Question

Solve: (x-1)/(x-2) + (x-3)(x-4) = 3 1/3 (mixed fraction]
​

Answer 1

Answer:

$\text{ Given} = \dfrac{(x-1)}{(x-2)} + \dfrac{(x-3)}{(x-4)} = 3\:\dfrac{1}{3}$

Converting the mixed fraction to improper fraction form, we get:

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

Taking LCM on the RHS, we get:

$\implies \dfrac{(x-1)(x-4) + (x-3)(x-2)}{(x-2)(x-4)} = \dfrac{10}{3}\\\\\\\implies \dfrac{(x^2 -4x - x + 4) + (x^2-2x-3x+6)}{(x^2 - 4x - 2x + 8)} = \dfrac{10}{3}\\\\\\\implies \dfrac{(x^2 - 5x + 4 + x^2 - 5x + 6)}{(x^2 - 6x + 8)} = \dfrac{10}{3}\\\\\\\implies \dfrac{(2x^2 -10x + 10)}{(x^2 - 6x +8)} = \dfrac{10}{3}\\\\\\\text{Cross multiplying we get:}$

$\implies 3 ( 2x^2 - 10x + 10 ) = 10 ( x^2 - 6x + 8 )\\\\\\\implies 6x^2 - 30x + 30 = 10x^2 - 60x + 80\\\\\\\implies 10x^2 - 6x^2 -60x + 30x + 80 - 30 = 0\\\\\\\implies 4x^2 - 30x + 50 = 0\\\\\\\implies 4x^2 - 20x - 10x + 50 = 0\\\\\\\implies 4x ( x - 5 ) - 10 ( x - 5 ) = 0\\\\\\\implies ( 4x - 10 ) ( x - 5 ) = 0\\\\\\\boxed{\bf{\implies x = \dfrac{10}{4} = \dfrac{5}{2} \:\:(and)\;\: x = 5}}$

Answer 2

$[tex]\text{ Given} = \dfrac{(x-1)}{(x-2)} + \dfrac{(x-3)}{(x-4)} = 3\:\dfrac{1}{3} Given= </p><p>(x−2)</p><p>(x−1)</p><p> </p><p> + </p><p>(x−4)</p><p>(x−3)</p><p> </p><p> =3 </p><p>3</p><p>1</p><p>$

[/tex]

$[tex]</p><p>Converting \: the \: mixed \: fraction \: to \: improper \: fraction \: form, \: we \: get:</p><p></p><p>\implies \dfrac{(x-1)}{(x-2)} + \dfrac{(x-3)}{(x-4)} = \dfrac{10}{3}⟹ </p><p>(x−2)</p><p>(x−1)</p><p> </p><p> + </p><p>(x−4)</p><p>(x−3)</p><p> </p><p> = </p><p>3</p><p>10</p><p>$

[/tex]

$\begin{gathered}\implies \dfrac{(x-1)(x-4) + (x-3)(x-2)}{(x-2)(x-4)} = \dfrac{10}{3}\\\\\\\implies \dfrac{(x^2 -4x - x + 4) + (x^2-2x-3x+6)}{(x^2 - 4x - 2x + 8)} = \dfrac{10}{3}\\\\\\\implies \dfrac{(x^2 - 5x + 4 + x^2 - 5x + 6)}{(x^2 - 6x + 8)} = \dfrac{10}{3}\\\\\\\implies \dfrac{(2x^2 -10x + 10)}{(x^2 - 6x +8)} = \dfrac{10}{3}\\\\\\\text{Cross multiplying we get:}\end{gathered} </p><p>⟹ </p><p>(x−2)(x−4)</p><p>(x−1)(x−4)+(x−3)(x−2)</p><p>$

$</p><p> = </p><p>3</p><p>10</p><p> </p><p> </p><p>⟹ </p><p>(x </p><p>2</p><p> −4x−2x+8)</p><p>(x </p><p>2</p><p> −4x−x+4)+(x </p><p>2</p><p> −2x−3x+6)</p><p> </p><p> = </p><p>3</p><p>10</p><p> </p><p> </p><p>⟹ </p><p>(x </p><p>2</p><p> −6x+8)</p><p>(x </p><p>2</p><p> −5x+4+x </p><p>2</p><p> −5x+6)</p><p> </p><p> = </p><p>3</p><p>10</p><p> </p><p> </p><p>⟹ </p><p>(x </p><p>2</p><p> −6x+8)</p><p>(2x </p><p>2</p><p> −10x+10)</p><p> </p><p> = </p><p>3</p><p>10</p><p>$

$[tex]Taking LCM on the RHS, we get:</p><p></p><p>\begin{gathered}\implies \dfrac{(x-1)(x-4) + (x-3)(x-2)}{(x-2)(x-4)} = \dfrac{10}{3}\\\\\\\implies \dfrac{(x^2 -4x - x + 4) + (x^2-2x-3x+6)}{(x^2 - 4x - 2x + 8)} = \dfrac{10}{3}\\\\\\\implies \dfrac{(x^2 - 5x + 4 + x^2 - 5x + 6)}{(x^2 - 6x + 8)} = \dfrac{10}{3}\\\\\\\implies \dfrac{(2x^2 -10x + 10)}{(x^2 - 6x +8)} = \dfrac{10}{3}\\\\\\\text{Cross multiplying we get:}\end{gathered} </p><p>⟹ </p><p>(x−2)(x−4)</p><p>(x−1)(x−4)+(x−3)(x−2)</p><p> </p><p> = </p><p>3</p><p>10$

$⟹ </p><p>(x </p><p>2</p><p> −4x−2x+8)</p><p>(x </p><p>2</p><p> −4x−x+4)+(x </p><p>2</p><p> −2x−3x+6)</p><p> </p><p> = </p><p>3</p><p>10</p><p> </p><p> </p><p>⟹ </p><p>(x </p><p>2</p><p> −6x+8)</p><p>(x </p><p>2</p><p> −5x+4+x </p><p>2</p><p> −5x+6)</p><p> </p><p> = </p><p>3</p><p>10</p><p> </p><p> </p><p>⟹ </p><p>(x </p><p>2</p><p> −6x+8)</p><p>(2x </p><p>2</p><p> −10x+10)</p><p> </p><p> = </p><p>3</p><p>10</p><p> </p><p> </p><p>Cross multiplying we get:</p><p> .$

[/tex]