Help to prove equation has no solutions

[copper crane]

Hello! See picture added. I have an equation that has no real solutions. Which i need to prove by. I wrote:
The expression has no solution because a fraction is 0 only if the numerator is 0. And here the numerator is the constant -3. He said it was wrongly motivated. Can someone explain where I went wrong?
Thanks for all help!

[plain fiber]

Maybe he wanted you to mention the case where the denominator is 0
Since then it's undefined and doesn't evaluate to 0

[copper crane]

maybe! should i just rewrite it as 3b-20 = 3/0 then?

[plain fiber]

What did you to get that

You can't divide both sides by 0

And even if you did wouldn't that be a -3/0

[copper crane]

idk im tired lol but thats what im saying its not possible

studying math for 12+ hours makes you create ur own rules hehe

[plain fiber]

I see
Though you're not allowed to use a step that's not allowed to show the equality doesn't hold
Your original justification is fine, it's just "a fraction can only evaluate to 0 if the numerator is 0" isn't good enough since that's not true if the denominator is 0
So just explain that if the denominator is 0 then the equality doesn't hold because the right hand side is undefined
And then you can say when the denominator isn't 0, if the numerator isn't equal to 0 then the fraction still won't evaluate to p

[plain fiber]

Consider 4x=0
Divide both sides by x
4=0/x
that gives us one case where x could be 0 and one case where jt isn't
So if x is not equal to 0
Then 4=0, so the equality doesn't hold
If x is equal to 0, then 4=0/0 so the equality doesn't hold either

But x=0 is the solution

[copper crane]

i see so i just need to add when the denominator is 0 its undefined and if the denominator isnt 0 then the fraction still doesnt work

[plain fiber]

"If the denominator isn't 0 then the fraction still doesn't work" because the numerator is not 0 so it cannot evaluate to 0
You have to say allat

For the second half

But I think then it should make sense

[copper crane]

my teacher is very strict so i think ill write all that xD

but thank you, you made my life a whole lot easier

[loud cobalt]

if b is infinity

It will also right

[sharp sorrel]

So firstly, as has been pointed out, you can't divide by 0. That doesn't mean that there's no solution.
The analogue is saying you can't cut an apple with a spoon - that doesn't mean you can't cut the apple, but it means the spoon isn't really the right tool

The natural thing to consider is that, since this is a fraction, we want to remove that division as fast as possible - consider multiplying both sides by the denominator of the fraction to get $0(3b-20) = 3$

[maiden finch]

Waes (Wires)

[sharp sorrel]

It should then be pretty clear why this has no solution

[loud cobalt]

i know that you are right

but if b is infinity

-3/3b-20 will approaching to 0

[sharp sorrel]

We can't allow that - you're a little confused by what infinity is

It's a common misconception to have something equal infinity

What we really mean is to talk about what happens when something approaches/tends to infinity

[sharp sorrel]

at which point, this would be correct; but the key point as per your wording here is that this is NOT equal to 0

We're looking for strict equalities, not mere tendencies

[plain fiber]

b cannot be infinity

b can only be a real number

[loud cobalt]

ok

thanks

[vapid socket]

it doesn't make sense why your teacher isn't accepting your proof

[copper crane]

yeah. now i wrote: that if the denominator is 0 then the equality doesn't hold because the right hand side is undefined and when the denominator isnt 0, if the numerator isnt equal to 0 then the fraction still wont evaluate to p. Then i also multiplied both sides to get 0(3b-20) = 3 which then obviously tells that anything multiplied with 0 cant be 3. If he doesnt accept this answer i dont know what lol

-3*

[sharp sorrel]

The part from "Then I also..." is the answer

The denominator being 0 isn't practical - there's no logic to test why that should be 0