Question

How to subtract a fraction with another fraction?

Answer 1

[3/(x-1)] - [(2x+10)/(x^2+2x-3)] = 1/3

Answer 2

what is x?

Answer 3

this question is solved yours other subtittle.

Answer 4

we say x=0

Answer 5

lcd = 3(x^2 +2x-3) = 3(x-1)(x+3)
multiplying through by this gives:
9(x+3) - 3(2x+10) = (x-1)(x+3)
9x +27 - 6x - 30 = x^2 +2x - 3
x^2 +x = 0
x(x+1)=0

x = 0 or x = -1

Answer 6

yeah cnsu90, thanks for ur help :)

Answer 7

oops look like an error in last two steps!
9x +27 - 6x - 30 = x^2 +2x - 3
-x^2 +x = 0
x (1 - x) = 0
x = 0 or x = 1

Answer 8

hmm..

Answer 9

@jimmy..denominator cant be 0..so u cant say x=1..

Answer 10

yes - you are right

Answer 11

http://www.khanacademy.org/video/subtracting--fractions?playlist=Developmental+Math&exid=subtracting_fractions

Answer 12

jimmy, whats lcd? lowest common denominator?

Answer 13

yes

Answer 14

this is given me a real headache - cant understand how i got erroneous root

Answer 15

the anwser is x=0

Answer 16

yes - i know thats correct - i just trying to find out why my method also gave x=1

Answer 17

your solution is true jimmy.but then you must try your root in function.

Answer 18

there must be some reason for this

Answer 19

at x=1 function is undefined.

Answer 20

yea i know

Answer 21

- i just dont like mysteries...

Answer 22

ahh - i'd forgot a very important fact about multiplying the whole equation by the lcd!!
- this will sometimes give extraneous roots - when x = 1 you are effectively multiplying by zero.
you guys solved it differently - you simplified the lhs and left rhs alone - thats why you didn't get the extra root

Answer 23

its ok when you multiply by a constant - it only applies when you you multiply by a variable

Answer 24

hey jimmy, so which method is better? :)