The equation (2x+3)/(x+2)=(x-1)/(x-2) has to be solved for x.

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

=> (2x+3)(x-2)=(x-1)(x+2)

=> 2x^2 + 3x - 4x - 6 = x^2 + 2x - x - 2

=> x^2 - 2x - 4 = 0

x1 = 2/2 + sqrt( 4 + 16)/2

=> 1 + sqrt 20 / 2

=> 1 + sqrt 5

x2 = 1 - sqrt 5

**The solutions of the equation are 1 + sqrt 5 and 1 - sqrt 5**

We'll impose constraints of existence of the given ratios: their denominators must be different from zero.

The range of admissible values for x is R - {-2 ; +2}.

We'll solve the equation by cross multiplying:

(2x+3)(x-2) = (x+2)(x-1)

We'll remove the brackets:

2x^2 - 4x + 3x - 6 = x^2 - x + 2x - 2

We'll move all terms to the left side:

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

x^2 - 2x - 4 = 0

We'll apply quadratic formula:

x1 = [2+sqrt(4 + 16)]/2

x1 = [2+2sqrt5]/2

**x1 = 1 + sqrt5**

**x2 = 1 - sqrt5**

**Since the solutions are real and different from the excepted values, we'll validate them.**