Where does (2 + √π)y come from in this algebra step?

[powp]

Hello

I am doing this problem in my textbook and I am not sure what is happing in this one step.

$$2y = 360 \sqrt{\pi} - \sqrt{\pi}y$$

Trying to solve for y and this is what they show as the next step

$$(2 + \sqrt{\pi})y = 360 \sqrt{\pi}$$

Where does this $$(2 + \sqrt{\pi})$$ come from?? where did the other y go and the $$-\sqrt{\pi}$$?

 

[Nylex]

The $$(2 + \sqrt{\pi})y$$ came from adding $$\sqrt{\pi}y$$ to both sides and then factoring out a y from the left hand side. Have you not learned about collecting terms??

$$2y = 360 \sqrt{\pi} - \sqrt{\pi}y$$

$$2y + \sqrt{\pi}y = 360\sqrt{\pi} + \sqrt{\pi}y - \sqrt{\pi}y$$

$$(2y + \sqrt{\pi})y = 360\sqrt{\pi}$$

 

[Jameson]

They added $$\sqrt{\pi}y$$ to both sides.

Now on the left side of the equation we get $$2y + \sqrt{\pi}y$$

and on the right we get $$360\sqrt{\pi}$$

Now on the left side, they factored out the y so you can solve for it.

$$2y + \sqrt{\pi}y = y(2 + \sqrt{\pi})$$

Now putting all of this info together we get

$$(2 + \sqrt{\pi})y = 360 \sqrt{\pi}$$

and $$y = \frac{(360 \sqrt{\pi})}{(2 + \sqrt{\pi})}$$

Jameson