Question

2x/x-4+2x-5/x-3=25/3

Answer 1

I hope this will help you. ....

Answer 2

$Given \:\frac{2x}{(x-4)} + \frac{2x-5}{(x-3)} = \frac{25}{3}$

$\implies \frac{2x(x-3)+(2x-5)(x-4)}{(x-4)(x-3)} = \frac{25}{3}$

$\implies \frac{2x^{2}-6x+2x^{2}-5x-8x+20}{x^{2}-7x+12} = \frac{25}{3}$

$\implies \frac{4x^{2}-19x+20}{x^{2}-7x+12} = \frac{25}{3}$

$\implies 3(4x^{2}-19x+20) = 25(x^{2}-7x+12)$

$\implies 12x^{2} - 57x + 60 = 25x^{2}-175x+300$

$\implies 0 = 25x^{2}-175x+300 - 12x^{2} + 57x - 60$

$\implies 13x^{2}-118x+240 = 0$

/* Compare this equation with ax²+bx+c=0 , we get*/

$a = 13 ,\: b = -118 \:and \: c = 240$

$Discreminant (D) = b^{2} - 4ac \\= (-118)^{2} - 4\times 13 \times 240 \\= 13924 - 12480\\= 1444$

$\underline { \blue { Using \: Quadratic\: Formula:}}$

$x = \frac{-b\pm \sqrt{D}}{2a}$

$= \frac{-(-118)\pm \sqrt{1444}}{2\times 13} \\= \frac{118\pm 38}{26}$

$\implies x = \frac{118+38}{26} \:Or \:x = \frac{118-38}{26}$

$\implies x = \frac{156}{26} \:Or \:x = \frac{80}{26}$

$\implies x = 6 \:Or \:x = \frac{40}{13}$

Therefore.,

$\green { x = 6 \:Or \:x = \frac{40}{13}}$

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