Question: Solve $\dfrac{y-25}{5+\sqrt{y}} = 4 + \dfrac{\sqrt{y} -5 }{5}$

Solution:
Given,

$\dfrac{y-25}{5+\sqrt{y}} = 4 + \dfrac{\sqrt{y} -5 }{5}$

$or, \dfrac{(\sqrt{y})^2 - 5^2}{\sqrt{y} + 5} = 4 + \dfrac{\sqrt{y} -5 }{5}$

$or, \dfrac{(\sqrt{y} +5)(\sqrt{x} -5)}{\sqrt{y} + 5} = 4 + \dfrac{\sqrt{y} -5 }{5}$

$or, \sqrt{y} - 5 = 4+ \dfrac{\sqrt{y} -5 }{5}$

$or, \sqrt{y} -5 - \dfrac{\sqrt{y} -5 }{5} = 4$

$or, \dfrac{5(\sqrt{y}-5) - (\sqrt{y} - 5)}{5} = 4$

$or, 5\sqrt{y} - 25 - \sqrt{y} + 5 = 4*5$

$or, 4\sqrt{y} -20 = 20$

$or, 4\sqrt{y} = 20 +20$

$or, 4\sqrt{y} = 40$

$or, \sqrt{y} = \dfrac{40}{4}$

$or, \sqrt{y} = 10$
squaring both sides

$or, (\sqrt{y})^2 = 10^2$

$\therefore y = 100$