1.

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

Answer»

To Solve:-

\rm{ \dfrac{x - 1}{x - 2} + \dfrac{x - 3}{x - <klux>4</klux>} = \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


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