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1. 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.
2. I may be wrong but this looks like partial fractions to me, show us what you've done so far(if anything)
3. 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.
4. (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:

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:

5. (Original post by 0-))
There are different methods for this but here's a quick one which I've always used:

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:

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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Updated: December 27, 2010
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