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/sci/ - Science & Math

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9655477 No.9655477 [Reply] [Original]

Is it possible to solve for x, y, and z if you already know a and b? Explain why or why not.

>> No.9655480

>>9655477
>2 equations
>3 unknowns

>> No.9655488

>>9655480
thats what makes it tricky, I know in theory there should be infinitely many solutions, but im wondering since it isn't linear algebra that there might be a workaround

>> No.9655491

>>9655488
Have you ever been in high school?

>> No.9655496

>>9655491
Lmao, im in college for CS. Currently working at a start up and my boss asked me if this question was possible, so im trying to see if anyone else knows because I personally believe its unsolvable

>> No.9655497

>>9655496
>im in college for CS

makes sense.

>> No.9655498

>>9655497
r00d

>> No.9655502

>>9655497
wanted to say this desu kek

>> No.9655503

>>9655497
Nvm, he just solved it assuming x^2 + y^2 + z^2 = 1, a = tan(theta), and b = tan(phi)

>> No.9655504

>>9655502
just curious but are cs majors a meme here? First time ive been on this board but im assuming more than half the people that go on this are in cs of some sort

>> No.9655515

>>9655503
>makes something up
>it works now wahoo
Are you sure you're not a physics major?

>> No.9655518

>>9655515
haha yeah cuz everybody who solves a differential equation is a physicist

>> No.9655519

>>9655515
lol, i didn't solve it. Shits solved on some whiteboard in front of me. Tbh i have no clue how he did it because my trig skills are lacking. I'm not even sure what its for, he just asked for my help because i took calc 2 more recently then he did

>> No.9655529

>>9655518
>missing the point this hard
le CS major

also
>differential equation

>> No.9655539

>>9655529
no apparently you missed the point
assuming something is present in a huge amount of fields and especially including almost every differential equation
if you didn't think about differential equations as assuming the solution you know almost nothing about them

>> No.9655546

>>9655518
>differential

>> No.9655563

>>9655497
Oof

>> No.9655616

>>9655539
>being a CS brainlet
I love it.

YOU missed the point: randomly assuming some piece of information (like x^2+y^2+z^2=1) to make your model consistent is a common criticism of physicists.
Also, these are not even differential equations (i.e. an equation that relates a function to its derivatives).

>> No.9655621

>>9655616
thanks tips i had no idea that wasn't a differential equation
thank you for enlightening me
>tfw take unsolvable equation
>make assumption and can do something with it
>alternatively leave it alone as unsolvable

hmmmmmmmmmm
still not a cs major but alright i guess engiequeer :^)

>> No.9655713

>>9655480
>>9655491
>>9655616
>all these faggots not knowing how to solve equations with number ranges
and i thought pol was retarded, kek

9 possibilities with a specific solution for each
1. A<0 B<0
2. A<0 B=0
3. A<0 B>0
4. A=0 B<0
5. A=0 B=0
6. A=0 B>0
7. A>0, B<0
8. A>0, B=0
9. A>0, B>0
1.
y=b*sqrt(((a^2+1)x^2)/a^2)
z=-x/a
2. y=0, z=-x/a
3.same as 1
4. not solvable since im assuming we are working with the real numbers range because im a lazy fuck
5. x=0, y=0, z can be anything in the real range except 0
6. x=0, y>0, z=+ or -sqrt(y^2/b^2)
7. same as 1.
8. x!=0, y=0, z=-x/a
9. see 1. motherfuckers
see wasnt that hard

>> No.9655720

>>9655713
You need to solve for x as well, brainlet

>> No.9655727
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9655727

>>9655720
i did retard
just because it isnt always a constant doesnt mean it isnt solved for

>> No.9655749

fpbp

>> No.9655754

>>9655713
>>9655727
Real brainlet for taking this shitty bait

>> No.9655755

>>9655727
X IS NOT SOLVWD ON PDOBLEM 1

>> No.9655762

>>9655755
>x=x
solve this equation buddy
1 equation one variable kek

>> No.9655763

>>9655755
x!=0

>> No.9655780

>>9655762
Then it's solvable.

