Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs
Community Proposals
Community Proposals
tag:snake search within a tag
answers:0 unanswered questions
user:xxxx search by author id
score:0.5 posts with 0.5+ score
"snake oil" exact phrase
votes:4 posts with 4+ votes
created:<1w created < 1 week ago
post_type:xxxx type of post
Search help
Notifications
Mark all as read See all your notifications »
Q&A

Comments on Missing a solution: Are A and B always equal if A-B=0

Parent

Missing a solution: Are A and B always equal if A-B=0

+4
−0

I just came across this problem on an online learning app:

4x(5−x)−12(5−x)+100=100

I tried to solve it by subtracting 100 from each side then inferring if something minus something else equals zero then those two things must be equal.

4x(5−x)=12(5−x)

Divide both sides by (5-x) and x must be 3.

But brilliant had a different way of solving it that revealed there are two possible values for x.

Where did I make a mistake in my approach to solving the problem?

Is it incorrect to assume that if a minus b equals 0 then a must equal b?

History
Why does this post require moderator attention?
You might want to add some details to your flag.
Why should this post be closed?

0 comment threads

Post
+4
−0

Dividing by 5-x is valid only if x is not 5. But x=5 is another solution to the original equation.

History
Why does this post require moderator attention?
You might want to add some details to your flag.

2 comment threads

Rule against adding uncertain denominators? (5 comments)
So if 4ab=12b, and you want to know what a is, is it a bad idea to divide both sides by b? Is this e... (2 comments)
Rule against adding uncertain denominators?
re89j‭ wrote about 2 years ago · edited about 2 years ago

Is it accurate to say it is "not allowed" to divide both sides by something that includes a variable unless I can prove some other way that the denominator I'm creating can not result in 0?

In other words, I could divide both sides by a finite number or pi or something, but probably not by the variable I'm trying to solve for?

Is there a name for this rule perhaps?

whybecause‭ wrote about 2 years ago · edited about 2 years ago

To answer this, it is important to remember what any of this even means. What do we mean by any equation, like x-2=3?

Actually, without any context, this doesn't mean anything at all. But usually when we write "x-2=3" this is like a short-hand for a question. Namely "What are all of the values of x which satisfy the equation x-2=3?" Clearly in these very simple contexts, there is only one answer, and therefore we do not need to be very careful.

But now consider the example equation x*(x-1)=0. If we say "what are all of the values of x which satisfy the equation?" then you will lose a solution if you divide by x. Because if you do that, you get x-1 = 0 which has only one solution.

So the reason why it is invalid to divide by a quantity that "might" be zero is that it causes you to lose solutions, when the question you are trying to answer is "What are ALL of the values of the variable which satisfy the equation?"

whybecause‭ wrote about 2 years ago · edited about 2 years ago

To my knowledge this rule does not have a name. Rather, to talk about it, we merely point out that division by zero could be a problem, so we proceed to separate the solution method into cases. Each case avoids the issue of dividing by zero.

Alternately you can characterize this by the zero-product property: https://en.wikipedia.org/wiki/Zero-product_property

Note that in the zero-product property, you do not know which of the two factors is zero. Therefore to include all possible solutions to the equation, you must consider the possibility that the first factor is zero, and the possibility that the second factor is zero. If you do not consider both then you may lose solutions.

re89j‭ wrote about 2 years ago

"it is invalid to divide by a quantity that "might" be zero [because] it [can cause] you to lose solutions"

^^ +1 THIS ^^

I wonder if it's just zero, or if it's any unknown value....

Dividing by an indeterminate value [can | does always] collapse a dimension of the problem, and so presents fewer solutions.

Also I will admit that my reptile brain sees simple equations like that and thinks what "IS" x, not "what ARE ALL POSSIBILITIES for X". The latter is completely my shortcoming. Maybe replacing what IS TRUE with what ARE ALL POSSIBLE TRUTHS is something I should do by default in life.

As an experiment, I've asked a few random adults x*x=4, what is x ,and only one of them acknowledged the possibility of -2 without prodding. I wish more ticktocks exposed mathematical truths like this than people trying to get a towel out of a running sink without getting wet...

whybecause‭ wrote about 2 years ago · edited about 2 years ago

It's only zero. If ax=b and a is not zero, then x=b/a. This effectively guarantees that division by a nonzero number preserves any solution, regardless of what x is (or what a and b are, again assuming a is nonzero).

This comes down to the definition of division. What does 15/5 = 3 mean? It means THE UNIQUE number x which satisfies 3x=5. There is only one number corresponding to division by a nonzero number.

What about 15/0? Well, we would need a number satisfying 0x=15. There is no such number and this is why so-to-speak "you cannot divide by zero".