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    Show that g(x) = 2x-1/x-3 can be expressed as g(x) = a/x-3 +b where a and b are real numbers and =/= 3.

    My brain can't work it out.

    Help is welcomed.
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    I may be wrong but this looks like partial fractions to me, show us what you've done so far(if anything)
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    Read the sticky at the top - the way you've written that function g could mean anything! I'm going to assume you mean (2x - 1)/(x -3)

    If that is the case, then divide it through using long division.
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    (Original post by CoffeeStinks)
    Show that g(x) = 2x-1/x-3 can be expressed as g(x) = a/x-3 +b where a and b are real numbers and =/= 3.

    My brain can't work it out.

    Help is welcomed.
    There are different methods for this but here's a quick one which I've always used:

    \displaystyle \frac{2x-1}{x-3}=\frac{2x-6+5}{x-3} = \frac{2x-6}{x-3}+\frac{5}{x-3}=...

    Can you finish it? The trick of changing the numerator in this way is very useful for simplifying improper fractions like this. Here's another example:

    \displaystyle \frac{4x-2}{2x-3} = \frac{4x-6+4}{2x-3} = \frac{4x-6}{2x-3}+\frac{4}{2x-3}=2+\frac{4}{2x-3}
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    (Original post by 0-))
    There are different methods for this but here's a quick one which I've always used:

    \displaystyle \frac{2x-1}{x-3}=\frac{2x-6+5}{x-3} = \frac{2x-6}{x-3}+\frac{5}{x-3}=...

    Can you finish it? The trick of changing the numerator in this way is very useful for simplifying improper fractions like this. Here's another example:

    \displaystyle \frac{4x-2}{2x-3} = \frac{4x-6+4}{2x-3} = \frac{4x-6}{2x-3}+\frac{4}{2x-3}=2+\frac{4}{2x-3}
    I'm starting to understand, but is there a reason why those two numbers the -6 and 4 in the example are used? Or are they just two numbers that add up to the original -2, so could be anything possible pairs?

    Edit: Oh I get it you do it so that the numerator is double the denominator plus the other number to make it the same fraction.

    Thanks
 
 
 
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