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Basic inequality problem

  1. Aug 6, 2013 #1
    1. The problem statement, all variables and given/known data

    ##\frac{a}{4}>\frac{a}{2}+6##

    3. The attempt at a solution

    ##\frac{2a}{2}>\frac{4a+48}{2}##

    ##a>2a+24##

    So do I just plug random numbers in and see what I get? I realised right away that it has to be a negative number so I stuck in -30 and got

    ##-30>-60+24## Well that's true.

    ##-24>-48+24## This is false.

    So ##a<-24## is this correct? Am I going about these the right way?
     
    Last edited: Aug 6, 2013
  2. jcsd
  3. Aug 6, 2013 #2
    Yes you are correct, but here is a different approach to the problem:

    a>2a+24

    Your could simply treat the > sign as an = sign and thus, algebra plays its role;

    a-2a>24
    -a>24

    Then divide both sides by -1 but when you multiply or divide an inequality by a negative number the sign switches from > to < or vice versa...



    [itex]\frac{-a}{-1}[/itex] < [itex]\frac{24}{-1}[/itex]

    [itex]\Rightarrow[/itex] a<-24

    Its the same answer but its avoids the random number input.

    Hope this helped.
     
  4. Aug 6, 2013 #3

    mfb

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    No, you can continue with the simplification. For example, subtract a on both sides, and think about the 24 afterwards.

    Your intermediate step looks more complicated than necessary.
     
  5. Aug 6, 2013 #4
    Here is another:

    ##\frac{y-3}{5}<\frac{y+2}{10}##

    ##\frac{10y-30<5y+10}{5y+10}##

    ##2y-3<1##

    ##y<2##

    is this correct?
     
  6. Aug 6, 2013 #5

    mfb

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    That does not make sense.

    The result is wrong. Just try y=3, for example. Is the original inequality satisfied? Is your result satisfied?
     
  7. Aug 6, 2013 #6
    How about this:

    ##\frac{y-3}{5}-\frac{y+2}{10}##

    ##4y-2y-12+8##

    ##2y<8##

    ##y<4## ?

    I just checked and now it doesn't work... how can I solve these without having to try lots of numbers?
     
    Last edited: Aug 6, 2013
  8. Aug 6, 2013 #7
    No, because its erroneous [itex]\frac{10y-30}{5y+10}[/itex] ≠ 2y-3

    10y-5y<10+30 is what you should be doing. (Remember treat the inequality like an = sign, with the exception of dividing/multiplying with a negative number during simplification)
     
  9. Aug 6, 2013 #8
    oh right I see so I can just treat them like normal fractions.

    ##10y-30=5y+10##

    ##5y=40##

    ##y=8## so the correct solution is ##y<8## ? On wolfram is specifically says ##y<8## and not ##y=8## does this matter?
     
  10. Aug 6, 2013 #9

    mfb

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    y<8 is the correct solution.
    There is no = in your inequality.
     
  11. Aug 6, 2013 #10


    Yes, it does. Remember it's an inequality not a linear equation. I said treat it LIKE an = sign in the sense that algebraic rules are still valid for inequalities. The < remains because its saying [itex]\frac{y-3}{5}[/itex]<[itex]\frac{y+2}{10}[/itex] stays true for all values y<8 and it does.

    Once you plug in y=8 you get 1<1 Which is false because 1 is not less than 1 its 1=1. And when you plug another number out of the inequality like say 9 you get 1.2<1.1 which again is false.

    The logic is really shown in the term inequality. Your trying to find what values of y will this relationship stay true and those that produce an inequality.

    Again, its not a linear equation you can't just replace the < with = its a relationship between those two sides.
     
  12. Aug 6, 2013 #11

    haruspex

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    A word of warning with this approach of treating < and > as like =. Suppose you had something more complicated, like ##\frac{x-3}{x+5} < 2x##. If you multiply through by x+5, as you would happily do with an equality, you do not know whether x+5 is positive or negative. So you do not know whether you should be reversing the inequality. Generally, all you can do here is treat the two possibilities separately.
     
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