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Thread: Problems solving some inequalities questions

  1. #1
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    Problems solving some inequalities questions

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

    I did this:

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

    $\displaystyle 6x+9 \leq 3x-15$

    $\displaystyle 6x-3x \leq -15-9$

    $\displaystyle 3x \leq -24$

    $\displaystyle x \leq -8$

    $\displaystyle (inf , -8]$

    however the answer should be : $\displaystyle [18,inf)$

    b) another one

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

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

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

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

    $\displaystyle 16x+64 \geq 16x-32$

    $\displaystyle 16x-16x \geq -64 -32$

    $\displaystyle 0 \geq -96$

    however the answer should be $\displaystyle [-10, inf)$

    c) last one

    $\displaystyle 3x \leq x+1 \leq x-1$

    $\displaystyle 3x-1 \leq x \leq x-1-1$

    don't know what to do after this step
    Last edited by Devi09; Mar 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.
    $\displaystyle \dfrac{x+3}{3}<x-5$ gives $\displaystyle x+3<3x-15$.
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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.
    $\displaystyle \dfrac{x+3}{3}<x-5$ gives $\displaystyle x+3<3x-15$.
    $\displaystyle x-3x \leq -3-15$

    $\displaystyle -2x \leq -18$

    $\displaystyle x \geq 9$

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

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


    I did this:

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

    why didn't you multiply BOTH sides by 3 ?

    $\displaystyle 6x+9 \leq 3x-15$

    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.


    $\displaystyle 6x-3x \leq -15-9$

    $\displaystyle 3x \leq -24$

    $\displaystyle x \leq -8$

    $\displaystyle (inf , -8]$

    however the answer should be : $\displaystyle [18,inf)$

    b) another one

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

    $\displaystyle \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.


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

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


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

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


    $\displaystyle 16x+64 \geq 16x-32$

    $\displaystyle 16x-16x \geq -64 -32$

    $\displaystyle 0 \geq -96$

    however the answer should be $\displaystyle [-10, inf)$

    c) last one

    $\displaystyle 3x \leq x+1 \leq x-1$

    $\displaystyle 3x-1 \leq x \leq x-1-1$

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

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

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

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

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

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

    $\displaystyle 4x +16 \geq 8x-16$

    $\displaystyle 4x-8x \geq -16-16$

    $\displaystyle -4x \geq -32$

    $\displaystyle x \leq -32/-4$

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

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

    Here is simple algebra.
    $\displaystyle \dfrac{\ x+4}{2} \geq \dfrac{\ x-2}{4}$
    Then we get $\displaystyle 4x+16\ge 2x-4.$ 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 $\displaystyle x=-14$ works.
    Plug it in an see that it does not.

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

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

    So its:

    $\displaystyle 4x+16 \geq 2x-4$

    $\displaystyle 4x -2x \geq -16-4$

    $\displaystyle 2x \geq -20$

    $\displaystyle x \geq -10$
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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:

    $\displaystyle 4x+16 \geq 2x-4$

    $\displaystyle 4x -2x \geq -16-4$

    $\displaystyle 2x \geq -20$

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

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

    this is what I did:

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

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

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