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# Solve the Following Quadratic Equations by Factorization: - Mathematics

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#### Question

Solve the following quadratic equations by factorization:

$\frac{x - 2}{x - 3} + \frac{x - 4}{x - 5} = \frac{10}{3}; x \neq 3, 5$

#### Solution

$\frac{x - 2}{x - 3} + \frac{x - 4}{x - 5} = \frac{10}{3}$

$\Rightarrow \frac{x - 2}{x - 3} - \frac{10}{3} = - \frac{x - 4}{x - 5}$

$\Rightarrow \frac{3\left( x - 2 \right) - 10\left( x - 3 \right)}{3\left( x - 3 \right)} = - \frac{x - 4}{x - 5}$

$\Rightarrow \frac{3x - 6 - 10x + 30}{3x - 9} = - \frac{x - 4}{x - 5}$

$\Rightarrow - \frac{7x - 24}{3x - 9} = - \frac{x - 4}{x - 5}$

$\Rightarrow \left( 7x - 24 \right)\left( x - 5 \right) = \left( 3x - 9 \right)\left( x - 4 \right)$

$\Rightarrow 7 x^2 - 59x + 120 = 3 x^2 - 21x + 36$

$\Rightarrow 4 x^2 - 38x + 84 = 0$

$\Rightarrow 2 x^2 - 19x + 42 = 0$

$\Rightarrow 2 x^2 - 12x - 7x + 42 = 0$

$\Rightarrow 2x(x - 6) - 7(x - 6) = 0$

$\Rightarrow (2x - 7)(x - 6) = 0$

$\Rightarrow 2x - 7 = 0 \text { or } x - 6 = 0$

$\Rightarrow x = \frac{7}{2} \text { or } x = 6$

Hence, the factors are 6 and $\frac{7}{2}$.

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