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jdjrferg4 Group Title

Find a and b. If |x-3|<7, then a<x+4<b ?

  • 10 months ago
  • 10 months ago

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  1. surjithayer Group Title
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    -7<x-3<7 -7+3<x<7+3 -4<x<10 add4 in each inequality and get the value of a and b.

    • 10 months ago
  2. jdjrferg4 Group Title
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    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?

    • 10 months ago
  3. SnuggieLad Group Title
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    Are we doing piece wise functions here?

    • 10 months ago
  4. jdjrferg4 Group Title
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    I do not understand? Piece wise functions?

    • 10 months ago
  5. SnuggieLad Group Title
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    You haven't learned that? Then I got my answer.

    • 10 months ago
  6. jdjrferg4 Group Title
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    LOL! Probably not, still in the beginning of College Algebra

    • 10 months ago
  7. SnuggieLad Group Title
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    This is not college algebra.

    • 10 months ago
  8. SnuggieLad Group Title
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    This is Algebra 2.

    • 10 months ago
  9. jdjrferg4 Group Title
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    That makes senses, but it is called College Algebra at my school, and towards end of semester it gets in to Trig

    • 10 months ago
  10. SnuggieLad Group Title
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    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.

    • 10 months ago
  11. jdjrferg4 Group Title
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    Thank you

    • 10 months ago
  12. jdoe0001 Group Title
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    \(\bf |x-3|<7 \implies \begin{cases} +(x-3)<7\\ \quad \\ \bf -(x-3)<7 \end{cases}\) if you were to solve those 2 cases, or scenarios, what would be the "values" of "x"?

    • 10 months ago
  13. jdjrferg4 Group Title
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    I am so lost

    • 10 months ago
  14. jdoe0001 Group Title
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    hmm ok, what part?

    • 10 months ago
  15. jdjrferg4 Group Title
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    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

    • 10 months ago
  16. jdjrferg4 Group Title
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    So would I have multiplied both sides by -1 and then added three to both sides?

    • 10 months ago
  17. jdoe0001 Group Title
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    those answers are correct

    • 10 months ago
  18. jdoe0001 Group Title
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    now to make each of the inequalities look like " x + 4 " just add 4 to each side :)

    • 10 months ago
  19. jdoe0001 Group Title
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    hmm... the 2nd one would be -4l

    • 10 months ago
  20. jdoe0001 Group Title
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    other than that, is ok... so \(\large {|x-3|<7 \implies \begin{cases} +(x-3)<7 \implies x < 10\\ \quad \\ -(x-3)<7 \implies x > -4 \end{cases}\\ \quad \\ x < 10 \implies x +4 < 14\\ \quad \\ x > -4 \implies x+4 > 0}\)

    • 10 months ago
  21. jdoe0001 Group Title
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    so you can see what "a" and "b" are

    • 10 months ago
  22. jdjrferg4 Group Title
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    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 ?

    • 10 months ago
  23. jdoe0001 Group Title
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    \(\large {\begin{cases} x < 10 \implies x +4 < 14\\ \quad \\ x > -4 \implies x+4 > 0 \end{cases}\implies \bf 0 < x +4 < 14}\)

    • 10 months ago
  24. jdjrferg4 Group Title
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    So I would add 4 to both sides and not subtract?

    • 10 months ago
  25. jdoe0001 Group Title
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    well, they want \(\large x \color{red}{+}4\) so you'd need to add, yes

    • 10 months ago
  26. jdjrferg4 Group Title
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    -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+4-4<10-4 = -8<x<6, how would it cancel the +4 unless you subtract?

    • 10 months ago
  27. jdoe0001 Group Title
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    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

    • 10 months ago
  28. surjithayer Group Title
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    -4<x<10 -4+4<x+4<10+4 0<x+4<14 hence a=0,b=14

    • 10 months ago
  29. surjithayer Group Title
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    \[or we can say a \le 0,b \ge 14 .\]

    • 10 months ago
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