# Homework Help: Simp Simultaneous Equation

1. Mar 24, 2007

### Delzac

Hi all,

Say i have 2 equations, s =t and 1-s = t-1.

Normally one would solve the equations simply but subbing in S=t into the other equation and obtain t = s = 1.

However, if i were to slightly rearrange this equation:
1-s = t-1
I can obtain this :
-1(s-1) = (t-1)
-1 = (t-1)/(s-1), wouldn't this not make sense if i sub s=t into it?

And i am anticipating that some would point out that by rearranging the equation i divided by 0. So the whole thing is not valid.

However what if i rewrote my whole post and state first that initially i have 2 equations, s=t and 1 = (t-1)/(1-s) and i wish to solve this.

Then i would have solve it like this :

1 = (t-1)/(1-t)
1-t = t-1
t=1 = s

However if this is the case then wouldn't the original equation 1 = (t-1)/(1-s) is self is invalid?

What is wrong here? or did i go something wrongly?

By the way, i came across this thing as i am doing my vectors -Planes tutorial.

Any help will be greatly appreciated.

2. Mar 24, 2007

### Delzac

what in the world do you mean? why has solving simultaneous equation got to do with calculus?

3. Mar 24, 2007

### AlephZero

Correct

That doesn't mean anything when s = 1, because you divided by 0. The value of t is irrelevant.

If 1-s = t-1, then when s = 1, you can write

A(s-1) = (t-1) = 0 for any value of A.

If you substitute to get 1 = (s-1)/(1-s), you can say there are no solutions when s is NOT equal to 1. The equation 1 = (t-1)/(1-s) doesn't mean anything when s = 1.

What this demonstrates is that the connection between "mathematical equations" and "physical reality" is more subtle that you might have thought.

FWIW I don't see the relevance of taking limits here, either.

4. Mar 24, 2007

### arildno

The equation of two variables $1=\frac{t-1}{1-s}$ is a perfectly valid equation. However, s cannot take the value 1.

Thus, the system of equations $s=t, 1=\frac{t-1}{1-s}$ has NO solutions, whereas the system $s=t, 1-s=t-1$ has a unique solution, nameley s=t=1

5. Mar 24, 2007

### Delzac

hmmm.......kk i got it. Thanks for the help.

6. Mar 24, 2007

### arildno

No, there is no "calculus" form or "limit form" in which what you state is correct.

You are wrong, on all accounts.

When we are solving equations, or systems of equations, we are interested in determining the solution set of those equations, i.e, those elements in our general number set that satisfy the given equations.

Any element (s,t) in R^2 of the form (1,t) is not part of our greatest possible domain, since we can't divide by zero.

7. Mar 24, 2007

### arildno

There is no "approaching" when we solve equations.

8. Mar 24, 2007

### interested_learner

Aplogize to the original poster for my silly posts and thank those you corrected me. I need to think before I post. Thanks!