Guys help how do I solve this?

For multiple reasons
>Novelty is fun
>Veeky Forums isn't here to help you with your homework
>most of the people here don't know how to do kinematic equations either.

So basically this board is full of STEM illiterates who will you call you a brainlet as a response to a tough question but once an easy question gets asked eveyone all of a sudden becomes and expert and answers it.

Welcome to the internet at population: Everyone.

kinematics equations are also highschool level

welcome to Veeky Forums, i hope you'll enjoy your stay!

Let x=10
2 + 10 = 3
12 = 3

After spending the last 7 hours doing hardcore simulation on java (my CS nerds know what I'm about!), I got the solution 1.0000000014200023 (thankfully I used a double to get the precise answer)

Substract 2 from both difes of the ewquation

2+x=3

first you remove the two

+x=3

so now we know +x is equal to 3

next step is to move the plus to the other side

x=+3

now we bring back the 2, but on both sides

2x=2+3

dont forget to make it a minus on the left side

-2x=2+3

now we just solve for x and we get -2.5

-x = 2.5

Yeah, I think this is too advanced. You should probably ask the nearest math professor (probably an expert on number theory by the looks of this equation) for help on solving this really advanced equation. Perhaps Terry Tao will have a solution on this, you should ask him on his blog.

Wolfram Alpha might also have a solution (though they don't always have answers to really hard problems like these) like suggested.

what does the sideways "x" mean?

2 + x = 3
(2 + x) = 3
(2 + x )^2 = 9
4 + x^2 = 9
x^2 = (9-5)^2
x^2 = 81 - 90 + 25
x= 9- 5
x= 4

what if the plus sign is the variable and the x is just a multiplication sign?

...

x=((i^2)/-1)

Doesn't anybody read the sticky? Requests for help with homework go to /wsr/..

>Veeky Forums
>reading

So we want to solve
[math]2 + x = 3[/math]
All we need to do is add -2 to both sides.
[math]-2+2 + x = -2 + 3[/math]
[math]\implies 0 + x = 1[/math]
[math]\implies x = 1[/math]

...

I'm guessing those faces are because I forgot to check my answer.
[math]x = 1[/math]
[math]\implies 2 + 1 = 3[/math]
[math]\implies 3 = 3[/math]
Now, of course 3 = 3 so we can see that x = 1 is the correct solution.

[math]2 + x = 3\\
1 + x = 2\\
0 + x = 1[/math]

Took me only 5 minutes.

Lol

For situations of the type
[math] Ax + B = C[/math]
[math] A \in \mathbb{R}\setminus\{0\}, B \in \mathbb{R}, C \in \mathbb{R}[/math]
We use the equation
[math]x = \frac{C-B}{A}[/math]
In your question we have
[math]A = 1, B = 2, C = 3[/math]
And so subbing in these values to the formula gives us
[math]x = \frac{3 - 2}{1}[/math]
[math]\implies x = \frac{1}{1}[/math]
[math]\implies x = 1[/math]

What happened on the right side between the first and second lines? I get the rest

I wasn't that poster, so I wouldn't want to put words in his mouth, but what I believe happened is he added -3 to both the right and left hand sides.
When he used |-3 on the very right of the second line, he was indicating what he was doing to both sides.
He also does this on the fourth line if you look closely.
As I said before, this is my best guess as to what he was doing, so take it with a grain of salt as only he can say for sure.

2 + x = 3
Then (2 + x) - 2 = 3 - 2 = 1
Now all you have to do is prove that addition is both associative and commutative and you have x = 1.

I got 0.99999999999999999 but I'm not sure how to prove that 0.9999 eventually leads to 1

this is an extremely high quality post, thank you user.

this is the best thread i've seen in a while

pomf =3

maybe you should stop asking babby questions and ponder real shit like OP is

you didnt even define any of the symbols, or what they belong to

>[math]0 + x = 1 \implies x = 1[/math]
fucking prove it, nerd
who the fuck told you to assume we were in a Monoid??

2+x=3
e^log(2)+sqrt(x^2)=(3^2)/(1^2)
2=1+1
log(a+b)=log(a)+log(b)
log(1)*2 = log(1^2) = log(1)
1=e^(2pi*i)
e^log(2) = e^log(1) = e^log(e^(2pi*i)) = e^(2pi*i)
e^(ab)=e^a*e^b
e^(2pi*i) = e^2 * -1
sqrt(x^2) = sqrt((-x)^2) = -x
-e^2 - x = (3^2)/(1^2)
-e^2 - (x^.5)^2 = 3^2/1^2
-e - sqrt(x) = 3
-sqrt(x)=3+e
sqrt(x)=-3-e
x=sqrt(-3-e) ~ 2.39129292 i

pretty easy

Why the fuck is this still here

Addition

Not him but a variable by itself is considered 1

0-x=1
-0 -0

x=1

x/1=1
x=1

wrong
your equations should read
[math]0 + x = 1 \implies x = 1 - 0[/math]
and again you haven't shown [math]1 - 0 = 1[/math]

>a variable by itself is considered 1
what the FUCK kind of wonky math are they teaching you there in india, boy?

well i'll try again without wonky math.

