Can you resolve this equation : (2x-5)/(x-1)=(x-1)/(x+1) ?

first, multiply through the denominators to obtain:

(2x-5)(x+1)=(x-1)(x-1)

multiply each side:

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

collect terms:

x^2-x-6=0
(x-3)(x+2)=0

x=3, x=-2

(2x-5)/(x-1)=(x-1)/(x+1)

cross multiply

2x^2 - 3x - 5 = x^2 - 2x + 1
x^2 - x - 6 = 0
(x - 3) (x + 2) = 0

therefore,

x = 3; x = -2

Cross multiply, collect like terms and solve:

(2x - 5)(x + 1) = (x - 1)(x - 1)
2x^2 -3x - 5 = x^2 - 2x + 1
x^2 - x - 6 = 0
(x + 2)(x - 3) = 0
x = -2 or x = 3

(2x - 5) (x + 1) = (x - 1)(x - 1)

2x² - 3x - 5 = x² - 2x + 1

x² - x - 6 = 0

(x - 3)(x + 2) = 0

x = 3 , x = - 2

(2x - 5)/(x - 1) = (x - 1)/(x + 1)
(2x - 5)(x + 1)/(x - 1) = x - 1
(2x - 5)(x + 1) = (x - 1)(x - 1)
2x*x - 5*x + 2x*1 - 5*1 = x*x - 1*x - x*1 + 1*1
2x^2 - 5x + 2x - 5 = x^2 - x - x + 1
2x^2 - 3x - 5 = x^2 - 2x + 1
2x^2 - x^2 - 3x + 2x - 5 - 1 = 0
x^2 - x - 6 = 0
x^2 + 2x - 3x - 6 = 0
(x^2 + 2x) - (3x + 6) = 0
x(x + 2) - 3(x + 2) = 0
(x + 2)(x - 3) = 0

x + 2 = 0
x = -2

x - 3 = 0
x = 3

∴ x = -2 , 3

(2x-5)/(x-1)=(x-1)/(x+1)

(2x-5)/(x-1)=(x-1)/(x+1)

=> (2x-5)*(x+1) = (x-1)^2

=> 2x^2 - 3x - 5 = x^2 - 2x + 1

=> x^2 - x - 6 =0

=> x^2 -3x + 2x - 6 =0

=> x(x-3) + 2(x-3) = 0

=> (x-3)*(x+2) =0

Either x = 3 or x = -2

what you have to do is first multiply both sides by (x-1) to get rid of the denominator on the right, so you get:
(2x-5)=((x-1)((x-1))/(x+1)

next you want to get rid of the denominator on the left, so you multiply both sides by (x+1) which is:
(2x-5)(x+1)=(x-1)(x-1)
next just distribute everything:
2x^2-3x-5=x^2-2x+1

and now combine like terms:
x^2-x-6=0

now, factor:
(x-3)(x+2)

and your answer is:
x=3 or -2

(2X-5)/(X-1)=(X-1)/(X+1)
(X+1)(2X-5)=(X-1)(X-1)
2X^2 - 3X -5 = X^2 -2X +1
X^2 - X - 6 = 0
(X-3)(X+2) = 0
X= 3 OR -2

(2x-5)÷(x-1) = (x-1)÷(x+1)

Multiply both sides by x-1:

2x-5 = (x-1)²÷(x+1)

Multiply both sides by x+1:

(2x-5)(x+1) = (x-1)²

Multiply out both sides.

Simplify.

Use quadratics to finish.