Equation Equality: Demonstrating the Sameness of Two Equations

[forcefield]

Homework Statement

Demonstrate that two equations are same.

Homework Equations

$$\frac {1} {x+z} + \frac {1} {y+z} = \frac {1} {z} ~~~~~~~~(1)$$
$$xy = z^2 ~~~~~~~~~(2)$$

The Attempt at a Solution

I take z in terms of x and y from (2) and replace z in (1) by that to get:
$$\frac {\sqrt{xy}} {x + \sqrt{xy}} + \frac {\sqrt{xy}} {y + \sqrt{xy}} = 1 ~~~~~~~~~(3)$$
I guess I can interpret that as a sum of probability amplitudes where the unnormalized probabilities are x and y but I don't think that qualifies as a demonstration and I don't know how to rearrange (3) to make it any simpler from that.

I guess there is some mathematical method that can be used here but what is it ?

(This is from a physics text which I hope is a good enough excuse for not having a clue.)

 

[PeroK]

If $z^2 = xy$ it does not follow that $z = \sqrt{xy}$. Why not?

I would start with equation (1) here and avoid square roots.

 

[forcefield]

If $z^2 = xy$ it does not follow that $z = \sqrt{xy}$. Why not?

Right, z can be negative too.

I would start with equation (1) here and avoid square roots.

Don't you need to solve for x or y in (2) first then ?

 

[PeroK]

Right, z can be negative too.

Don't you need to solve for x or y in (2) first then ?

No, it's just a bit of algebra, actually. Easier than it looks.

 

[forcefield]

No, it's just a bit of algebra, actually. Easier than it looks.

Ok, I'm not sure what exactly "algebra" is, but I managed to rearrange (1) to get (2). Thanks.

 

[PeroK]

Ok, I'm not sure what exactly "algebra" is, but I managed to rearrange (1) to get (2). Thanks.

I think that counts as algebra!

 

[fresh_42]

$(2)$ has more solutions than $(1)$.

 

[Mark44]

Homework Statement

Demonstrate that two equations are same.

Homework Equations

$$\frac {1} {x+z} + \frac {1} {y+z} = \frac {1} {z} ~~~~~~~~(1)$$
$$xy = z^2 ~~~~~~~~~(2)$$

The two equations are not "the same," (not equivalent, meaning that they don't have the same set of solutions for x, y, and z).
The first equation is not defined if any of the denominators happens to be zero, which means that $x \ne -z$, and $y \ne -z$, and $z \ne 0$.
Some of the solutions of the second equation violate one or more of these restrictions.
Here are just a few solutions of the second equation that are not also solutions of the first equation:
x = 0, y = 0, z = 0 (violates $z \ne 0$)
x = 0, y = 1, z = 0 (ditto)
x = 1, y = 1, z = 1 (violates $y \ne -z$ and $x \ne -z$)

 

[forcefield]

$(2)$ has more solutions than $(1)$.

The two equations are not "the same," (not equivalent, meaning that they don't have the same set of solutions for x, y, and z).
The first equation is not defined if any of the denominators happens to be zero, which means that $x \ne -z$, and $y \ne -z$, and $z \ne 0$.
Some of the solutions of the second equation violate one or more of these restrictions.
Here are just a few solutions of the second equation that are not also solutions of the first equation:
x = 0, y = 0, z = 0 (violates $z \ne 0$)
x = 0, y = 1, z = 0 (ditto)
x = 1, y = 1, z = 1 (violates $y \ne -z$ and $x \ne -z$)

Right, I now realize that I was wrong to think that this problem is pure math. The origin is the "thin lense formula", so one should also specify the allowed range of possible values which was not done in my physics text.

 

[PeroK]

Right, I now realize that I was wrong to think that this problem is pure math. The origin is the "thin lense formula", so one should also specify the allowed range of possible values which was not done in my physics text.

As long as you assume (1) is a valid equation then the two are equivalent. But, technically, you need $z \ne 0$ etc.