Question

x-2/x-3 + x-4/x-5 = 3⅓, x ≠ 3, 5

Lesson :- Quadratic equations​

Answer 1

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Answer 2

Answer

Given that,

$\sf \dashrightarrow \dfrac{x - 2}{x - 3} + \dfrac{x - 4}{x - 5} = \dfrac{10}{3}$

$\sf \dashrightarrow \dfrac{(x - 2)(x - 5) + (x - 4)(x - 3)}{(x - 3)(x - 5)} = \dfrac{10}{3}$

$\sf \dashrightarrow \dfrac{({x}^{2} - 7x + 10) + ({x}^{2} - 7x + 12)}{({x}^{2} - 8x + 15)} = \dfrac{10}{3}$

$\sf \dashrightarrow \dfrac{({2x}^{2} - 14x + 22)}{({x}^{2} - 8x + 15)} = \dfrac{10}{3}$

Cross multiply the fractions.

$\sf \dashrightarrow 3 \: ({2x}^{2} - 14x + 22) = 10 \: ({x}^{2} - 8x + 15)$

$\sf \dashrightarrow {6x}^{2} - 42x + 66 = {10x}^{2} - 80x + 150$

$\sf \dashrightarrow {4x}^{2} - 38x + 84 = 0$

$\sf \dashrightarrow {2x}^{2} - 19x + 42 = 0$

$\sf \dashrightarrow {2x}^{2} - 12x - 7x + 42 = 0$

$\sf \dashrightarrow 2x \: (x - 6) - 7 \: (x - 6) = 0$

$\sf \dashrightarrow (x - 6)(2x - 7) = 0$

$\sf \dashrightarrow x - 6 = 0 \: \: or \: \: 2x - 7 = 0$

$\sf \dashrightarrow x = 6 \: \: or \: \: x = \dfrac{7}{2}$

Thus, 6 and $\sf \dfrac{7}{2}$ are the roots of the given equation.