#### Question

Solve for *x*

`(x-1)/(x-2)+(x-3)/(x-4)=3 1/3`; x ≠ 2, 4

#### Solution

We have been given,

`(x-1)/(x-2)+(x-3)/(x-4)=3 1/3`; x ≠ 2, 4

Now we solve the above equation as follows,

`((x-1)(x-4)+(x-3)(x-2))/((x-2)(x-4))=10/3`

`(x^2-5x+4+x^2-5x+6)/(x^2-6x+8)=10/3`

6x^{2} - 30x + 30 = 10x^{2} - 60x + 80

4x^{2} - 30x + 50 = 0

2x^{2} - 15x + 25 = 0

Now we also know that for an equation ax^{2} + bx + c = 0, the discriminant is given by the following equation:

D = b^{2} - 4ac

Now, according to the equation given to us, we have,a = 2, b = -15 and c = 25.

Therefore, the discriminant is given as,

D = (-15)^{2} - 4(2)(25)

= 225 - 200

= 25

Now, the roots of an equation is given by the following equation,

`x=(-b+-sqrtD)/(2a)`

Therefore, the roots of the equation are given as follows,

`x=(-(-15)+-sqrt25)/(2(2))`

`=(15+-5)/4`

Now we solve both cases for the two values of *x*. So, we have,

`x=(15+5)/4`

= 5

Also,

`x=(15-5)/4`

`=5/2`

Therefore, the value of `x = 5, 5/2`