this post was submitted on 27 Nov 2025
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The first textbook only gets 5(17) by not doing what the second textbook says to do with 5(3+14).
First image says 'always simplify inside,' and shows that.
Second image says 'everything inside must be multiplied,' and shows that.
You're such an incompetent troll that you proved yourself wrong within the same post.
Because the first textbook is illustrating do brackets from the inside out, which the second textbook isn't doing (it only has one set of brackets, not nested brackets like the first one). They even tell you that right before the example. They still are both Distributing. You're also ignoring that they actually wrote 5[3+(14)], so they are resolving the inner brackets first, exactly as they said they were doing. 🙄 The 5 is outside the outermost brackets, and so they Distribute when they reach the outermost brackets. This is so not complicated - I don't know why you struggle with it so much 🙄
And then says to Distribute, and shows that 🙄 "A number next to anything in brackets means the contents of the brackets should be multiplied".
Yep, that's right, same as I've been telling you the whole time 😂
Ah, no, you did, again - you even just quoted that the second one also says to Distribute! BWAHAHAHAHAHAHAHA! 😂 I'll remember that you just called yourself an incompetent troll going forward. 😂
I think I know what you're missing - perhaps intentionally 🙄 - in a(b+c), c can be equal to 0. It can be any number, not just positive and negative, leaving us with a(b)=(axb), which is also what I've been saying all along (not sure how you missed it, other than to deliberately ignore it)
Nope! I've said a(b+c)=(ab+ac) is correct.
You mean I know that, because it disobeys The Distributive Law 🙄 The expression you're looking for is 2x(3+5)², which is indeed not subject to Distribution, since the 2 is not next to the brackets.
Instead I've stuck to one actual law of Maths, a(b+c)=(ab+ac).
The Distributive Law, including c=0 🙄 Not sure why you would think c=0 is somehow an exception from a law
No, the rules of Maths is the point
Says person who thinks c=0 is somehow an exception that isn't allowed,🙄but can't cite any textbook which says that
Dude you're not even hitting the right reply buttons anymore. Is that what you do when you're drunk? It'd explain leading with 'nope! I've said exactly what you accused me of.'
You keep pretending distribution is different from multiplication:
And then posting images that explicitly say the contents of the brackets should be multiplied. Or that they can be simplified first. I am not playing dueling-sources with you, because your own sources call bullshit on what you keep hassling strangers about.
Yes I am
Is that why you think I'm hitting the wrong buttons?
I have no idea what you're talking about. Maybe stop drinking
No pretending - is is different - it's why you get different answers to 8/2(1+3) (Distribution) and 8/2x(1+3) (Multiplication) 😂
B 8/2(1+3)=8/(2+6)=8/8
E
DM 8/8=1
AS
B 8/2x(1+3)=8/2x4
E
DM 8/2x4=4x4=16
AS
That's right.
The "contents OF THE BRACKETS", done in the BRACKETS step , not the MULTIPLICATION step - there you go quoting proof that I'm correct! 😂
That's right, you can simplify then DISTRIBUTE, both part of the BRACKETS step, and your point is?
B 8/2(1+3)=8/2(4)=8/(2x4)=8/8
E
DM 8/8=1 <== same answer
AS
No, because you haven't got any 😂
says person failing to give a single example of that EVER happenning 😂
I'll take that as an admission of being wrong then. Thanks for playing
This is your own source - and it says, juxtaposition is just multiplication. It doesn't mean E=mc^2^ is E=(mc)^2^.
Throwing other numbers on there is like arguing 1+2 is different from 2+1 because 8/1+2 is different from 8/2+1.
inside brackets. Don't leave out the inside brackets that they have specifically said you must use - "Parentheses must be introduced"! 🤣 BTW, this is a 19th Century textbook, from before they started calling them PRODUCTS 🙄
No, it means E=mc² is E=mcc=(mxcxc)
I have no idea what you're talking about 🙄
But you understand E=mc^2^ does not mean E=(mxc)^2^.
This is you acknowledging that distribution and juxtaposition are only multiplication - and only precede other multiplication.
In your chosen Introduction To Algebra, Chrystal 1817, on page 80 (page 100 of the PDF you used), under Exercises XII, question 24 reads (x+1)(x-1)+2(x+2)(x+3)=3(x+1)^2^. The answer on page 433 of the PDF reads -2. If 3(x+1)^2^ worked the way you pretend it does, that would mean 3=9.
I already answered, and I have no idea what your point is.
Nope. It's me acknowledging they are both BRACKETS 🙄
E=mcc=(mxcxc) <== BRACKETS
a(b+c)=(ab+ac) <== BRACKETS
everything 😂
Then why doesn't the juxtaposition of mc precede the square?
In your chosen book is the example you're pestering moriquende for, and you can't say shit about it.
Another: Keys To Algebra 1-4's answer booklet, page 19, upper right: "book 2, page 9" expands 6(ab)^3^ to 6(ab)(ab)(ab), and immediately after that, expands (6ab)^3^ to (6ab)(6ab)(6ab). The bullshit you made up says they should be equal.
For starters stop calling it "juxtaposition" - it's a Product/Term. Second, as I already told you, c²=cc, so I don't know why you're still going on about it. I have no idea what your point is.
You know I've quoted dozens of books, right?
Again I have no idea what you're talking about.
Ah, ok, NOW I see where you're getting confused. 6ab²=6abb, but 6(ab)²=6abab. Now spot the difference between 6ab and 6(a+b). Spoiler alert - the latter is a Factorised Term, where separate Terms have been Factorised into 1 term, the former isn't. 2 different scenario's, 2 different rules relating to Brackets, the former being a special case to differentiate between 6ab² and 6a²b²=6(ab)²
P.S.
this is correct - 2+1 is different from 1+2, but (1+2) is identically equal to (2+1) (notice how Brackets affect how it's evaluated? 😂) - but I had no idea what you meant by "throwing other numbers on there", so, again, I have no idea what your point is
Juxtaposition is key to the bullshit you made up, you infuriating sieve. You made a hundred comments in this thread about how 2*(8)^2^ is different from 2(8)^2^. Here is a Maths textbook saying, you're fucking wrong.
Here's another: First Steps In Algebra, Wentworth 1904, on page 143 (as in the Gutenberg PDF), in exercise 54, question 9 reads (x-a)(2x-a)=2(x-b)^2^. The answer on page 247 is x=(2b^2^-a^2^)/(4b-3a). If a=1, b=0, the question and answer get 1/3, and the bullshit you've made up does not.
You have harassed a dozen people specifically to insist that 6(ab)^2^ does not equal 6a^2^b^2^. You've sassed me specifically to say a variable can be zero, so 6(a+b) can be 6(a+0) can just be 6(a). There is no out for you. This is what you've been saying, and you're just fucking wrong, about algebra, for children.