Solving an equation with respect to y, where y is twice [hard]

1. Jan 26, 2014

Science4ver

1. The problem statement, all variables and given/known data

Given the following equation

X(y) = y/(t*(sqrt(b^2+y^2)) - 1/p

How would I go solving that equation X(y) = 0 with respect to y?

3. The attempt at a solution

I can choose a commen dominator called p*t*(b^2+y^2)

But I when end up with

((y*p - t*p*(sqrt(y^2+b^2))/(p*t(b^2+y^2)) = 0

How would you guys surgest I proceed from where in order to isolate y?

2. Jan 26, 2014

Dick

Now that you have it over a common denominator, if f(y)/g(y)=0, then f(y)=0.

3. Jan 26, 2014

Science4ver

That implies that I need to solve the equation

(y*p - t*p*(sqrt(y^2+b^2)) = 0

Which allows me to arrive at the solution (using my graphical calculator)

y = -b * p *(sqrt(-1/(p^2-1)

But solution is suppose to be:

y = b/(sqrt(p/t +1) * sqrt(p/t-1))

I can't quite comprehend which step I need to use arrive at that solution. But cause to the best of my knowledge there aren't any variables which I substitute in order to arrive at that solution.

Any idears?

4. Jan 26, 2014

Dick

You aren't going to get there by being sloppy and relying on a calculator.

There's an extra 'p' in that equation.

Extra 'p' here also, and what happened to the 't'? I think you should fix those up and figure out how your calculator found that expression before continuing.

5. Jan 26, 2014

Science4ver

You are right. I see now the equation I am suppose to solve is y/(t*(sqrt(y^2+b^2))-1 = 0

Is it correct now?

6. Jan 26, 2014

Dick

No, the p is gone. Your first step of putting it over a common denominator is wrong. It happened when you expressed the term 1/p with that common denominator.

Last edited: Jan 26, 2014
7. Jan 26, 2014

Science4ver

So I am lost here :( What would you surgest I do as a first step with original equation? I can add 1/p on both sides I see that and next add the denominator on both sides. But what next?

8. Jan 26, 2014

Dick

The strategy is to do what you were doing. Express X(y) over a common denominator. But do it right.

9. Jan 26, 2014

Science4ver

I get that now but the only commen denominator I can deduce is my pee size brain is :)

p*t*(b^2+y^2))

But what I get from what you are saying is that denominator is wrong?

10. Jan 26, 2014

Dick

I would use p*t*sqrt(b^2+y^2) as a common denominator. Actually that's what I thought you were using and just forgetting the 'sqrt' part. If you have two fraction a/b-c/d you can always use c*d as a common denominator. What do you get expressing a/b-c/d over a common denominator?

11. Jan 26, 2014

Science4ver

Okay, Glad I am not totally stupid then :)

So any if I take the original equation : y/(t*(sqrt(b^2+y^2)) - 1/p

and use the commen denominator p*t*sqrt(b^2+y^2)

I arrive at

py/(p*t*sqrt(b^2+y^2) - (t*sqrt(b^2+y^2)/(p*t*sqrt(b^2+y^2) =

y/(t*sqrt(b^2+y^2)) - 1 = 0 ?

But if that the right cause of action I still have y twice with the term y^2 and y, and no matter what I do I can't arrive at the result that

y = b/(sqrt(p/t +1) * sqrt(p/t-1)) :( So I understand I must be doing something wrong and there possibly a term which can be rewritten to arrive that the right result for y, but I cant see which one :(

12. Jan 26, 2014

Dick

Sorry, I've been assuming that some of the things you were writing were typos, when they are actually algebra mistakes. I revised the post you quoted. I'm asking you to show me how to put a/b-c/d over a common denominator.

13. Jan 26, 2014

haruspex

You confused me dropping the '0' at the end of the first of those two lines.
You have dropped a p in getting to the last line.
Don't collect everything over on the left like that. Leave it as py = (t*sqrt(b^2+y^2)).
You next step is to get rid of the square root. How can you do that?

14. Jan 26, 2014

Science4ver

But regarding the other expression I attempted now 10 times if I take my equation

X(y) = y/(t*(sqrt(b^2+y^2)) - 1/p = 0

and use the commen denominator (p*t(b^2+y^2))

I end up with the expression:

py/(pt*sqrt(b^2+y^2)) - (t(b^2+y^2))
/(p*t(b^2+y^2)) = ?

I must have messed up somewhere :( ?

15. Jan 26, 2014

Dick

a/b-c/d=(ad-bc)/bd is correct. You are messing up the second part. Actually, looking at it, it's not messed up - but you don't have a common denominator. Just apply the correct pattern with a=y, b=t*(sqrt(b^2+y^2)), c=1 and d=p.

Last edited: Jan 26, 2014
16. Jan 26, 2014

Ray Vickson

$$\frac{y}{t\sqrt{b^2+y^2}}-\frac{1}{p} = 0 \Rightarrow \frac{y}{t\sqrt{b^2+y^2}} = \frac{1}{p}$$
Square both sides to get
$$\frac{y^2}{t^2(b^2+y^2)} = \frac{1}{p^2}$$
Solve for $y^2$; in other words, let $y^2 = z$ and solve for $z$ from the simple equation
$$\frac{z}{b^2 t^2 + t^2 z} = \frac{1}{p^2}$$

17. Jan 26, 2014

Dick

Yes, I know there is an easier way to attack this. You can also do it by putting everything over a common denominator and going from there. I was trying to diagose what was going so wrong with the OP's attempt to put things over a common denominator. Sometimes you have to do that. Here, it's optional.

18. Jan 27, 2014

Science4ver

thank and then I should be able to arrive at the solution ?

But if I solve the rewritten equation with respect to z I get z=-b^2*t^2/(t^2-p^2)?

I am no closer to y = b/(sqrt(p/t +1) * sqrt(p/t-1)) which is suppose to be the solution for original equation.

Which is still no closer to y = b/(sqrt(p/t +1) * sqrt(p/t-1))

Last edited: Jan 27, 2014
19. Jan 27, 2014

haruspex

Yes you are... almost there in fact. Just a little more juggling.

20. Jan 27, 2014

Science4ver

I got it now :D

Last edited: Jan 27, 2014