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    (X-1)/(x-3) greater than 1

    I know how to solve it but im not sure what to do with the 1 on the right side. If someone could point me in the right direction would be great. Thanks!

    Basically how to get it in a form where i can use a table of inequalities
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    (Original post by Mitul106)
    (X-1)/(x-3) greater than 1

    I know how to solve it but im not sure what to do with the 1 on the right side. If someone could point me in the right direction would be great. Thanks!

    Basically how to get it in a form where i can use a table of inequalities
    You want to get rid of the fraction so that you can get a polynomial and then use your method to solve it.

    Remember that with inequalities if you multiply by a negative number it flips the sign, and we don't want that. So to avoid that, what do we multiply both sides of the equation by?
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    (Original post by Mitul106)
    (X-1)/(x-3) greater than 1

    I know how to solve it but im not sure what to do with the 1 on the right side. If someone could point me in the right direction would be great. Thanks!

    Basically how to get it in a form where i can use a table of inequalities
    subtract the 1 to the left
    then put everything over a single fraction
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    or draw the graph of y = (x-1)/(x-3) and the graph of y=1 on the same axis to see whrre the two intersect and where one will ne bigger than the other
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    But what apporach would i use to get it there?
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    I would say drawing the graphs is easier..
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    (Original post by Mitul106)
    (X-1)/(x-3) greater than 1

    I know how to solve it but im not sure what to do with the 1 on the right side. If someone could point me in the right direction would be great. Thanks!

    Basically how to get it in a form where i can use a table of inequalities
    You have several, valid approaches, I urge you to become familiar with all of them so that you can use the most efficient approach for any inequality you come across.

    First approach (I dislike this one somewhat)

    Graph (sketch) \displaystyle y = \frac{x-1}{x-3} and y=1, find their intersections and where the intervals where the rational function is greater than the line y=1.

    Second approach:

    Consider \displaystyle \frac{x-1}{x-3} - 1 >  0 \iff \frac{2}{x-3} > 0 and graph this, finding the intervals where the graph lies above the x-axis.

    It should be relatively easy to graph by transforming the (standard) graph of \frac{1}{x}.

    It can also be noted that from here, it's quite easy to note that your function is a constant positive number divided by x-3, and the whole thing will only be positive when the denominator is positive as well, so x-3 > 0, from which...

    Third approach:

    Multiply both sides of the inequality by a quantity that is always positive, so in this case, we find that we can do \displaystyle \frac{(x-1)(x-3)^2}{x-3} >(x-3)^2 \iff (x-1)(x-3) > (x-3)^2 where you can expand, collect all the terms to one side and solve it as an (easy) quadratic inequality.

    I have to admit that in this scenario, the second approach seems to be the most elegant one and is what I would choose.
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    (Original post by Mitul106)
    (X-1)/(x-3) greater than 1

    I know how to solve it but im not sure what to do with the 1 on the right side. If someone could point me in the right direction would be great. Thanks!

    Basically how to get it in a form where i can use a table of inequalities
    Multiply by  (x-3)^2 and rearrange. Then all you have to do is solve it like normal.
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    (Original post by Zacken)
    You have several, valid approaches, I urge you to become familiar with all of them so that you can use the most efficient approach for any inequality you come across.

    First approach (I dislike this one somewhat)


    Second approach:


    Third approach:
    Fourth approach

    Note that:

    if x-3 >0 then \frac{x-1}{x-3} > 1 \Rightarrow x-1 > x-3 with x>3

    if x-3 <0 then \frac{x-1}{x-3} > 1 \Rightarrow x-1 < x-3 with x<3

    The first inequality gives -1 > -3 which is true for all values of x where x >3
    The second gives -1 < -3 which is false for all values of x where x < 3

    So the inequality is true for all x > 3
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    (Original post by atsruser)
    Fourth approach

    Note that:

    if x-3 >0 then \frac{x-1}{x-3} > 1 \Rightarrow x-1 > x-3 with x>3

    if x-3 <0 then \frac{x-1}{x-3} > 1 \Rightarrow x-1 < x-3 with x<3

    The first inequality gives -1 > -3 which is true for all values of x where x >3
    The second gives -1 < -3 which is false for all values of x where x < 3

    So the inequality is true for all x > 3
    Nice, I use to use that approach, many thanks for including it.

    For you to be able to make your assertions that -1 > -3 implies that x > 3, shouldn't your arrows be iff ones?

    Edit to clarify, shouldn't you have: if x-3 >0 then \frac{x-1}{x-3} > 1 \iff x-1 > x-3 with x>3?

    I'm aware that this is a little pedantic, but I'm trying to get my head around logic and implications.
 
 
 
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