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jdjrferg4
 2 years ago
Find a and b. If x3<7, then a<x+4<b ?
jdjrferg4
 2 years ago
Find a and b. If x3<7, then a<x+4<b ?

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surjithayer
 2 years ago
Best ResponseYou've already chosen the best response.07<x3<7 7+3<x<7+3 4<x<10 add4 in each inequality and get the value of a and b.

jdjrferg4
 2 years ago
Best ResponseYou've already chosen the best response.0If 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?

SnuggieLad
 2 years ago
Best ResponseYou've already chosen the best response.0Are we doing piece wise functions here?

jdjrferg4
 2 years ago
Best ResponseYou've already chosen the best response.0I do not understand? Piece wise functions?

SnuggieLad
 2 years ago
Best ResponseYou've already chosen the best response.0You haven't learned that? Then I got my answer.

jdjrferg4
 2 years ago
Best ResponseYou've already chosen the best response.0LOL! Probably not, still in the beginning of College Algebra

SnuggieLad
 2 years ago
Best ResponseYou've already chosen the best response.0This is not college algebra.

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

SnuggieLad
 2 years ago
Best ResponseYou've already chosen the best response.0That 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.

jdoe0001
 2 years ago
Best 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"?

jdjrferg4
 2 years ago
Best ResponseYou've already chosen the best response.0When 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

jdjrferg4
 2 years ago
Best ResponseYou've already chosen the best response.0So would I have multiplied both sides by 1 and then added three to both sides?

jdoe0001
 2 years ago
Best ResponseYou've already chosen the best response.0those answers are correct

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

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

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

jdoe0001
 2 years ago
Best ResponseYou've already chosen the best response.0so you can see what "a" and "b" are

jdjrferg4
 2 years ago
Best ResponseYou've already chosen the best response.0So 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 ?

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

jdjrferg4
 2 years ago
Best ResponseYou've already chosen the best response.0So I would add 4 to both sides and not subtract?

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

jdjrferg4
 2 years ago
Best ResponseYou've already chosen the best response.04<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?

jdoe0001
 2 years ago
Best ResponseYou've already chosen the best response.0yeap 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

surjithayer
 2 years ago
Best ResponseYou've already chosen the best response.04<x<10 4+4<x+4<10+4 0<x+4<14 hence a=0,b=14

surjithayer
 2 years ago
Best ResponseYou've already chosen the best response.0\[or we can say a \le 0,b \ge 14 .\]
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