Question

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

Answer 1

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To Solve:-

$\rm{ \dfrac{x - 1}{x - 2} + \dfrac{x - 3}{x - 4} = \dfrac{10}{3} }x−2x−1+x−4x−3=310$

Required Answer:-

For solving for x, we have to take the LCM i.e. (x - 2)(x -4). Then,

$⇒ \rm{ \dfrac{(x - 1)(x - 4) + (x - 3)(x - 2)}{(x - 2)(x - 4)} = \dfrac{10}{3} }(x−2)(x−4)(x−1)(x−4)+(x−3)(x−2)=310$

Cross multiplying for easier calculation:

$⇒ \rm{3 \{(x - 1)(x - 4) + (x - 3)(x - 2) \}= 10(x - 2)( - 4)}3{(x−1)(x−4)+(x−3)(x−2)}=10(x−2)(−4)$

Multiplying the terms inside the curly brackets:

$⇒ \rm3 {\{ {x}^{2} - 5x + 4 + {x}^{2} - 5x + 6 \} = 10( {x}^{2} - 6x + 8}3{x2−5x+4+x2−5x+6}=10(x2−6x+8$

Opening the parentheses,

$⇒ \rm{3(2 {x}^{2} - 10x + 10) = 10{x}^{2} - 60x + 80}3(2x2−10x+10)=10x2−60x+80$

$⇒ \rm{6 {x}^{2} - 30x + 30 = 10 {x}^{2} - 60x + 80}6x2−30x+30=10x2−60x+80$

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

$⇒ \rm{3{x}^{2} - 15x+15 = 5{x}^2 - 30x+40}3x2−15x+15=5x2−30x+40$

Now shifting the terms to one side of the equation:

$⇒ \rm{5{x}^2 - 3{x}^2 - 30x+15x+40 - 15=0}5x2−3x2−30x+15x+40−15=0$

$⇒ \rm{2x^2 -15x+25=0}2x2−15x+25=0$

Now find the possible values of x by Middle term factorisation,

$⇒ \rm{2x^2 -10x-5x+25=0}2x2−10x−5x+25=0$

$⇒ \rm{2x(x - 5) -5(x-5) =0}2x(x−5)−5(x−5)=0$

$⇒ \rm{(2x-5)(x-5)=0}(2x−5)(x−5)=0$

Equating to 0, We get

$⇒ \rm{x = \dfrac{5}{2} \: or \: 5}x=25 or 5$

Hence The value of x after solving: 5/2 or 5.

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Answer 2

Answer:

refer to the attachment