alright, so once again I've confused myself. This time about the meaning of something as simple as "minus/negative". I have multiple questions, so answer what you can.

Question 1) So as I saw it before was that when I have, say $5-4x$, or just $-4x$ it would mean the exact same (in regards to the $-4x$): "minus four, times x", but Ive come to realize today that apparently one should interpret $5-4x$ as: "five, minus, four times x)" aka basically $5-(4×x)$. Which one is it?

Question 2 which basically expands on question 1) The idea of "adding a negative". It is obvious to me that $5+(-3)=5-3=2$ since the 3 definitely is "negative". However, $5+(-5×3)$ or $5+(-x×y)$ ?? What should I think about the $-5×3$ or the $-x×y$? Adding a negative is negative, but are we really adding a "negative"? Because as I see it now, it is just minus x, times y, and minus 5, times 3. That is to ask, why is $+(-x×y)=-x×y$.

Let me make this question clearer. Let's have a term with a minus negative variable inside, but let's not simplify the term: $$5+(-x×(-y)$$ You see, to simplify this, should one think of the latter term as $-(x×(-y)$? Can you do that, and is it the same thing? My books and the internet tell me to keep the sign negative WHENEVER a plus is followed by a minus,and change to positive WHENEVER 2 negatives come after each other. Is it best to just accept this rule? Why does it work? For example $+(-a×5×4×f×(-c))$ simplifies to (not opening the (-c)) $-a×5×4×f×(-c))$ based purely on the logic that the sign immediately after the + is minus. conversely $-(-a×5×4×f×(-c))$ simplifies to $-a×5×4×f×(-c))$ on the logic that "the sign after the minus is a minus"

question 3) Somewhat related as well. For as long as I've done algebra, and when I've had to open brackets (use the distributive property) or just multiply both sides of an equation, I've gone with this logic without realizing: Say I have $-5x(2-y+c)$ Then I multiply each term with the $-5x$ so I get $-10y, 5xy, -5cx$ and now I just look at the signs, 2 of them have minuses so I wont touch them, but since the $5xy$ doesn't have a sign I assume it means $+5xy$ just as 5 means +5. Then I place them next each other (basically add them?) and get $-10x+5xy-5cx$. Is this valid? It certainly has always given the right answer.

I'll try to respond if you don't understand my questions. Thank you in advance.

  • 1
    $\begingroup$ $a-b = a+(-b)$ That's what you need to remember. So $5-4x = 5 + (-4)x.$ $-4x = (-4)x$ $\endgroup$
    – amWhy
    Mar 19 '18 at 18:41
  • $\begingroup$ $(- x\cdot y) =( -(1) \cdot x\cdot y)= (-1) \cdot (x\cdot y) = -(x\cdot y)$ $$ $$ $5+(-x\cdot (-y)) = 5 + (-1)(-1)(x\cdot y) = 5+ x\cdot y.$ $\endgroup$
    – amWhy
    Mar 19 '18 at 18:45
  • $\begingroup$ $$ $$ $5+(-x\cdot (-y)) = 5 + (-1)(-1)(x\cdot y) = 5+ x\cdot y.$ $$ $$ Note that $+(-a×5×4×f×(-c)) = ((-1)\times a \times 5 \times 4 \times f\times (-1) \times c) = (-1)(-1)(a\times 5\times 4 \times f \times c) = a\times 5\times 4 \times f\times c$. $\endgroup$
    – amWhy
    Mar 19 '18 at 18:54
  • $\begingroup$ Could I have some clarification regarding your first comment? Did you "modify" the $-4$ into $+(-4)$ (,which is then multiplied by the x)? $\endgroup$ Mar 19 '18 at 18:59
  • $\begingroup$ $a-b = a +(-b)$, so $5-4x= 5+(-4)x = 5 + -x+-x+-x+-x$. $\endgroup$
    – amWhy
    Mar 19 '18 at 19:02

This is why we have the rules of operator precedence. Multiplication is an operator even it's not written out. Multiplication and division have higher precedence than addition and subtraction.

