Question

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

To Solve:-

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

Required Answer:-

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

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

Cross multiplying for easier calculation:

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

Multiplying the terms inside the curly brackets:

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

Opening the parentheses,

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

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

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

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

Now shifting the terms to one side of the equation:

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

⇒ $\rm{2x^2 -15x+25=0}$

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

⇒ $\rm{2x^2 -10x-5x+25=0}$

⇒ $\rm{2x(x - 5) -5(x-5) =0}$

⇒ $\rm{(2x-5)(x-5)=0}$

Equating to 0, We get

⇒ $\rm{x = \dfrac{5}{2} \: or \: 5}$

Hence:-

• The value of x after solving: 5/2 or 5