Question

2y/y-4+2y-5/y-3=25/3 and it is solve plz reply fast plzzzz

Answer 1

Answer is in the attachment

Answer 2

y = 40/13; 6

Step-by-step explanation:

$\frac{2y}{y - 4} + \frac{2y - 5}{y - 3} = \frac{25}{3}$

Taking L.C.M.

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

$\frac{2y^{2} - 6y + 2y^{2} -5y -8y + 20 }{y^{2} - 3y -4y + 12} = \frac{25}{3}$

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

$13y^{2} - 118y + 240 = 0$

$13y^{2} - 78y - 40y + 240 = 0$

13y (y-6) - 40(y-6) = 0

(13y - 40) (y - 6) = 0

y = 40/13; 6

Learn more: Quadratic Equation

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