Given the equation $x-y = 1$, I want to rearrange it to solve for $y$. The answer in the learning materials I have is $y = x-1$.

When I try to rearrange the equation myself I go through the following thought process:

$x-y =1$ --- first I think I want to get the x over to the right hand side of the equals sign so that I'm left with the $-y$ on the left side. So I do minus $x$ on both sides:

$-y = 1-x$

---Then I'm thinking that I want to change the minus $y$ to positive $y$. So then I multiply both sides by minus $1$

$y = -1 + x$

--Then I assume I can just switch around the position of $-1$ and $x$ to give $y = x-1$

My question is: Is this the same thought process that you would go through to solve this simple equation or am I taking too many unnecessary steps?

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    $\begingroup$ This is $\textbf{exactly}$ the thought process you should have. This should develop your intuition as to why the shortcuts taken in converting these two forms of the same equation are valid. $\endgroup$ – Don Thousand Aug 3 '18 at 21:37

The only way to do it more simply is this: It doesn't matter whether we solve for $y$ on the right or on the left. Thus, since $y$ is subtracted on the right, add it to both sides, and then subtract $1$ from each side to get $y$ alone over there: $$x-1=y.$$

There you have $y=x-1$, written the other way around.

I guess you could look at the original equation as telling you that taking $y$ away from $x$ leaves a remainder of $1$, which means that $y$ must be just $1$ less than $x$. Translate that sentence from words to symbols, and you have the same answer.

Is this what you were looking for?

  • $\begingroup$ I honestly don't think an answer is needed for this question... $\endgroup$ – Don Thousand Aug 3 '18 at 21:58
  • $\begingroup$ I think there's profit in looking at the detailed structure of our intuitions about arithmetic. If you find it fruitless, then I guess we differ in that regard. :) $\endgroup$ – G Tony Jacobs Aug 3 '18 at 22:02
  • $\begingroup$ I think you misunderstand my point ... I believe OP was looking for confirmation in his approach, not a simplification of methodology. Hence, an affirmation was what he desired. $\endgroup$ – Don Thousand Aug 3 '18 at 22:03
  • $\begingroup$ Ok, I read the question and came to a different conclusion. Thanks, though. $\endgroup$ – G Tony Jacobs Aug 3 '18 at 22:04
  • $\begingroup$ Thanks both for you comments. I guess I was after both confirmation as well as alternative possible simplifications in methodology. I hadn't thought of how the original equation is telling you that taking y away from x leaves a remainder of 1, which means that y must be just 1 less than x. This was useful :) $\endgroup$ – axiom111 Aug 4 '18 at 2:44


  • $x$ is on the left with a plus sign,

  • $y$ is on the left with a minus sign,

  • $1$ is on the right with a plus sign.


  • $x$ is on the $\color{red}{right}$ with a plus sign,

  • $y$ is on the left with a $\color{red}{plus}$ sign,

  • $1$ is on the right with a $\color{red}{minus}$ sign.

So chances are low that you can do in less than three elementary transformations... (Though you can reduce to two transformations if you allow the reversal $x-1=y$.)

My own way to think:

Isolate $y$ in two moves

$$x-y=1\to x=y+1\to x-1=y.$$

($y$ goes to the right to get a positive sign.)


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