solve for the positive value of x?

Mutiply both sides by 5x,
x(x-4) = 5
x^2-4x-5 = (x-5)(x+1) = 0
x = 5, the positive root.

(x - 4)/5 = 1/x
x² - 4x = 5
x² - 2x = 5 + (- 2)²
x² - 2x = 5 + 4
(x - 2)² = 9
x - 2 = 3

x = 3 + 2, x = 5
x = - 3 + 2, x = - 1

Answer: x = 5, - 1

Proof (x = 5):
(5 - 4)/5 = 1/5
1/5 = 1/5

Proof (x = - 1):
(- 1 - 4)/5 = 1/ - 1
- 5/5 = - 1
- 1 = - 1

These questions are always suspect due to lack of brackets.

Do you actually mean :-

(x - 4) / 5 = 1 / x__________????????

x² - 4x = 5

x² - 4x - 5 = 0

(x - 5)(x + 1) = 0

x = 5

PS
Should not have to GUESS the question !

(x - 4)/5 = 1/x
x - 4 = 5/x
x(x - 4) = 5
x^2 - 4x = 5
x^2 - 4x - 5 = 0
x^2 + x - 5x - 5 = 0
(x^2 + x) - (5x + 5) = 0
x(x + 1) - 5(x + 1) = 0
(x + 1)(x - 5) = 0

x + 1 = 0
x = -1

x - 5 = 0
x = 5

∴ x = -1 , 5

x-4/5 = 1/x
LCM is 5x

x(x-4) = 5
distributive property

x^2 - 4x - 5 = 0
factor

(x-5)(x+1) = 0

x= -1 and 5

That's it Maths rock

this equation has no solution:

http://www.mathway.com/answer.aspx?p=trig?p=x-4SMB105SMB011SMB10xSMB15?p=1?p=?p=?p=?p=?p=?p=?p=0?p=?p=

x-4/5=1/x

multiply the whole equation by "x"
x^2 - (4/5)*x = 1

transpose "1"
x^2 - (4/5)*x -1 = 0

solving equation:
2/5+1/5*29^(1/2)
2/5-1/5*29^(1/2)

positive value of x:
2/5+1/5*29^(1/2) = 1.4770



THERE ARE DIFFERENT BETWEEN
x-4/5=1/x
AND
(x-4)/5=1/x

First step:
Multiply everything by x
= X^2 - (4x/5) = 1
Second Step:
Transpose to a quadratic
= X^2 - (4x/5) - 1 = 0
Third Step:
Multiply by 5
= 5x^2 - 4x - 5 = 0
Fourth step:
Solve for x

We want to get everything on the left side of the equation...

(x - 4) / 5 = 1 / x

Multiply throughout by 5

(x - 4) = 5 / x

Multiply throughout by x

x (x - 4) = 5

Expand the brackets

x^2 - 4x = 5

Bring the 5 to the left side

x^2 - 4x -5 = 0

Factor it

(x + 1)(x - 5) = 0

x = -1 or x = 5

Multiply by x
Then line two is Xsquared -0.8X =1
Then third line is Xsquared -0.8X - 1 = 0

You've got a formual for that?

All I did now, was to manipulate it into the style the formula can solve, you noticed this?