Question

Solve the following.
$\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3}(x\ne 2,4)$

Answer 1

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

$3\left[\right(x^2-5x+4)+(x^2-5x+6\left)\right]=10[x^2-6x+8]$

$3[2x^2-10x+10]=10x^2-60x+80$

$6x^2-30x+30=10x^2-60x+80$

$4x^2-30x+50=0.$

$2x^2-15x+25=0.$

$2x^2-10x-5x+25=0.$

$\Rightarrow 2x(x-5)-5(x-5)=0$

$(2x-5)(x-5)=0$

$(2x-5)=0$ or $x-5=0$

$x=5/2$ or $x=5$

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