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Math Help - Help solving for set builder notation

  1. #1
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    Help solving for set builder notation

    I need help solving for {X|X ?}

    3/8 (5+6x)+3 greater than or equal to 21

    I appreciate help working through this!
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  2. #2
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    If I understand what you're asking, you want:

    \{x\ :\ \frac{3}{8}(5+6x) + 3 \geq 21\}

    Technically that set definition works (though, we might put a condition on the left-hand side by x\in \mathbb{R}, the real numbers). You should, however, simplify the specification on the right-hand side. That is just algebra, though. You should be able to do that. It's straight-foward. The answer is 6x \geq 43.
    Last edited by bryangoodrich; May 26th 2011 at 03:59 PM. Reason: opps, messed up my own algebra!
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  3. #3
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    Thank you! I am looking for the solution in a set builder notation form {x|x ?}

    So, to solve a different one: 3/8 (3+2x) +1 greater than or equal to 13 is the answer {x|x greater than or equal to 25}?
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  4. #4
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    No, that's not correct. It might be more helpful if you showed us your working so we can see where you are going wrong.
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  5. #5
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    Also, try formatting your problem in latex, just so we can be sure that we are understanding your problem correctly. So, is the statement you are trying to solve the following?:

    \frac{3(3+2x)}{8} + 1 \geqslant 13
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  6. #6
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    Solve for x: subtract 1 from both sides, cross multiple, ....
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  7. #7
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    Quote Originally Posted by borophyll View Post
    Also, try formatting your problem in latex, just so we can be sure that we are understanding your problem correctly. So, is the statement you are trying to solve the following?:

    \frac{3(3+2x)}{8} + 1 \geqslant 13
    No, it is exactly like my first problem, that Bryan quoted, but with different numbers... 3/8 (3+2x) +1 greater than or equal to 13. Thanks for any help!
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  8. #8
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    Quote Originally Posted by jay1 View Post
    No, it is exactly like my first problem, that Bryan quoted, but with different numbers... 3/8 (3+2x) +1 greater than or equal to 13. Thanks for any help!
    \frac{3(3+2x)}{8}+1=\frac{3}{8}(3+2x)+1\geq 13
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  9. #9
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    Here are the steps to your problem:

    (1) subtract
    (2) cross multiply
    (3) subtract
    (4) divide
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  10. #10
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    The way I have written it and the way Bryan wrote it are the same, just formatted in different ways.
    Last edited by borophyll; May 26th 2011 at 11:06 PM. Reason: spelling
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  11. #11
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    Do I begin by subtractin 3/8 from 13?
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  12. #12
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    It sounds like you are having some troubles with basic algebraic manipulation. You start by subtracting the 1 from both sides, to get {\frac{3}{8}}(3+2x) \geqslant 12
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  13. #13
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    When I multiply 3/8 by 2 I get a decimal. Is the answer a fraction?
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  14. #14
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    Your goal is to get x by itself. To do this first you subtract the constant '1' from each side to get

    \frac{3}{8}(3 + 2x) \geq 12

    Now you cross-multiply both sides by \frac{8}{3}. Do you see why this was chosen? It removes the fraction from the left side. The result is then

    3+2x \geq 12\times \frac{8}{3} = 4\times 8 = 32

    Then again we subtract the free constant 3 from both sides to get

    2x \geq 29

    And finally we divide both sides by 2 to get a result simplified to an inequality regarding x:

     x \geq \frac{29}{2} = 14.5

    Did each step make sense to you? Also, when did the equation change? Didn't we originally start with \frac{3}{8}(5+6x)+3 \geq 21? In any case, the steps would be identical to the above, just with different values, and I already reported the result almost completely simplified.
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