So in $5 - 4x$, you should probably first figure out what $4x$ is before subtracting it from 5. You're aware that you can use parentheses to override precedence. For example, $(5 - 4)x$ you evaluate 5 minus 4 first, leaving you with 1 and so $(5 - 4)x = x$.

This is very basic, but remember that $x$ is a variable that can stand for any number (though there might be some constraints stated at the beginning, like maybe it's a real rational number).

Suppose for a moment that $x$ stands for a dozen. Then 5 singles minus four dozens is $-43$ singles. Or suppose that $x = -\frac{3}{2}$. Then $5 - 4x = 11$. I chose that to lead to your question 2.

Which has a problem of mismatched parentheses: $5 + (-x \times (-y)$... did you mean $5 + (-x \times (-y))$? Or $5 + (-x) \times (-y)$? Or maybe $5 + ((-x) \times (-y))$? I guess that when you asked the Internet, you didn't ask Wolfram Alpha.

I'm going to assume you meant $5 + (-x)(-y)$. Then $(-x)(-y) = (-1)(-1)xy = xy$, so $5 + (-x)(-y) = 5 + xy$.

For question 3, I think you meant "it has always given me the right answer." But I don't quite see how it could give the wrong answer. Unless maybe there is some kind of style guide being enforced. I want you to go Wolfram Alpha and put this in: -5x(2 - y + c)

  • $\begingroup$ Thank you, I think I understand now. It is minus, 4x, and not minus 4, times x. So just to be clear since I have my matriculation exam In a couple of days. The classic sign rules (positive times a negative is negative, subtracting is the same as adding the thing, but with a minus in front, etc.) work for all defined algebraic expressions? In other words, $2-(\frac {x+1}{x^2}-x)=2+(-(\frac {x+1}{x^2}-x)$ for example? (I realize it could be simplified further easily, but is my deduction correct) $\endgroup$ Mar 20 '18 at 9:15
  • $\begingroup$ As for my question 3, well I guess I just wanted to check if it was correct to do what I did, aka deducing that no sign in front of the product means you add a plus in front of the product $\endgroup$ Mar 20 '18 at 9:21
  • $\begingroup$ And lastly to be honest I totally forgot that the order of operations even exists....when I simplified equations etc. I just took every term (minus included in the term), calculated the product, and if I ended up with no minus, I added a plus. Seems to work anyways. That is that I did not even think that you could calculate the product first, as is said bu the order of operations, and then try to subtract (which you more often than not, could not) $\endgroup$ Mar 20 '18 at 9:25
  • $\begingroup$ That's right, they work for all defined algebraic expressions. And when they don't, it's a problem of misunderstanding, either on the part of the author or the part of the reader. $\endgroup$ Mar 20 '18 at 14:22

Part of the problem is that the $``-"$ symbol has two completely different meanings: an opposite symbol and a subtraction symbol.

Every number, $x$, has an opposite written $-x$ such that $x + (-x) = (-x) + x = 0$.

In the old days the opposite of $x$ was written as ${}^-x$.

These are the old rules of subtraction and opposites

  • $x+{}^{-}x = 0$
  • ${}^{-}x + x = 0$
  • ${}^{-}({}^{-}x)=x$
  • $x-y = x+{}^-y$
  • $x-{}^-y = x+{}^-({}^{-}y) = x+y$

Nowadays this is written as

  • $x+(-x) = 0$
  • $(-x) + x = 0$
  • $-(-x)=x$
  • $x-y = x+(-y)$
  • $x-(-y)= x+(-(-y)) = x+y$


  • $5-4x = 5+(-4)x$
  • $-xy = (-x)y = -(xy)$
  • $-x(-y) = (-(-(xy))=xy$
  • $5+(-x(-y)) = 5+(-x)(-y) = 5+(-(-(xy)) = 5+xy$
  • Or $5+(-x(-y)) = 5-(x(-y)) = 5-(-xy)=5+(-(-xy))=5+xy$
  • $-5x(2-y+c)=(-5x)(2-y+c)=(-5x)(2)+(-5x)(-y)+(-5x)(c)=-10x+5xy-5xc$

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