Question
Solve the following quadratic equations by factorization:
$\frac{\text{x}-2}{\text{x}-3}+\frac{\text{x}-4}{\text{x}-5}=\frac{10}{3},$ $\text{x}\neq3,5$

Answer

$\frac{\text{x}-2}{\text{x}-3}+\frac{\text{x}-4}{\text{x}-5}=\frac{10}{3}$
$\Rightarrow\frac{\text{x}-2}{\text{x}-3}-\frac{10}{3}=-\frac{\text{x}-4}{\text{x}-5}$
$\Rightarrow\frac{3(\text{x}-2)-10(\text{x}-3)}{3(\text{x}-3)}=-\frac{\text{x}-4}{\text{x}-5}$
$\Rightarrow\frac{3\text{x}-6-10\text{x}+30}{3\text{x}-9}=-\frac{\text{x}-4}{\text{x}-5}$
$\Rightarrow-\frac{7\text{x}-24}{3\text{x}-9}=-\frac{\text{x}-4}{\text{x}-5}$
$\Rightarrow (7x - 24)(x - 5) = (3x - 9)(x - 4)$
$\Rightarrow 7x^2 - 59x + 120 = 3x^2 - 21x + 36$
$\Rightarrow 4x^2 - 38x + 84 = 0$
$\Rightarrow 2x^2 - 19x + 42 = 0$
$\Rightarrow 2x^2 - 12x - 7x + 42 = 0$
$\Rightarrow 2x(x - 6) - 7(x - 6) = 0$
$\Rightarrow (2x - 7)(x - 6) = 0$
$\Rightarrow 2x - 7 = 0 or x - 6 = 0$
$\Rightarrow\text{x}=\frac{7}{2}$ or x = 6
Hence, the factors are 6 and $\frac{7}{2}$

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