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anonymous

  • one year ago

Solve for x. −ax + 3b > 5 x > the quantity 3 times b minus 5 all over a x > the quantity 5 minus 3 times b all over negative a x < the quantity 3 times b plus 5 all over a x < the quantity negative 3 times b plus 5 all over negative a

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  1. misty1212
    • one year ago
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    hi!!

  2. misty1212
    • one year ago
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    i will provide the answer for you so long as you tell me where this question comes from (my guess FLVS)

  3. gibbs
    • one year ago
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    x<−5−3b/a

  4. misty1212
    • one year ago
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    because in fact you cannot solve an absolute value inequality with variable coefficients

  5. misty1212
    • one year ago
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    @gibbs is straight up wrong

  6. anonymous
    • one year ago
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    Its from FLVS @misty1212

  7. misty1212
    • one year ago
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    ok lets solve it the way they want you to, even though it is wrong

  8. anonymous
    • one year ago
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    ok

  9. misty1212
    • one year ago
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    \[−ax + 3b > 5\] subtract \(3b\) get \[-ax>5-3b\] then divide by \(-a\)

  10. misty1212
    • one year ago
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    now here is the mistake they want you to change the inequality, because they want you to think that \(-a\) is negative that is sheer nonsense you do not know if \(-a\) is negative or positive

  11. misty1212
    • one year ago
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    in any case they want you to write \[x<\frac{5-3b}{-a}\] or \[x<\frac{3b-5}{a}\]

  12. misty1212
    • one year ago
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    it is totally incorrect, you should write them and tell them that anyone who thinks \(-a\) is negative (because it has a minus sign) does not understand what a variable is and has no business trying to teach math or write math questions

  13. anonymous
    • one year ago
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    My answers ar this though... x > the quantity 3 times b minus 5 all over a x > the quantity 5 minus 3 times b all over negative a x < the quantity 3 times b plus 5 all over a x < the quantity negative 3 times b plus 5 all over negative a @misty1212

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