View Full Version : Algerbraic fractions to partial fractions
bayley86
Jan 7, 2010, 02:55 PM
reduce the following algerbraic fractions to partial fractions
7x-4
(x-2)(2x-3)
5x-7
(3x+2)^2
(x-3)(x+4)
x^2-12x
can anyone help me with these
thanks
Andy
galactus
Jan 7, 2010, 05:52 PM
reduce the following algerbraic fractions to partial fractions
7x-4
(x-2)(2x-3)
I will show you this one, then you try the others.
Rewrite as \frac{A}{x-2}+\frac{B}{2x-3}=7x-4
The idea is to find A and B.
A(2x-3)+B(x-2)=7x-4... [1]
What makes 2x-3=0? x=3/2. Sub this into [1]
\frac{-1}{2}B=\frac{13}{2}
B=-13
What makes x-2=0, x=2 of course. Sub this into [1]
A(2(2)-3)=7(2)-4
A=10
There it is:
\fbox{\frac{10}{x-2}-\frac{13}{2x-3}}
bayley86
Jan 9, 2010, 05:14 AM
hi this is the bit I don't understand or were you got it from
-1 B=13
2 2
B=-13
please could you explain what I have to do step by step
thanks
Andy
Unknown008
Jan 9, 2010, 10:00 AM
Well, this was a good explanation... I'll do deeper then. We started with
\frac{7x-4}{(x-2)(2x-3)} = \frac{A}{x-2}+\frac{B}{2x-3}
Multiplied everything by the denominator:
A(2x-3)+B(x-2)=7x-4
You need to have A removed completely to find B. That means A has to be multiplied by 0. In turn, that means that 2x-3 = 0, or x = 3/2.
Well, OK, replace all the 'x' by 3/2.
A(2(\frac32)-3)+B((\frac32)-2)=7(\frac32)-4
A(3-3)+B(-\frac12)=\frac{21}2-4
A(0)+B(-\frac12)=\frac{13}2
-\frac12 B=\frac{13}2
B=-13
Then, you find how to make B disappear by making x-2 = 0.
galactus
Jan 9, 2010, 10:17 AM
\frac{(x-3)(x+4)}{x^{2}-12x}
\frac{A}{x}+\frac{B}{x-12}=(x-3)(x+4)
\frac{5x-7}{(3x+2)^{2}}
\frac{A}{3x+2}+\frac{B}{(3x+2)^{2}}=5x-7