Question

Decompose (3x - 1)/(x^2 - x) into partial fractions

Answer 1

(3x-1) / (x^2 - x) = (3x-1)/x(x-1) = (A/x) + (B/ x-1). So, A(x-1) + Bx = 3x-1 => Ax-A+Bx= 3x-1 => (A+B)x - A = 3x-1 => A=1. So, A+B =3 => B=2. Finally (3x-1)/ (x^2-x) = [1/x ] + [2/ (x-1)]

Answer 2

wow umm can u explain??

Answer 3

which part?

Answer 4

the whole part like idk how to decompose it what is the first step?

Answer 5

first factor the denominator

Answer 6

okay

Answer 7

so did you get any roots? for this problem?

Answer 8

you don't need to find roots for this question

Answer 9

or solutions either?

Answer 10

[1/x] + [2/ (x-1)] is the answer

Answer 11

ok

Answer 12

partial fraction means you're breaking down fractions so it's easier to integrate later

Answer 13

okay

Answer 14

any more question?

Answer 15

i'm sorry but that to me looks confusing :/ maybe i need to try it again on my own. thanks tho

Answer 16

do it step by step. write it down. it's hard at first

Answer 17

okay i will thanks

Answer 18

@hcmathkim ok so i did it, can you tell me if its right?

1-3x/x - x^2 = 0
3x - 1/ (x - 1) x = 0
3x - 1/(x - 1) x (multiply both by (x - 1)x

3x - 1 = 0
+1 +1
= 3x = 1
divide 3 both side

x = 1/3

Answer 19

you never said they were equal to zero

Answer 20

yea its equal to zero...sorry

Answer 21

then just set the numerator equal to zero. 3x-1=0 => 3x =1 => x=1/3

Answer 22

okay so what i did up there is correct?

Answer 23

1) you copied the question wrong 2) you dont need to factor the denominator in this case

Answer 24

ok

Answer 25

only way for a fraction to be zero is if the numerator is zero. so just worry about the numerator.

Answer 26

okay

Answer 27

the bottom half of the work is fine

Answer 28

ok i'll try it again