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Question Number 173620 by eiverzurita last updated on 14/Jul/22

	Answered by FelipeLz last updated on 14/Jul/22

	$2 b^{2} - 2 b x + 2 a^{2} x + a b x + {a b}^{2} = 2 a b x + 2 a^{2} b$ $2 b^{2} - 2 b x + 2 a^{2} x + {a b}^{2} = a b x + 2 a^{2} b$ $2 b (b - x) - 2 a^{2} (b - x) + a b (b - x) = 0$ $(b - x) (2 b - 2 a^{2} + a b) = 0$ $x = b$
	Commented by Tawa11 last updated on 15/Jul/22

	$Great sirs$
	Answered by Rasheed.Sindhi last updated on 15/Jul/22

	$AnOther Way$ $\frac{b - x}{a} + \frac{a x}{b} + \frac{x + b}{2} = x + a; x = ?$ $\frac{b}{a} - \frac{x}{a} + \frac{a x}{b} + \frac{x}{2} + \frac{b}{2} - x = a$ $- \frac{x}{a} + \frac{a x}{b} + \frac{x}{2} - x = a - \frac{b}{a} - \frac{b}{2}$ $x (- \frac{1}{a} + \frac{a}{b} + \frac{1}{2} - 1) = \frac{2 a^{2} - 2 b - a b}{2 a}$ $x (\frac{- 2 b + 2 a^{2} + a b - 2 a b}{2 a b}) = \frac{2 a^{2} - 2 b - a b}{2 a}$ $x = \frac{2 a^{2} - 2 b - a b}{2 a} \times \frac{2 a b}{- 2 b + 2 a^{2} - a b}$ $= \frac{b (2 a^{2} - 2 b - a b)}{- 2 b + 2 a^{2} - a b} = b$
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