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anonymous

  • one year ago

Solve the following inequality. write the solution set using interval notation.

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    thats the question and how the answer must be.

  3. jim_thompson5910
    • one year ago
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    Multiply both sides by 40 to clear out the fractions \[\Large \frac{6x+2}{8}-\frac{2x-8}{5} \le -4\] \[\Large 40*\left(\frac{6x+2}{8}-\frac{2x-8}{5}\right) \le 40*(-4)\] \[\Large 40*\left(\frac{6x+2}{8}\right)+40*\left(-\frac{2x-8}{5}\right) \le 40*(-4)\] \[\Large 5*\left(6x+2\right)-8*\left(2x-8\right) \le -160\] Do you see how to finish up?

  4. anonymous
    • one year ago
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    why is it 40? and i would distribute the 5 out into the 6X+2 then the 8 into the 2x-8 and keep going until i can take the left side and either divide the right by it or the left by the right.

  5. jim_thompson5910
    • one year ago
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    because 8*5 = 40 is the LCD

  6. jim_thompson5910
    • one year ago
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    if you multiply both sides by the LCD, then you clear out the fractions

  7. anonymous
    • one year ago
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    ohhhhhhhhhhh ok. am i correct in the final steps?

  8. jim_thompson5910
    • one year ago
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    ` i would distribute the 5 out into the 6X+2 then the 8 into the 2x-8` correct `keep going until i can take the left side and either divide the right by it or the left by the right` these steps seem a bit vague to me

  9. anonymous
    • one year ago
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    if the number on the left of the less than or equal to sign is less than 160 divide 160 by it. if it is more than 160 than divide that number by 160?

  10. jim_thompson5910
    • one year ago
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    when you distributed, what did you get?

  11. anonymous
    • one year ago
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    30x+10-16x-64≤-160

  12. jim_thompson5910
    • one year ago
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    the -64 should be +64

  13. anonymous
    • one year ago
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    14x-54≤-160

  14. jim_thompson5910
    • one year ago
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    since -8 times -8 = 64

  15. anonymous
    • one year ago
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    +54**

  16. anonymous
    • one year ago
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    didnt change the sign

  17. jim_thompson5910
    • one year ago
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    after distributing you should get `30x+10-16x+64≤-160` which simplifies to `14x+74≤-160` after you combine like terms

  18. anonymous
    • one year ago
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    14x≤-214

  19. anonymous
    • one year ago
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    oh i messed up somewhere

  20. anonymous
    • one year ago
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    had to do the +64

  21. anonymous
    • one year ago
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    14x≤234?

  22. jim_thompson5910
    • one year ago
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    close but no

  23. jim_thompson5910
    • one year ago
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    you lost a sign somewhere, it should be `14x≤-234`

  24. anonymous
    • one year ago
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    ah i put it in my calculator and forgot the negative!

  25. anonymous
    • one year ago
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    x=-16.4142857?

  26. jim_thompson5910
    • one year ago
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    leave it as a fraction and reduce as much as possible

  27. anonymous
    • one year ago
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    is it a fraction? i got a decimal when i divided

  28. jim_thompson5910
    • one year ago
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    234/14 can be reduced to ???

  29. jim_thompson5910
    • one year ago
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    hint: find the GCF of 234 and 14

  30. anonymous
    • one year ago
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    GCF=2?

  31. jim_thompson5910
    • one year ago
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    yes

  32. anonymous
    • one year ago
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    117/7

  33. jim_thompson5910
    • one year ago
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    divide each part of the fraction by 2 234/2 = 117 14/2 = 7 so 234/14 = 117/7

  34. jim_thompson5910
    • one year ago
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    good

  35. anonymous
    • one year ago
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    so what would the solution set be?

  36. jim_thompson5910
    • one year ago
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    so -234/14 = -117/7

  37. jim_thompson5910
    • one year ago
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    in the end, the solution would be \[\Large x \le -\frac{117}{7}\] what is this in interval notation?

  38. anonymous
    • one year ago
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    [-117,7]?

  39. jim_thompson5910
    • one year ago
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    draw a number line with -117/7 on it |dw:1443394703025:dw|

  40. jim_thompson5910
    • one year ago
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    the inequality \(\LARGE x \le -\frac{117}{7}\) means that we shade to the left of -117/7 on the number line |dw:1443394759997:dw|

  41. jim_thompson5910
    • one year ago
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    so the interval would start at -infinity, then stop at -117/7 giving us \[\Large \left. (-\infty, -\frac{117}{7}\right]\]

  42. anonymous
    • one year ago
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    (-inf, -117/7)

  43. jim_thompson5910
    • one year ago
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    |dw:1443394823713:dw|

  44. jim_thompson5910
    • one year ago
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    |dw:1443394850880:dw|

  45. jim_thompson5910
    • one year ago
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    |dw:1443394878344:dw|

  46. anonymous
    • one year ago
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    is it a ] because -117/7 is included?

  47. jim_thompson5910
    • one year ago
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    |dw:1443394904462:dw|

  48. jim_thompson5910
    • one year ago
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    `is it a ] because -117/7 is included?` correct

  49. jim_thompson5910
    • one year ago
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    and there is a closed circle at -117/7 on the number line to tell readers to include the endpoint |dw:1443394959903:dw|

  50. anonymous
    • one year ago
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    ok so that should be my answer than. Thanks! i understand it all now. and i know to use the inf.

  51. jim_thompson5910
    • one year ago
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    you're welcome

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