Two equations with three variables isn't always solvable. In this case it isn't.

>> No.9655787

>>9655497
Other CS major here, you can find manlets in any major.

>> No.9655790

>>9655780
>then its solvable
Then what is the value of x fucking brainlet

>> No.9655791

>>9655790
x

>> No.9655793
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9655793

>>9655780
>In this case it isn’t

>> No.9655794

>>9655791
then why is this solution wrong double brainlet
>>9655713

>> No.9655796

>>9655780
>what are functions

>> No.9655805

>>9655794
Because they are expressed in terms of the other variable. The reason why x = x is solvable is because it is the only independent variable in the equation, so plugging anything in solves it. In the case of the other equations though, a function like (>>9655796 ,>>9655793) these brainlets would use is based on the other variables and just simply would not work. In fact, the dependent variables are supposed to be A and B, while the independent variables are the three other letters that are the inputs for the two functions. I can put whatever I want for the dependent variable, you see, but it brings me no closer to defining what the independent variables are.

In fact let me give you an example.

In the case of two equations and two variables, I could say

A = x/y
B = 2y + x

In this case, two independent variables and two F(x,y)'s. The only reason it's absolutely solvable 100% of the time is because it is discrete (meaning the system of equations has the amount of equations equal to the amount of independent variables). It's why sociologists/economists like to try to do this also. To the point where you can read Pareto, and he will have a system of thousands of equations, but he doesn't even need to solve for the variables to know that it is a discrete system of equations, because the equations equal the number of unknowns

>> No.9655812

>>9655805
>babbys first linear algebra lesson
>I can put whatever I want for the dependent variable, you see, but it brings me no closer to defining what the independent variables are
holy kek the absolute state of /sci/

>> No.9655815

>>9655812
Just saying, you've defined it through other variables. That's all. You've not even given me an approximation. That other anon posted variables that were functions of other variables, within parameters defined for the dependent variables.

That's about as useful as punching yourself in the nutsack to get over a cold.

>> No.9655819
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9655819

>>9655815
>ranges are useless, if you dont have a constant for a variable you should just give up
just stop and think for a minute

>> No.9655827

>>9655819
In this case, the range is only useful for the independent variables, not the dependent variables, brainlet. The only situations in which there is a range are those in which you've defined an independent value.

Maximum pseud.

>> No.9655829

>>9655815
>>9655827

holy fuck
let me demonstrate to you why what you said is absolutely retarded

You have to build a block of x*y*c with following properties
1. the volume has to be 1000
2. the base has to be 50

so what does the engineer do? he makes an equation system
I. x*y*c=1000
II. x*y=50
so what does he do? he fucking solves it because he is not a retard like you
x=50/y
z=20
from there on he can decide whatever the fuck the variables but still meeting the requirements

>> No.9655831

>>9655477
Let x, z be any numbers which satisfy b=-x/z. Any other solution pair x', z' must be Z-multiples of x and z. In fact, there are infinitely many x', z' pairs, since ax/az=x/z.

When trying ax, az in the top equation, you get (ax)^2+(az)^2=a^2(x^2+z^2). So y must be different by multiples of a as well.

There are infinitely many solution x,y, and z but they are scalar multiples of each other. So there's in a sense a unique solution.

>> No.9655841

>>9655831
This seems correct, but I wouldn't call this system solvable by any means.

>>9655829
In your example there are two equations and two unkowns, hence the system is discrete and can be solved. You have proven nothing.

Why would x*y be two variables if it only shows up in the system as x*y.

Premium pseud.

>> No.9655845

>>9655841
>Why would x*y be two variables if it only shows up in the system as x*y
Because width and length are teo different values you absolute moron

>> No.9655856

>>9655841
Yes, it's not solvable in the typical sense but the solution is a linear subspace, which is pretty nifty. It depends on what the equations are for that determines where that set of solutions is still useful or not

>> No.9655862

>>9655856
it does not depend on that actually.
if the solution subspace is the size of the vectorspace it operates in, its still a solution by any means. just because it is theoretically useless, doesnt mean it isnt a solution.

>> No.9655864

>>9655845
Not in this case. But lets assume you are saying that x and y are two different independent variables.

