Question

need help ( 1+3/x) /(1-1/x)+(1-1/x)/(1+2/x)=2

Answer 1

\[\frac{1+\frac{3}{x}}{1-\frac{1}{x}}+\frac{1-\frac{1}{x}}{1+\frac{2}{x}}=2 ?\]

Answer 2

Is that right @chogaz ? What I have above?

Answer 3

this no call underfine rather indetermine

Answer 4

dude call 1/x=y
then the problem becomes fairly easy looking
(1+3y)(1-y)+(1-y)(1+2y)=2 => (1-y)(2+5y)=2
=>5y^2+3y=0
=>y=-3/5 as y cant be zero
so x=-5/3

Answer 5

indeterminate

Answer 6

y=1/x (I'm using @vamgadu 's little sub-y :) )
\[\frac{1+3y}{1-y}+\frac{1-y}{1+2y}=2\]
By the way x cannot be 0
''''''''''''''''''''y cannot be 1 or -1/2

So anyways lets solve this puppy for y:

Clear the denominators by multiplying both sides by (1-y)(1+2y)


\[(1+3y)(1+2y)+(1-y)(1-y)=2(1-y)(1+2y)\]

Now we subtract 2(1-y)(1+2y) on both sides

\[(1+3y)(1+2y)+(1-y)(1-y)-2(1-y)(1+2y)=0\]

\[(1+2y)[(1+3y)-2(1-y)]+(1-y)(1-y)=0\]
\[(1+2y)(1+3y-2+2y)+(1-y-y+y^2)=0\]
\[(1+2y)(-1+5y)+(1-2y+y^2)=0\]
\[-1+5y-2y+10y^2+1-2y+y^2=0\]
Collect like terms:

\[11y^2+y-1+1=0\]

\[11y^2+y=0\]

\[y(11y+1)=0\]

\[y=0, y=\frac{-1}{11}\]

But y=1/x

But 1/x cannot be 0

But 1/x=-1/11 when x=-11
:)

Answer 7

\[\frac{1-\frac{1}{x}}{1+\frac{2}{x}}+\frac{1+\frac{3}{x}}{1-\frac{1}{x}}-2=0\]Combine the above fractions.\[\frac{x+11}{(x-1) (x+2)}=0 \]\[x=-11 \]

Answer 8

x=-11/5 .

Answer 9

You guys if you feel like you don't see the right explanation then please post the explanation. Thanks. Don't just post answers only.

Answer 10

\[\frac{1-\frac{1}{x}}{1+\frac{2}{x}}+\frac{1+\frac{3}{x}}{1-\frac{1}{x}}-2 \]\[\frac{1-\frac{1}{-11}}{1+\frac{2}{-11}}+\frac{1+\frac{3}{-11}}{1-\frac{1}{-11}}-2 \]\[\frac{\frac{12}{11}}{\frac{9}{11}}+\frac{\frac{8}{11}}{\frac{12}{11}}-2 \]\[\frac{4}{3}+\frac{2}{3}-2 \]\[2-2 \]

Answer 11

thank you so much