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anonymous

  • 5 years ago

ok this one is simple looking to figure out if the answer would be negative or positive Y 6 - 13y ≤ 4 - 12y 6 -13y +12y ≤ 4 - 12y + 12y 6 - y ≤ 4 6 - 6 - y ≤ 4 - 6 -y ≤ -2

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  1. bahrom7893
    • 5 years ago
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    i would do this in another way..

  2. bahrom7893
    • 5 years ago
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    instead of adding 12y to both sides, add 13 y to both sides

  3. bahrom7893
    • 5 years ago
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    6 ≤ 4 +y

  4. bahrom7893
    • 5 years ago
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    2 ≤ y or y is more than or equal to two

  5. bahrom7893
    • 5 years ago
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    Please click on become a fan if I helped, I really want to get to 100!! Thanks =)

  6. bahrom7893
    • 5 years ago
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    sorry meant to the next level lol

  7. anonymous
    • 5 years ago
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    well what lost me was line 3 as it should read 6-(-y) shouldnt it ?

  8. bahrom7893
    • 5 years ago
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    okay let me try this step by step

  9. bahrom7893
    • 5 years ago
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    6 - 13y ≤ 4 - 12y 6 -13y +13y ≤ 4 - 12y + 13y 6 ≤ 4 + y 6 - 4 ≤ y 2 ≤ y y \[\ge\] 2

  10. anonymous
    • 5 years ago
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    ok so the entire problem was backwords i thought normal operations was to put the variable to the left with numbers on the right ok thanks i think lol

  11. bahrom7893
    • 5 years ago
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    lol sorry, im pretty bad at -y>-x i am never sure if its y<x or anything.. i just try it backwards lol

  12. anonymous
    • 5 years ago
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    raj, you did the problem right. the last step is y >= 2. when you multiply or divide by -1 in an inequality, you have to switch the direction of the inequality.

  13. anonymous
    • 5 years ago
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    ok thank you its making a little mo anymore its been a few to many years lol been out of school for to long

  14. anonymous
    • 5 years ago
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    if you are ever in doubt of whether you have the correct answer, you can test it by plugging in an x-value and see if the answer you got makes sense. since we got x>2, you could plug in x = 3 to the inequality and see if it is true.

  15. anonymous
    • 5 years ago
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    y that is

  16. anonymous
    • 5 years ago
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    sorry got booted off one system had to log the other in

  17. anonymous
    • 5 years ago
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    thank you for your help

  18. anonymous
    • 5 years ago
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    no problem

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