Solving for y in a Textbook Problem: Understanding the Next Step | Helpful Tips

In summary, the (2 + sqrt(pi)) comes from adding sqrt(pi)y to both sides of the equation and then factoring out a y from the left hand side. This allows for easier solving for y in the next step.
  • #1
powp
91
0
Hello

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

[tex]2y = 360 \sqrt{\pi} - \sqrt{\pi}y [/tex]

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

[tex](2 + \sqrt{\pi})y = 360 \sqrt{\pi}[/tex]

Where does this [tex](2 + \sqrt{\pi})[/tex] come from?? where did the other y go and the [tex]-\sqrt{\pi}[/tex]?
 
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  • #2
The [tex](2 + \sqrt{\pi})y[/tex] came from adding [tex]\sqrt{\pi}y[/tex] to both sides and then factoring out a y from the left hand side. Have you not learned about collecting terms??

[tex]2y = 360 \sqrt{\pi} - \sqrt{\pi}y [/tex]

[tex]2y + \sqrt{\pi}y = 360\sqrt{\pi} + \sqrt{\pi}y - \sqrt{\pi}y[/tex]

[tex](2y + \sqrt{\pi})y = 360\sqrt{\pi}[/tex]
 
Last edited:
  • #3
They added [tex]\sqrt{\pi}y[/tex] to both sides.

Now on the left side of the equation we get [tex]2y + \sqrt{\pi}y[/tex]

and on the right we get [tex]360\sqrt{\pi}[/tex]

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

[tex]2y + \sqrt{\pi}y = y(2 + \sqrt{\pi})[/tex]

Now putting all of this info together we get

[tex](2 + \sqrt{\pi})y = 360 \sqrt{\pi}[/tex]

and [tex] y = \frac{(360 \sqrt{\pi})}{(2 + \sqrt{\pi})}[/tex]

Jameson
 
Last edited by a moderator:

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It is always best to double-check the answers you find online for your textbook questions. Make sure the source is reliable and credible, and compare the answers to your own understanding of the material. It is always better to confirm with your teacher or professor to ensure accuracy.

4. How can I improve my understanding of textbook answers?

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