Question

What is wrong?
From \(1/x -x) = (1/2x)+ (1/2)x\)

Answer 1

How to solve?

Answer 2

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

Is that the right equation?

Answer 3

yes

Answer 4

Oh it is? I thought that first fraction would be 1/(2x)

Answer 5

oh, @johnweldon1993 is correct. I am sorry @wmj259

Answer 6

So the x is in the denominator.

Answer 7

so it is:
\[\frac{1}{x}-x=\frac{1}{2x}+\frac{1}{2}x\]
is this an equation you are suppose to be solving?

Answer 8

@freckles, that looks correct.

Answer 9

yes, it is

Answer 10

So I get \(\large \pm \frac{1}{\sqrt{3}}\) solving for 'x'

Answer 11

did you try to get rid the fractions first
multiply both sides by 2x

Answer 12

*I know no direct answers...but theres not much to do but simplify

Answer 13

and it is also @Loser66
I think she knows how to solve this...

Answer 14

You cant multiply both sides by 2x since the fractions denominators aren't the same. I believe you first have to get a common denominator on the right side first.

Answer 15

1st statement should be x cannot equal zero

Answer 16

Feel free to give me the direct answer. hehehe...

Answer 17

Lol but it's easy to do

\[\large \frac{1}{x} - x = \frac{1}{2x} + \frac{1}{2}x\]
\[\large \frac{1}{x} - \frac{x^2}{x} = \frac{1}{2x} + \frac{x^2}{2x}\]
\[\large \frac{1 - x^2}{x} = \frac{1 + x^2}{2x}\]
\[\large 2x(1 - x^2) = x(1 + x^2)\]
\[\large 2x - 2x^3 = x + x^3\]
\[\large x = 3x^3\]
\[\large 3x^3 - x = 0\]
\[\large x(3x^2 - 1) = 0\]

Then just

\[\large 3x^2 - 1 = 0\]

Unless I'm doing this wrong lol

Answer 18

Your solution is perfect. Thank you so much.

Answer 19

I thought it was that the 2nd to last step, you can solve it so either x's = 0 so...... x=0 or x=+-sqrt(1/3)

Answer 20

\[x=0 , x=\sqrt{1/3}\]

Answer 21

Well x cannot = 0 is the thing

As you will not get an answer in the original problem

*1/x, 1/(2x)

Answer 22

Okay so plus or minus sqrt(1/3)