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Express the equation $\frac{x-2}{x-3} + \frac{x-4}{x-5} = \frac{10}{3}$; $(x\neq3,5)$ as a quadratic equation in standard form. Hence, find the roots of the equation so formed.
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$\frac{x-2}{x-3} + \frac{x-4}{x-5} = \frac{10}{3}$
$\frac{(x-2)(x-5)+(x-4)(x-3)}{(x-3)(x-5)} = \frac{10}{3}$
Simplifying, we get $2 x^2 - 19x + 42 = 0$
$\Rightarrow (x – 6)(2x – 7) = 0$
$\Rightarrow x = 6$ or $x = \frac{7}{2}$
$\frac{(x-2)(x-5)+(x-4)(x-3)}{(x-3)(x-5)} = \frac{10}{3}$
Simplifying, we get $2 x^2 - 19x + 42 = 0$
$\Rightarrow (x – 6)(2x – 7) = 0$
$\Rightarrow x = 6$ or $x = \frac{7}{2}$