Math answered urgently required. :(

ac1d1ty

ac1d1ty

ac1d1ty

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Joined
26 Jan 2007
Posts
2,462
Any chance of some help?

20rtnxs.jpg
 
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.
 
Duff-Man said:
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.
Click to expand...

Cheers mate, i'll check the question to see if I posted everything. Thanks for taking the time though to respond positively.
 
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.
 
Why is maths so urgent on a sunday? Text the problem to Any Question Answered and ask them to show their work. Costs £1
 
Darg said:
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.
Click to expand...

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.
 
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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