Question

Solve for x: (x-1)/(x-2 )+( x-3)/ (x-4)=10/3​

Answer 1

As given,

To solve (x-1)/(x-2) + (x-3)/(x-4) =10/3 we have to ttake the LCM

L.C.M. of (x-2) & (x-4) = (x-2)(x-4)

Then,

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

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

= 3( 2x^2–10x+10) = 10(x^2–6x+8)

= 6x^2–30x +30 = 10x^2–60x+80

Taking 2 as a common from both sides & cancelling it, we have,

= 3x^2–15x+15 = 5x^2–30x+40

=5x^2–3x^2–30x+15x+40–15=0

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

Now, by the middle term splitting method,we have,

= 2x^2 -10x-5x+25=0

= 2x(x–5) -5(x-5) =0

= (2x-5)(x-5)=0

= 2x-5 = 0

So, x= 5/2

Or, x-5= 0

So, x=5

Hence, x=5 or 5/2

Answer 2

Answer:

As given,

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

L.C.M. of (x-2) & (x-4) = (x-2)(x-4)

Then,

=> [(x-1)(x-4)+(x-3)(x-2)]/(x-2)(x-4) = 10/3

=> 3(x^2 -5x +4 + x^2 -5x +6) = 10(x-2)(x-4)

=> 3( 2x^2–10x+10) = 10(x^2–6x+8)

=> 6x^2–30x +30 = 10x^2–60x+80

Taking 2 as a common from both sides & cancelling it, we have,

=> 3x^2–15x+15 = 5x^2–30x+40

=>5x^2–3x^2–30x+15x+40–15=0

=> 2x^2 -15x+25=0

Now, by the middle term splitting method,we have,

=> 2x^2 -10x-5x+25=0

=> 2x(x–5) -5(x-5) =0

=> (2x-5)(x-5)=0

=> 2x-5 = 0

So, x= 5/2

Or, x-5= 0

So, x=5

Hence, x=5 or 5/2 Ans