Well a variable by itself could equal 1 by this statement
[math]x/x=1[/math]

[math]x=1[/math]

[math]x/1x[/math] [math]1/1x[/math]

[math]x/1x=x[/math]
[math]1/1x=1[/math] or [math] 1/1=1[/math]
[math]x=1[/math]

[math]2+x=3[/math]
[math]-2[/math] [math]-2[/math]
[math]x=1[/math]
[math]1+2=3[/math]

...

cheeky

(2+x)^2 doesnt equal 4+x^2. It equals x^2+4x+4

Remember PEMDAS (please excuse my dear aunt Sally.)
First, multiply (that's what x means) the + and the =, to create +=. What the += does, is it makes the thing on the left of it equal to itself plus the thing on the right.
So you get 2 += 3 which is the same as 5. 2 now means 5. Problem solved.

You're in what's called the "retard valley".

not

This is so stupid I nearly pissed myself of laughter.

Ban this mother fu and the guys who are giving the solution

QED

You can't add numbers and letters lol dummy

just put 2 on the other side, it will become minus
then 3-2=1 so x=1

x=3 and return value is 5

3-2=1
x=1
2+1=3

did i doo a gud job?

2+x=3
log2+logx=log3
logx=log3+log2
10^logx=10^log3+10^log2
x=3+2
x=5

Fuck off, troll

i hate this place so much

Is your post a joke?

In ANY algebraic problem you always do addition and subtraction FIRST, for example think of this problem here: [math]2+x/4=15[/math] first you need to do subtraction because we need to simplify the equation to find [math]x[/math]

[math]2=+x/4=15[/math]
[math]-2[/math] and [math]-2[/math]

Now we have [math]x/4=13[/math]
We continue towards the reciprocal of division(multiplication)
Since [math]x/4=13[/math] we multiply 13 by four to get [math]52[/math] and the
[math] 4's[/math] of the equation cancel each other out and we are left with:
[math]x=52[/math] This equation is true for [math]13(4)[/math][math]+2[/math] is equal to [math]54[/math]

I messed up the term 2=+x/4=15 in the text, what I meant to say:
[math]2+x/4=15[/math]

Divide over the t, so 2x = 3/t; then subtract the 2 and multiply by 3 so we get 3x = 1/t - 2; and finally, subtract the 3 so we get a final answer of x = 1/t + 1.

Fuck, I'm a mongoloid, if I subtract the 3 it keeps going negative. My bad, the final answer is x = 1/t - 5

this guy gets it

We shall use proof by contradiction.
Assume 2 + x is not 3, this implies in particular that 2 + x is not a multiple of 3. So 2 + x = 3k + 1 or 3k + 2 for some k. We can assume without loss of generality that k is a complex number with non-zero imaginary part, but then either 3k + 1 or 3k + 2 would have non-zero imaginary part as well. This implies that x has non-zero imaginary part, which is impossible since x is a letter. Hence, 2 + x = 3. QED

Veeky Forums is a place for value exchange - you create a thread that offers value for me (usually by making me laugh) to respond to it, and I in turn give back value by responding to it and bumping it. When you create some autistic homework question thread you're not offering me any value at all - you're only trying to leech value from me and have me waste my time solving your homework. When a guy makes a shitpost/joke thread, he's offering me value by making me smile so I return the favor by replying to it

>tl;dr see

Your x-axis us backwards.

I just throw the 2 over to the other side making it 3-2 and I'm done.

You can't do that. He can't do this, can he, Veeky Forums?

This was so retarded I burst out laughing

What are 2 and 3? What values is x allowed to range upon? How is the plus operation defined? How is the equals relationship defined?

throwing is specifically disallowed in the math literature
he must pass the two to an open and available receiver on the other side so long as the receiver is not passed the critical strip
all passes must also be done in the forward (left to right) direction

equals is always interpreted the same way no matter what model of set theory you're in

i'm p sure this problem will break it.

fucking math majors.
2 + x - 2 = 3 - 2
x = 1

Put 3 finger in your ass then out 2 finger, number fingers in you ass is the answer.

Define 2, +, x, = and 3.

I just learned to throw shit over and remember to change the sign when I do so.

Not sure but I think x equals 1.

This, my man.

given the form Ay'+By=C we start with a characteristic equation and solve a system

first assume that y' = 1 & y=1, so our equation looks like 2y'' + xy = 3

we rewrite as 2y'' + 3y = x due to the law of constant coefficients

characterisric 2s+3=0, s =-3/2

Ae^-3/2t + x = y =1
y(0)=1
y'(0)=1

d/dt y = d/dt Ae^-3/2t + X = y' = A+ x = 1

A + x = 1

unsolvable system, sorry bro

Everyone who replied here unirionically should consider suicide.

Instructions not clear, got my dick stuck in a car engine

Nailed Veeky Forums

Taking the natural definition of equality which should be considered here, we see that we want the ideal I = (2 - x) = (3) in Z[x].
Since Z[x] is not a PID, there doesn’t necessarily exist r in Z[x] such that (2-x, 3) = (r). Exercise to show it doesn’t exist in this case.
Thus 2-x != 3

Nigga is easy