Math answered urgently required. :(

Associate
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26 Jan 2007
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Any chance of some help?

20rtnxs.jpg
 
Soldato
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Not quite sure what's going on there, but if you're asking what I THINK you are then:

The denominators are equal, so as long as:

x != 1 and
x != -2 then:

you have

3x = x(a+b) -a +2b

so:

x(3-a-b) = -a +2b

hence:

x = (2b-a)/(3-a-b)


If x=1 or x=-2, then you have a singular problem, which has no solution.
 
Associate
OP
Joined
26 Jan 2007
Posts
2,462
Not quite sure what's going on there, but if you're asking what I THINK you are then:

The denominators are equal, so as long as:

x != 1 and
x != -2 then:

you have

3x = x(a+b) -a +2b

so:

x(3-a-b) = -a +2b

hence:

x = (2b-a)/(3-a-b)


If x=1 or x=-2, then you have a singular problem, which has no solution.

Cheers mate, i'll check the question to see if I posted everything. Thanks for taking the time though to respond positively.
 
Soldato
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6 Sep 2006
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6,275
Location
London
Next time maybe don't post a massive image, just the text. And try to explain what the actual question is asking, it's really not clear from your post.
 

alx

alx

Soldato
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Dubai, UAE
Next time maybe don't post a massive image, just the text. And try to explain what the actual question is asking, it's really not clear from your post.

This :)

In the first part, is it meant to be 3x/(x+2)(x-1) or 3x(x+2)(x-1)?

The layout/formatting isn't very clear atm.
 
Associate
OP
Joined
26 Jan 2007
Posts
2,462
Thanks for all the help. Turns out the answer was six?


Must have missed something out somewhere.


P.s no its not pasty smashing time, I thought you would have guessed this by the GCSE level maths :D :eek:
 
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