This is a very basic question that's been bugging me for some time.

I know "=" means equality: that is, the expression on other side of the equation is the same exact number, so doing anything to one side requires the same thing to the other side. This makes sense when we were with concrete numbers: if you have $5 = 5/1$ and multiply by 8 on both sides, you maintain the equality.

I get a bit confused (but also not really confused - this is definitely me overthinking since I've been "isolating the variable" since grade school) when we introduce a variable $x$.

Let's say we have $2x + 1 = 9$. Why can I just subtract $2x$ from both sides? It's because $x$ represents an unknown number, and so properties of numbers work right? Or like $\frac{2}{x} = 17$ why can I just multiply by $x$ on both sides? It's because $x$ is just a number and so behaves like a number - it doesn't matter that we don't know the value, right?

With two variabes, let's say we have $y + 3x = 17$. How do I know the relationship between $x,y$ is "preserved" when I subtract $3x$ from both sides to get $y = 17-3x$? Or should I not think in terms of "preserved relationships between x, y" and instead "maintaining equalities."

  • 2
    $\begingroup$ $a=b$ means you can substitute $a$ with $b$ in any expression without changing its value. For example, if the expression is $a+c$ and you substitute $a \to b$, the expression becomes $b+c$ but the value does not change, so $a+c=b+c$. Do this with $a=2x+1, b=9, c=-2x$, and you get $\require{cancel}\cancel{2x}+1-\cancel{2x}=9-2x$. $\endgroup$
    – dxiv
    Commented Feb 8, 2022 at 3:55
  • 3
    $\begingroup$ This is a very good question. When you subtract some number from both sides, you actually apply an injective function on both sides, preserving the full relationship your variables need to satisfy. This is also why you can multiply by some non-zero number on both sides. However, squaring both sides is not a injective function, and hence doing so will give you a weaker relationship that your variables need to satisfy. Applying an injective function may thus generatw some unwanted solutions. $\endgroup$
    – Zuy
    Commented Feb 8, 2022 at 4:01
  • $\begingroup$ I wonder whether the statement $y+3x=17$ qualifies as an equation involving unknowns; for I would not say it s true for a given value of $x$ of or $y$; in other words, claiming to have found the solution of this " equation" seems meaningless; rather, it looks like a definiton, namely, the definiton of a relation, that is of a set of couples $(x,y)$. $\endgroup$ Commented Feb 8, 2022 at 11:30
  • $\begingroup$ The $x$ in $2x+1=9$ may be unknown to you but it very much has a definite value. $\endgroup$ Commented Feb 8, 2022 at 12:02
  • $\begingroup$ Relevant, not quite a duplicate: math.stackexchange.com/questions/2738360/… $\endgroup$ Commented Feb 8, 2022 at 12:11

1 Answer 1


$y + 3x = 17\tag1$ How do I know the relationship between $x,y$ is "preserved" when I subtract $3x$ from both sides? Or should I not think in terms of "preserved relationships between $x, y$" and instead "maintaining equalities"?

The integer solution set of the conditional equation $(1)$ is $$\{(5-n,3n+2)\mid n\in\mathbb Z\}.$$ Not every operation performed on $(1)$ preserves this solution set; for example, $$xy + 3x^2 = 17x\tag2$$ has an additional solution $(0,1),$ while $$\frac yx + 3 = \frac{17}x\tag3$$ is missing the solution $(0,17).$ As such, the first operation is valid while the second isn't.

To show that two equations have the same solution set, choose an arbitrary solution of one equation and show it must be in the solution solution of the other equation and vice versa, so the proof would be along the lines of: To show eqA = eqB, choose an arbitrary solution of eqA. We got to eqB from eqA through a sequence of operations such as multi/divide/sub/add and each operation simply scaled both sides or added/subtracted a number from both sides, and so this chosen solution hasn't been changed by the operations and so it must satisfy eqB too. Do the same thing for the other direction; grab an arbitrary solution, and we know it must satisfy eqA because we simply scaled both sides of the eqn or added/subtracted

This is not wrong, but also not the usual practice, where the reverse implication is typically not required. Furthermore, framing an inference or an operation on an equation as being either valid or invalid frequently feels more natural than literally thinking about whether the solution set is being narrowed down.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .