Question

help!!

Answer 1

Solve for x

Answer 2

\[ \frac{ 8 }{ x^2-1 }=\frac{ 3 }{ x+1 }+\frac{ 4 }{ x-1 }\]

Answer 3

bro,
write the left hand side under a common denominator
it will be
(x + 1)*(x -1 ) which is equivalent to (x^2 -1)
so u can cut out the denominators of both fractions and u could get a equation like this
3( - - ) + 4( - - ) = 8

Then I think u won't have any problems in solving that ..... : -)

Answer 4

uhh, dont get it lol

Answer 5

let's take the left hand side
\[\frac{ 3 }{x +1 } +\frac{ 4 }{ x - 1 }\]

and now get a common denominator for that part
can u do that bro .... wt will it be ?

Answer 6

well, x+1 and x -1 is the same... i think?

Answer 7

no ,
x + 1 and x -1 r not the same

Answer 8

u cant get x + 1 by using x - 1 , even though u could take a negative sign as a common factor in (x - 1) and write it as -(1 - x ) still the r 2 different values

Answer 9

this is an example
if u have
1/2 + 1/5
u will write it as
5/10 + 2/10
and
( 5 + 2)/10

use the same theory here it's crucial for the next step

Answer 10

r u still messed up ?

Answer 11

pretty much :/

1

1/7

7

no solution



those are the answers...

Answer 12

Sorry. "those r the answers" for wt ?

Answer 13

the question

Answer 14

Just multiply the whole eq'n by \[x^2 - 1\]
Then it will appear like this
\[8 = 3 (x-1) + 4 (x+1)\]
Applying rules in axioms and distribution
you will have
8 = 3x - 3 + 4x + 4
Therefore,
8 = 7x +1
7 = 7x
And x is now? :) @charlieberzak

Answer 15

so the answer is 1 :)