Maths Help

[mshassy]

ela! rep + oni na machan!

 

[Tom Riddle]

thanks very much for the help machan. much much appreciated.

can i just clarify one thing

to get from this -6/(x-3)(x+3) > 0 to this 1/(x-3)(x+3) < 0 did u multiply both sides by-1/6 ?

Yes

 

[seyd2p]

1/(x+3) -1/(x-3) > 0

(x-3)-(x+3) > 0
__________
x^2-9

6 > 0 and x should not be +3 or -3
ans= x should not be +3 or -3

 

[~unknown~]

answer should be, -3 < x < 3

 

[flag au:]

1/(x+3)>1/(x-3)

=
x-3-(x+3)/(x-3)(x+3)>0------------uda kapenawa


-6/(x+3)(x-3)>0

then we can write (using basic concept )
(-) nesa > change wenawa)
(x+3)(x-3)<0------basic concept

then answere is

-3<x<3 (u can draw it like this)
Number line:

machan amaru math ? kiyapan.......help ekak dan nam