Question

(2x/x-5)²+10x/x-5-24=0

Answer 1

$let \: \: \frac{2x}{x - 5} = y \\ so { (\frac{2x}{x - 5}) }^{2} + \frac{10x}{x - 5} - 24 = 0 \\ {y}^{2} + 5y - 24 = 0 \\ {y}^{2} + 8y - 3y - 24 = 0 \\ y(y + 8) - 3(y + 8) = 0 \\ (y - 3)(y + 8) = 0 \\ y = 3 \: or \: - 8 \\ \\ now \: \: \frac{2x}{x - 5} = y \: \\ \frac{2x}{x - 5} = 3 \: or \: \frac{2x}{x - 5} = - 8 \\ 2x = 3x - 15 \: or \: 2x = - 8x + 40 \\ x = 15 \: or \: x = 4$

Answer 2

So,The value of $x$ are $4$ and $15$

Step-by-step explanation:

Given;

$(\frac{2x}{x-5} )^{2} +\frac{10x}{x-5} -24=0$

Let,

$\frac{2x}{x-5}=y$

Then,

$y^{2} +5y-24=0$

⇒$y^{2}+8y-3y-24=0$

⇒$y(y+8)-3(y+8)=0$

⇒$(y+8)(y-3)=0$

⇒$(y+8)=0$ or $(y-3)=0$

∴$y=-8$ or $y=3$

When,

$y=-8$ then,

$\frac{2x}{x-5}=-8$

⇒$2x=-8x+40$

⇒$10x=40$

∴$x=4$

And when,

$y=3$ then,

$\frac{2x}{x-5}=3$

⇒$2x=3x-15$

∴$x=15$

∴ The value of $x$ are $4$ and $15$