# Weird algebra

• Aug 12th 2013, 11:58 AM
Paze
Weird algebra
(x+1)/(x-2)=3/(x-2)+5 - Wolfram|Alpha

Why on earth can't I just multiply by (t-2) and get the solution t=2 ?!

To make it absolutely clear that I should be able to just multiply by (t-2), I can rewrite this equation as:

$\frac{t+1}{t-2}=\frac{5t-7}{t-2}$

I am VERY confused..

I do realize that t can't be 2 because of the denominator becoming 0, but why does it give me the algebraic solution t=2 and then it doesn't work?
• Aug 12th 2013, 02:54 PM
Paze
Re: Weird algebra
I'm finding out that this is due to a problem in algebra called "extraneous solutions".

Why does this problem arise? I'm having problems finding it... Is algebra flawed?
• Aug 12th 2013, 04:28 PM
Paze
Re: Weird algebra
The reason that extraneous solutions sometimes occur is because if you were to raise both sides of the equation to a higher power such as cubed, you would increase the number of the solution. This new number night not satisfy the original equation.
• Aug 12th 2013, 04:47 PM
Shakarri
Re: Weird algebra
Quote:

Originally Posted by Paze
I do realize that t can't be 2 because of the denominator becoming 0, but why does it give me the algebraic solution t=2 and then it doesn't work?

Well it wouldn't give you the solution t=2 if you followed the rules of algebra and did not take $\frac{0}{0}=1$ (which is what you are doing when you cancel $\frac{t-2}{t-2}$ to 1.
Simply put this equation has no solution because it is wrong to begin with, you can see this more clearly if you rearrange it.

$\frac{t+1}{t-2}=\frac{5t-7}{t-2}$

$\frac{t+1}{t-2}-\frac{5t-7}{t-2}=0$

$\frac{1}{t-2}\times((t+1)-(5t-7))=0$

$\frac{1}{t-2}\times(-4t+8)=0$

$\frac{-4t+8}{t-2}=0$

$\frac{-4(t-2)}{t-2}=0$

$\frac{(t-2)}{t-2}=0$

Nothing divided by itself can equal 0 so you can see that the equation is false to begin with.
• Aug 12th 2013, 05:23 PM
Paze
Re: Weird algebra
Quote:

Originally Posted by Shakarri
Well it wouldn't give you the solution t=2 if you followed the rules of algebra and did not take $\frac{0}{0}=1$ (which is what you are doing when you cancel $\frac{t-2}{t-2}$ to 1.
Simply put this equation has no solution because it is wrong to begin with, you can see this more clearly if you rearrange it.

$\frac{t+1}{t-2}=\frac{5t-7}{t-2}$

$\frac{t+1}{t-2}-\frac{5t-7}{t-2}=0$

$\frac{1}{t-2}\times((t+1)-(5t-7))=0$

$\frac{1}{t-2}\times(-4t+8)=0$

$\frac{-4t+8}{t-2}=0$

$\frac{-4(t-2)}{t-2}=0$

$\frac{(t-2)}{t-2}=0$

Nothing divided by itself can equal 0 so you can see that the equation is false to begin with.

I realize what the problem is now. However, I did not break any rules of algebra as you claim. Thanks though.
• Aug 12th 2013, 06:02 PM
Shakarri
Re: Weird algebra
Quote:

Originally Posted by Paze
(x+1)/(x-2)=3/(x-2)+5 - Wolfram|Alpha

Why on earth can't I just multiply by (t-2) and get the solution t=2 ?!

To make it absolutely clear that I should be able to just multiply by (t-2), I can rewrite this equation as:

$\frac{t+1}{t-2}=\frac{5t-7}{t-2}$

I am VERY confused..

I do realize that t can't be 2 because of the denominator becoming 0, but why does it give me the algebraic solution t=2 and then it doesn't work?

You were going to break a rule of algebra when you planned to multiply by t-2. If you look at it in steps you'll see that

$\frac{t+1}{t-2}=\frac{5t-7}{t-2}$

Multiply both sides by t-2

$\frac{t+1}{t-2}\times(t-2)=\frac{5t-7}{t-2}\times(t-2)$

$\frac{t-2}{t-2}\times(t+1)=\frac{t-2}{t-2}\times(5t-7)$

$(t+1)=(5t-7)$
At this point you have made the mistake of turning (t-2)/(t-2) into 1
• Aug 12th 2013, 06:18 PM
Paze
Re: Weird algebra
Quote:

Originally Posted by Shakarri
At this point you have made the mistake of turning (t-2)/(t-2) into 1

I fail to see how dividing something by itself should not equal 1?
• Aug 12th 2013, 06:38 PM
Shakarri
Re: Weird algebra
It is not true for all t. When t=2 it is not equal to 1. So when you are trying to find a solution to the equation when you make the step to turn (t-2)/(t-2) into 1 you must make a note that any conclusions you reach are under the assumption that t is not equal to 2. And when you continued to find a solution you reached the conclusion that t=2. As I said "any conclusions you reach are under the assumption that t is not equal to 2." so your conclusion that t=2 is under the assumption that t is not equal to 2.
• Aug 12th 2013, 06:43 PM
Paze
Re: Weird algebra
Quote:

Originally Posted by Shakarri
It is not true for all t. When t=2 it is not equal to 1. So when you are trying to find a solution to the equation when you make the step to turn (t-2)/(t-2) into 1 you must make a note that any conclusions you reach are under the assumption that t is not equal to 2. And when you continued to find a solution you reached the conclusion that t=2. As I said "any conclusions you reach are under the assumption that t is not equal to 2." so your conclusion that t=2 is under the assumption that t is not equal to 2.

That is correct, but I don't feel like I am breaking any "rules of algebra". I feel that the rules of algebra are leading me to a false conclusion due to the ALGEBRA expecting a higher degree equation...Is that crazy?