Rearrange $x-y=1$

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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?




algebra-precalculus




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  • 6




    This is $textbfexactly$ 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.
    – Rushabh Mehta
    Aug 3 at 21:37














up vote
1
down vote

favorite













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?




algebra-precalculus




share|cite|improve this question

















  • 6




    This is $textbfexactly$ 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.
    – Rushabh Mehta
    Aug 3 at 21:37












up vote
1
down vote

favorite









up vote
1
down vote

favorite












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?




algebra-precalculus




share|cite|improve this question














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?





algebra-precalculus





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 3 at 21:37









Math Lover

12.2k21132




12.2k21132









asked Aug 3 at 21:33









axiom111

93




93







  • 6




    This is $textbfexactly$ 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.
    – Rushabh Mehta
    Aug 3 at 21:37












  • 6




    This is $textbfexactly$ 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.
    – Rushabh Mehta
    Aug 3 at 21:37







6




6




This is $textbfexactly$ 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.
– Rushabh Mehta
Aug 3 at 21:37




This is $textbfexactly$ 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.
– Rushabh Mehta
Aug 3 at 21:37










1 Answer
1






active

oldest

votes

















up vote
1
down vote













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 $x$. Write this down in words, and you have the answer.



Is this what you were looking for?






share|cite|improve this answer





















  • I honestly don't think an answer is needed for this question...
    – Rushabh Mehta
    Aug 3 at 21:58










  • 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. :)
    – G Tony Jacobs
    Aug 3 at 22:02










  • 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.
    – Rushabh Mehta
    Aug 3 at 22:03










  • Ok, I read the question and came to a different conclusion. Thanks, though.
    – G Tony Jacobs
    Aug 3 at 22:04










  • 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 :)
    – axiom111
    Aug 4 at 2:44










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote













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 $x$. Write this down in words, and you have the answer.



Is this what you were looking for?






share|cite|improve this answer





















  • I honestly don't think an answer is needed for this question...
    – Rushabh Mehta
    Aug 3 at 21:58










  • 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. :)
    – G Tony Jacobs
    Aug 3 at 22:02










  • 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.
    – Rushabh Mehta
    Aug 3 at 22:03










  • Ok, I read the question and came to a different conclusion. Thanks, though.
    – G Tony Jacobs
    Aug 3 at 22:04










  • 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 :)
    – axiom111
    Aug 4 at 2:44














up vote
1
down vote













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 $x$. Write this down in words, and you have the answer.



Is this what you were looking for?






share|cite|improve this answer





















  • I honestly don't think an answer is needed for this question...
    – Rushabh Mehta
    Aug 3 at 21:58










  • 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. :)
    – G Tony Jacobs
    Aug 3 at 22:02










  • 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.
    – Rushabh Mehta
    Aug 3 at 22:03










  • Ok, I read the question and came to a different conclusion. Thanks, though.
    – G Tony Jacobs
    Aug 3 at 22:04










  • 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 :)
    – axiom111
    Aug 4 at 2:44












up vote
1
down vote










up vote
1
down vote









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 $x$. Write this down in words, and you have the answer.



Is this what you were looking for?






share|cite|improve this answer













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 $x$. Write this down in words, and you have the answer.



Is this what you were looking for?







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Aug 3 at 21:38









G Tony Jacobs

25.5k43483




25.5k43483











  • I honestly don't think an answer is needed for this question...
    – Rushabh Mehta
    Aug 3 at 21:58










  • 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. :)
    – G Tony Jacobs
    Aug 3 at 22:02










  • 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.
    – Rushabh Mehta
    Aug 3 at 22:03










  • Ok, I read the question and came to a different conclusion. Thanks, though.
    – G Tony Jacobs
    Aug 3 at 22:04










  • 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 :)
    – axiom111
    Aug 4 at 2:44
















  • I honestly don't think an answer is needed for this question...
    – Rushabh Mehta
    Aug 3 at 21:58










  • 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. :)
    – G Tony Jacobs
    Aug 3 at 22:02










  • 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.
    – Rushabh Mehta
    Aug 3 at 22:03










  • Ok, I read the question and came to a different conclusion. Thanks, though.
    – G Tony Jacobs
    Aug 3 at 22:04










  • 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 :)
    – axiom111
    Aug 4 at 2:44















I honestly don't think an answer is needed for this question...
– Rushabh Mehta
Aug 3 at 21:58




I honestly don't think an answer is needed for this question...
– Rushabh Mehta
Aug 3 at 21:58












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. :)
– G Tony Jacobs
Aug 3 at 22:02




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. :)
– G Tony Jacobs
Aug 3 at 22:02












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.
– Rushabh Mehta
Aug 3 at 22:03




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.
– Rushabh Mehta
Aug 3 at 22:03












Ok, I read the question and came to a different conclusion. Thanks, though.
– G Tony Jacobs
Aug 3 at 22:04




Ok, I read the question and came to a different conclusion. Thanks, though.
– G Tony Jacobs
Aug 3 at 22:04












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 :)
– axiom111
Aug 4 at 2:44




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 :)
– axiom111
Aug 4 at 2:44












 

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