You have created an unsolvable system. Because the amount of knowns are greater than the equations, he cannot directly determine the value of x or y, because x is a function of y and y is a function of x. You would see the problem? If you decided to try and solve for this sort of thing it would be a mathematical petito principii, you'd be saying that x is such a value because y is, or that y is such a value because x is.

That's not how you solve a system of equations in mathematics.

If you gave me what z equaled, I could not solve the system.

>> No.9655877

>>9655864
>if i use smart words maybe then i will be right
give me a simple system with more equations than variables and i will show you that there is a solution.

>> No.9655879

>>9655877
>he doesn't know what a petito principii is

It's pretty much a mathematical term. Charles Dodgson used it when justifying propositions from Euclid's first two books.

>> No.9655884

>>9655877
x=y
-x=-y
2x=2y

>> No.9655890

>>9655884
the solution is x=y. it is a linear subspace of the two dimensional R^2 space. how is this not a solution to you

>> No.9655900
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9655900

Nope, there's quite a lot of points that overlap.

>> No.9655903

>>9655900
thank you. now please explain to this retard >>9655864
that a solution set doesnt have to be a one dimensional point in space

>> No.9655905

>>9655884
>>9655890

That's not me (I'm the other guy). I'm sorry this brainlet tried to butt in...

Try this

x = y/2
x ^(2) = y
sqrt (y) = x/50

I've got you now, I'll take my money please.

>> No.9655906

>>9655903
>He says no which is what I said originally
>tries to say he's answered the question

The point is, neither he nor anyone else can solve this system, and it's because

F.
P.
B.
P.

>> No.9655908

>>9655905
x=0 and y=0 holy kek

>> No.9655909

>>9655906
>if i go by my retarded convention that a solution has always to be a onedimensional subspace of the solution space then im right

>> No.9655910

>>9655908
Fuck the zero. It's not even a number.

Fine, flip one of the fractions around. There. System is unsolvable.

>> No.9655913

>>9655909
You mean a single solution? That is how you solve problems, after all. This is basic algebra at this point, anon.

>> No.9655914

>>9655910
yes now that is true. the system has no solutions because no points overlap, not because there are more variables than equations like retard >>9655906 claims

>> No.9655915

>>9655913
>algebra doesnt have solution sets
holy fuck

>> No.9655916

>>9655914
>yes now that is true. the system has no solutions because no points overlap
Which is only true because....
>there are more variables than equations like retard >>9655906 (You) claims
Wooooooooo what do you know. You've proven me right by thinking you are proving me wrong. Incredible.

Maximum pseud achieved. No pseudery possible past this point.

>> No.9655918

>>9655916
but anon, does x=y have no solutions? do you even know what a FUCKING SOLUTION IS?!

>> No.9655919

>>9655915
A 'range' of solutions doesn't make something applicable. That's not what particle physics shoots for, that's not what economics shoots for. Nothing. No one wants a range of values for applicability.

I'm guessing you're some kind of mathematician who thinks he can use a system of equations because he can find a vague three dimensional space where the solution should be.

You've already achieved maximum pseud. We can go no further.

>> No.9655923

>>9655918
That is incorrect, since x and y are the same variable. Now if you tried to tell me that x and y were two different variables and not the same, then x = y would indeed not function, and it fulfills the stipulations we were going over before.

Please continue proving me right.

>> No.9655925

>>9655919
>No one wants a range of values for applicability.
holy kek how old are you /sci/ please help me from wanting to go to his house and kill him

>> No.9655926

>>9655480
Number of equations and variables don't matter shit for nonlinear equation.
For example the system
>f(x,y,z) = 0
>g(x,y,z) = 0
is equivalent to the system
>(f(x,y,z))^2 + (g(x,y,z))^2 = 0

>> No.9655927

>>9655926
Yeah that's why it only matters sometimes. Still, it does matter.

And if the number of unknowns is equal to the number of equations the entire system is solvable anyway, so this rule has some significance.

