've already chosen the best response.0
those answers are correct
 one year ago

jdoe0001 Group TitleBest ResponseYou've already chosen the best response.0
now to make each of the inequalities look like " x + 4 " just add 4 to each side :)
 one year ago

jdoe0001 Group TitleBest ResponseYou've already chosen the best response.0
hmm... the 2nd one would be 4l
 one year ago

jdoe0001 Group TitleBest ResponseYou've already chosen the best response.0
other than that, is ok... so \(\large {x3<7 \implies \begin{cases} +(x3)<7 \implies x < 10\\ \quad \\ (x3)<7 \implies x > 4 \end{cases}\\ \quad \\ x < 10 \implies x +4 < 14\\ \quad \\ x > 4 \implies x+4 > 0}\)
 one year ago

jdoe0001 Group TitleBest ResponseYou've already chosen the best response.0
so you can see what "a" and "b" are
 one year ago

jdjrferg4 Group TitleBest ResponseYou've already chosen the best response.0
So I did end up with 4<x<10, so if I supplement those in a<x+4<b = 4<x+4<10. Even though it says find a and b, would I go on to say 8<x<6 ?
 one year ago

jdoe0001 Group TitleBest ResponseYou've already chosen the best response.0
\(\large {\begin{cases} x < 10 \implies x +4 < 14\\ \quad \\ x > 4 \implies x+4 > 0 \end{cases}\implies \bf 0 < x +4 < 14}\)
 one year ago

jdjrferg4 Group TitleBest ResponseYou've already chosen the best response.0
So I would add 4 to both sides and not subtract?
 one year ago

jdoe0001 Group TitleBest ResponseYou've already chosen the best response.0
well, they want \(\large x \color{red}{+}4\) so you'd need to add, yes
 one year ago

jdjrferg4 Group TitleBest ResponseYou've already chosen the best response.0
4<x+4<10, I thought you would work it like a normal equation and to carry over the 4, you would have to subtract from all sides? 4+4<x+44<104 = 8<x<6, how would it cancel the +4 unless you subtract?
 one year ago

jdoe0001 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

surjithayer Group TitleBest ResponseYou've already chosen the best response.0
4<x<10 4+4<x+4<10+4 0<x+4<14 hence a=0,b=14
 one year ago

surjithayer Group TitleBest ResponseYou've already chosen the best response.0
\[or we can say a \le 0,b \ge 14 .\]
 one year ago
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