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Partial fractions

Hi everyone,

I have attached a question which I'm not sure about along with my workings, and I was wondering if they were correct?

If they're not, can I have some help as to where to go?

Thanks in advance
20161021_181537.jpg270.7KB
Reply 1
Avatar for jamestg
jamestg
21
Original post by Electrogeek
Hi everyone,

I have attached a question which I'm not sure about along with my workings, and I was wondering if they were correct?

If they're not, can I have some help as to where to go?

Thanks in advance


Looks good to me
Reply 2
Avatar for Electrogeek
Electrogeek
OP
8
Original post by jamestg
Looks good to me


So when I write the final answer, would I put the denominator on the bottom - For example the B fraction becomes 5/(8(3-x)) ?
Original post by Electrogeek
So when I write the final answer, would I put the denominator on the bottom - For example the B fraction becomes 5/(8(3-x)) ?


Yes. It's practically doing (58)(13−x)(\frac{5}{8})(\frac{1}{3-x})
Reply 4
Avatar for Electrogeek
Electrogeek
OP
8
Original post by RDKGames
Yes. It's practically doing (58)(13−x)(\frac{5}{8})(\frac{1}{3-x})


Thanks for the help. :smile:
Reply 5
Avatar for Electrogeek
Electrogeek
OP
8
I'm now stuck on another question - I don't know how to calculate A?
20161021_195339.jpg287.4KB
Original post by Electrogeek
I'm now stuck on another question - I don't know how to calculate A?


I'm not quite familiar with your method of doing these.

So you need to expand RHS fully and collected like terms. From that point you get simultaneous equations from equating the coefficients:

1=A+B+C1=A+B+C

−3=3A+2B+C-3=3A+2B+C

−8=2A-8=2A

which you can solve.
Reply 7
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Electrogeek
OP
8
Cool - Glad I got the right answer. Thanks for the help. :smile:

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