Solve for x-y

[kugalskaper]

Hi there people.

It's been a while since i did any math, but i used to be fairly good at it (college level). So when a friend asked me to show her how to solve a problem I had big confidence in that i would be able to help her. However i soon realized that its been too long since i used my brain for math. I don't quite know what my problem is, which is quite frustrating. The sticky on this forum says that i should include my current progress, but i've tried so many different things, and right now i wouldnt know what to include. Now the problem is not solving the equation. I would be able to solve for x and y and then either substitute or solve it graphically (i haven't tried though since i dont have my calculator here). However it need to be solved in the manner x-y=?. My guess is that the answer to my problem is easy, and that i've just forgotten about some basic mathematical rules. A step by step solution would be most welcome

So for the equation:

((x^2-y^2+2y-1)/(y^2-x^2+2x-1))=-2

The answer is supposed to be 3, but the answer is irrelevant for me

Thank you
Morten

 

[mmm4444bot]

Factor the numerator: (x - y + 1)(x + y - 1)

Factor the denominator: (y - x + 1)(x + y - 1)

Cancel the common factors:

(x - y + 1)/(y - x + 1) = -2

Multiply both sides by the denominator:

x - y + 1 = -2y + 2x - 2

This leads to: x - y = 3

It seems that any pair of (x,y) values where x is three more than y is a solution to the original equation.

However, there exists one such (x,y) pair of values that is NOT a solution.

Can you determine this exceptional pair?

 

[mmm4444bot]

Hint: Any pair of (x,y) values where x is 3 more than y that causes the left-hand side of the given equation to be undefined cannot be a solution.

 

[soroban]

Re:

Hello, mmm4444bot!

It seems that any pair \(\displaystyle (x,y)\) where \(\displaystyle x\) is three more than \(\displaystyle y\) is a solution to the equation.

However, there exists one such pair \(\displaystyle (x,y)\) that is NOT a solution.

Can you determine this exceptional pair?

\(\displaystyle \text{We already have: }\:x - y \:=\:3\)


\(\displaystyle \text{We had cancelled }\,(x+y-1) \quad \hdots \quad\text{Hence: }\:x+y-1 \:\neq\:0\)


\(\displaystyle \text{So we are }not\text{ allowed the system: }\;\begin{array}{ccc}x + y &=& 1 \\x - y &=& 3\end{array}\)

. . \(\displaystyle \text{which has the solution : }\:x=2,\;y=-1\)


\(\displaystyle \text{Therefore, the one such pair which is }{not\text{ a solution is: }\2,-1)\)

 

[mmm4444bot]

Correct, Soroban!

Excellent!

So, the solution to the original equation is not simply x - y = 3, as implied.

The solution is any xy pair where x - y = 3 except (2, -1).

 

[Denis]

((x^2-y^2+2y-1)/(y^2-x^2+2x-1))=-2

OK guys: suppose the above equation did NOT appear in the problem statement; instead, this appeared:
x^2 - y^2 + 2y - 1 = 2x^2 - 2y^2 - 4x + 2
What would you then say about (2,-1) ? :shock:

 

[mmm4444bot]

((x^2-y^2+2y-1)/(y^2-x^2+2x-1))=-2

What would you then say about (2,-1) ? :shock:

I would say that you were talking about a function that differs from the function given in the original post.

 

[Denis]

Re:

I would say that you were talking about a function that differs from the function given in the original post.

Ok, follow that, but WHY does it differ?
Does 10/2 = 5 differ from 10 = 10 ?

OK; I'm given this problem:
x^2 - y^2 + 2y - 1 = 2x^2 - 2y^2 - 4x + 2
Show that x - y = 3.

I do the work, which includes (2,-1).
Then I yell: HANG ON!!
Since x^2 - y^2 + 2y - 1 = 2x^2 - 2y^2 - 4x + 2
can be written as (x^2 - y^2 + 2y - 1) / (y^2 - x^2 + 2x - 1) = -2
then (2,-1) is not a solution....
I'm just curious...

 

[Deleted member 4993]

Re: Re:

I would say that you were talking about a function that differs from the function given in the original post.

Ok, follow that, but WHY does it differ?
Does 10/2 = 5 differ from 10 = 10 ?

OK; I'm given this problem:
x^2 - y^2 + 2y - 1 = 2x^2 - 2y^2 - 4x + 2
Show that x - y = 3.

I do the work, which includes (2,-1).
Then I yell: HANG ON!!
Since x^2 - y^2 + 2y - 1 = 2x^2 - 2y^2 - 4x + 2
can be written as (x^2 - y^2 + 2y - 1) / (y^2 - x^2 + 2x - 1) = -2
then (2,-1) is not a solution....
I'm just curious...

Denis in the second case (the original case) at (2,-1) the denominator is = 0 - so the function doesnot exist there.

Go stand in the corner for arguing ......

 

[mmm4444bot]

Ok, follow that, but WHY does it differ?

One contains an algebraic ratio, and the other does not.

 

[Denis]

Re: Re:

Go stand in the corner for arguing ......

I'm gonna tell my daddy on you; he's stronger than yours

Seriously, methinks yer both missing my point. But not important.

 

[mmm4444bot]

OK; I'm given this problem:
x^2 - y^2 + 2y - 1 = 2x^2 - 2y^2 - 4x + 2
Show that x - y = 3.

I do the work, which includes (2,-1).
Then I yell: HANG ON!!
Since x^2 - y^2 + 2y - 1 = 2x^2 - 2y^2 - 4x + 2
can be written as (x^2 - y^2 + 2y - 1) / (y^2 - x^2 + 2x - 1) = -2 Not for all (x, y), it can't.

then (2,-1) is not a solution... This line is ambiguous.

The distinction is important. We need to be wary of domain.

(2, -1) is not a solution to your new equation, if that's what you mean.But (2, -1) remains a solution to your original equation.

The two equations must both exist, before they are equivalent.

Sorta like proving some trig identities. When we're done, we can't always say that it's true for all angles. Trig identities are only true with angles for which both sides are defined.

We can't say that the original poster's equation is true for all solutions to x - y = 3 because their original equation does not even exist for (2,-1).

Kapish?