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Math Help - Problems solving some inequalities questions

  1. #1
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    Problems solving some inequalities questions

    a) \frac{\2x+3}{3} \leq x-5

    I did this:

    \frac{\3(2x+3)}{3} \leq 3(x-5)

    6x+9 \leq 3x-15

    6x-3x \leq -15-9

    3x \leq -24

    x \leq -8

    (inf , -8]

    however the answer should be : [18,inf)

    b) another one

    \frac{\ x+4}{2} \geq \frac{\ x-2}{4}

    \frac{\2(x+4)}{2} \geq \frac{\2(x-2)}{2(4)}

    2x+8 \geq \frac{\2x-4}{8}

    8(2x+8) \geq \frac{\8(2x-4)}{8}

    16x+64 \geq 16x-32

    16x-16x \geq -64 -32

    0 \geq -96

    however the answer should be [-10, inf)

    c) last one

    3x \leq x+1 \leq x-1

    3x-1 \leq x \leq x-1-1

    don't know what to do after this step
    Last edited by Devi09; March 29th 2011 at 09:38 AM.
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  2. #2
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    Your basic algebra skills are in need of some real work.
    \dfrac{x+3}{3}<x-5 gives x+3<3x-15.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Your basic algebra skills are in need of some real work.
    \dfrac{x+3}{3}<x-5 gives x+3<3x-15.
    x-3x \leq -3-15

    -2x \leq -18

    x \geq 9

    still not getting the answer
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  4. #4
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    Quote Originally Posted by Devi09 View Post
    x-3x \leq -3-15
    -2x \leq -18
    x \geq 9
    still not getting the answer
    The given answer is wrong. [9,\infty)
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  5. #5
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    Quote Originally Posted by Devi09 View Post
    a) \frac{\2x+3}{3} \leq x-5

    according to your next line,
    that should be 2x+3 in the numerator which will give the answer provided


    I did this:

    \frac{\3(2x+3)}{3} \leq 3(x-5)

    why didn't you multiply BOTH sides by 3 ?

    6x+9 \leq 3x-15

    Now you multiplied only the left side by 9.
    If you multiply only one side by something other than 1,
    you will have a different inequality.


    6x-3x \leq -15-9

    3x \leq -24

    x \leq -8

    (inf , -8]

    however the answer should be : [18,inf)

    b) another one

    \frac{\ x+4}{2} \geq \frac{\ x-2}{4}

    \frac{\2(x+4)}{2} \geq \frac{\2(x-2)}{2(4)}

    You halved the right and did nothing to the other side,
    so you now have a different inequality altogether.
    You could have multiplied both sides by 4.


    2x+8 \geq \frac{\2x-4}{8}

    You multiplied the left by 4
    and changed the denominator of the other side.
    Any reason ?


    8(2x+8) \geq \frac{\8(2x-4)}{8}

    You multiplied the left by 8 and made a mysterious modification
    to the numerator on the right again.
    The rest is also haphazard.


    16x+64 \geq 16x-32

    16x-16x \geq -64 -32

    0 \geq -96

    however the answer should be [-10, inf)

    c) last one

    3x \leq x+1 \leq x-1

    3x-1 \leq x \leq x-1-1

    don't know what to do after this step
    For the last one, you cannot have

    x+1\le\ x-1 since x+1 is 2 greater than x-1
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  6. #6
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    For b) 2nd try

    \frac{\ x+4}{2} \geq \frac{\ x-2}{4}

    \frac{\4(x+4)}{4(2)} \geq x-2

    \frac{\4x+16}{8} \geq x-2

    4x + 16 \geq 8(x-2)

    4x +16 \geq 8x-16

    4x-8x \geq -16-16

    -4x \geq -32

    x \leq -32/-4

    x \leq 8
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  7. #7
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    Quote Originally Posted by Devi09 View Post
    For b) 2nd try
    \dfrac{\ x+4}{2} \geq \dfrac{\ x-2}{4}
    x \leq 8
    Let's see why that solution is not correct.
    According to your solution x=-14 works.
    Plug it in an see that it does not.

    According to your solution x=10 does not works.
    Plug it in an see that it does work.

    Here is simple algebra.
    \dfrac{\ x+4}{2} \geq \dfrac{\ x-2}{4}
    Then we get 4x+16\ge 2x-4. HOW?
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  8. #8
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    Quote Originally Posted by Plato View Post
    Let's see why that solution is not correct.
    According to your solution x=-14 works.
    Plug it in an see that it does not.

    According to your solution x=10 does not works.
    Plug it in an see that it does work.

    Here is simple algebra.
    \dfrac{\ x+4}{2} \geq \dfrac{\ x-2}{4}
    Then we get 4x+16\ge 2x-4. HOW?
    So when you multiply by 4 you don't multiply the denominator being 2 in this case

    So its:

    4x+16 \geq 2x-4

    4x -2x \geq -16-4

    2x \geq -20

    x \geq -10
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  9. #9
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    Quote Originally Posted by Devi09 View Post
    So when you multiply by 4 you don't multiply the denominator being 2 in this case

    So its:

    4x+16 \geq 2x-4

    4x -2x \geq -16-4

    2x \geq -20

    x \geq -10
    Way to go! By George you have it.
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  10. #10
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    One more thing....

    233 + 6x \leq 8x - 75 \leq 475 + 6x

    this is what I did:

    233 + 75 + 6x \leq 8x \leq 475 + 75 + 6x

    308 + 6x \leq 8x \leq 550 + 6x

    What do you do with the x's? move them to the middle x?
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