Positive value of x help !?

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

1st interpretation (my best shot):
(x - 4)/5 = 1/x
x(x - 4) = 5
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

2nd interpretation:
x - 4/5 = 1/x
(5x - 4)/5 = 1/x
x(5x - 4) = 5
5x² - 4x = 5
x² - 4/5x = 1
x² - 2/5x = 1 + (- 2/5)²
x² - 2/5x = 25/25 + 4/25
(x - 2/5)² = 29/25
x - 2/5 = 1.0770329

x = 1.0770329 + 0.4, x = 1.4770329
x = - 1.0770329 + 0.4, x = - 0.6770329

Answer: x = 1.4770329, - 0.6770329

Proof (x = 1.4770329):
1.4770329 - 0.8 = 1/1.4770329
0.6770329 = 0.6770329

Proof (x = - 0.6770329):
- 0.6770329 - 0.8 = 1/- 0.6770329
- 1.477033 = - 1.477033

These question worry me because I don`t trust them !

Is it:-

(i) x - (4/5) = 1 / x ?????
OR
(ii) (x - 4) / 5 = 1 / x ????

Will do both :-
(i)
x² - (4/5) x = 1
5x² - 4x - 5 = 0
x = [ 4 ± √ (16 + 100) ] / 10
x = [ 4 ± √ (116) ] / 10
x = [ 4 ± 2√ (29) ] / 10
x = [ 2 ± √ (29) ] / 5

(ii)
(x - 4) / 5 = 1 / x
x² - 4x = 5
x² - 4x - 5 = 0
(x - 5)(x + 1) = 0
x = 5 , x = - 1

TIP
Make 2009 the year of the brackets !
Good luck.

(x - 4)/5 = 1/x
x - 4 = 5(1/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 x not = 0

x^2-4x=5

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

cross multiply: x^2-4x=5
x^2-4x-5=0
x=(-4+-sqrt(16-20))/2
there is no positive value because the roots are imaginary

x-4/5 = 1/x then

x-1/x = 4/5

x^2/x - 1/x = 4/5

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

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

x^2-4/5x - 1 = 0 solve for x you get 2 values pick the positive one

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

x+1=0
x= -1 you reject because not positive
x-5=0

X = 5

5

If you substitute 5 in your equation you can see it works.

working
x-4/5 = 1/x [Now cross multiply]
x(x-4) = 5
x^2 - 4x - 5 = 0
(x+1) (x-5) =0

Since you want the positive value of x
take x-5 =0
So, x=5