Question

solve for x and y , x-1 / x-2 + x-3 / x-4 = 10/3.n.​

Answer 1

$\frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = \frac{10}{3}$

$\frac{ (x - 1)(x - 4) + (x - 2)(x - 3)}{(x - 2)(x - 4)} = \frac{10}{3}$

$\frac{( {x}^{2} - 4x - x + 4) + ( {x}^{2} - 3x - 2x + 6)}{ {x}^{2} - 4x - 2x + 8 } = \frac{10}{3}$

$\frac{ {x}^{2} - 4x - x + 4 + {x}^{2} - 3x - 2x + 6 }{ {x}^{2} - 6x + 8 } = \frac{10}{3}$

$\frac{2 {x}^{2} -10x + 10}{ {x}^{2} - 6x + 8} = \frac{10}{3}$

$\frac{ {x}^{2} - 5x + 5 }{ {x}^{2} - 3x + 4 } = \frac{10}{3}$

$3( {x}^{2} - 5x + 5) = 10( {x}^{2} - 3x + 4)$

$3 {x}^{2} - 15x + 15 = 10 {x}^{2} - 3x + 4$

${x}^{2} - 5x + 5 = 10 {x}^{2} - 3x + 4$

${x}^{2} - 10 {x}^{2} - 5x + 3x + 5 - 4 = 0$

${ - 9x}^{2} - 2x + 1 = 0$

$9{x}^{2} + 2x - 1 = 0$

Answer 2

Step-by-step explanation:

As given,

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

LCM 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 & cancellinf 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.

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