>> No.9655935

>>9655927
>>9655926
do you not know what a solution is? even in linear equations there is "no solutions, 1 solution or infinite solutions"
you are trying to claim thst infinite solutions=no solutions

>> No.9655936

>>9655890
That's not a solution to x and y, it just restates one of the conditions.

You need more than 1 linearly independent relation to solve for each variable. Calling your solution a "solution" when you repeat old information is trivial and a waste of everybody's time.

P.s. a set of solutions is different from a solution, retard.

>> No.9655938

>>9655919
>That's not what particle physics shoots for, that's not what economics shoots for. Nothing. No one wants a range of values for applicability.

This is the dumbest fucking things ive ever read. Do you know what a confidence interval is?

>> No.9655943

>>9655936
>a set of solutions is different from a solution, retard
>Any and all value(s) of the variable(s) that satisfies an equation, inequality, system of equations, or system of inequalities. - Solution
>Thats not a solution to x and y
yes it is. x is y ind vice versa. they are dependent on each other retard
>just because one equation was automatically the solution doesnt mean its not one faggot
if i say x=5 is that not a solution for x?

>> No.9655946

>>9655935
>you are trying to claim thst infinite solutions=no solutions
How does THAT not make sense?

No, it's a little bit more specific than that, what we are trying to claim is that a function of solutions, the function being one of the independent variables, necessarily invalidates the solution, because at some point you have to plug in a variable to determine the other, instead of determining the variables from the system.

>> No.9655948

>>9655938
see
>>9655946

Captain, I see a pseudberg on the water. OH NO.

>> No.9655955

>>9655477
Select
z = anything not zero
x = -bz
y = a sqrt(x^2 + z^2)
and you have a solution.

>> No.9655957

>>9655948
>>9655946
"solution sets are useless and dont exist"-the post

>> No.9655967

>>9655943
A single numerical solution is different from a set of solutions. You're repeatedly equivocating on the meaning of solution, and are purposefully ignoring the distinction because you're just another buttblasted 4chan retard who can't accept being wrong. Any further discussion is just going to be you arguing semantics, so just accept you're stupid and move on.

>> No.9655970

>>9655967
>sets of different sizes are a very important distinction
ah yes, and if x^2=1 happens, what do you do? kill yourself because its not one solution?

>> No.9655977

>>9655970
1 is a solution. -1 is a solution. {-1, 1} is the set of solutions in [math]\mathbb{R}[/math].

>> No.9655978

>>9655977
x=1
{1} is the set of solutions faggot.
when anyone talks about "find the solution" no one assumes "find the solution and if there are multiple, just asusme its useless, because after all it said find THE solution not set of solutions"

>> No.9655979

Jesus Fuck this thread, why, who are you people, why are you so retarded, why OP sucks so much dick I don't understand let's go home please

>> No.9656019

>>9655900
based

>> No.9656077

what a shit fest this became. wasn't expecting so many faggots to get heated over this

>> No.9656088

>>9656077
stupidity breeds discussions
>>9655919

>> No.9656134

>>9655488
of course not
you want to know why?
just try the problem with
[eqn]a=\frac{1}{\sqrt{x^{2}+z^{2}}}[/eqn]
[eqn]b=\frac{-x}{z}[/eqn]
(assuming a is positive)
You can easily solve this system of equations with algebra, or trig identities.
Now, solve
[eqn]a=\frac{2}{\sqrt{x^{2}+z^{2}}}[/eqn]
[eqn]b=\frac{-x}{z}[/eqn]
easy, right?
now, do
[eqn]a=\frac{3}{\sqrt{x^{2}+z^{2}}}[/eqn]
[eqn]b=\frac{-x}{z}[/eqn]
you see what I mean?
In fact, because y is a variable, you can choose y so that it accounts for the sign of a
and the signs of x and z only effect the second equation, so they can be selected accordingly.
As long as y is the same sign as a, it doesn't matter what y is, you can always solve the system, which means you can't determine x, y, and z based on a and b.

>> No.9656141

>>9655762
Holds for any x.
Solved

>> No.9656156
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9656156

>>9655919

>> No.9656189

>>9655488
>since it isn't linear algebra
What exactly do you mean by this?

Interesting thread so far to be sure.

>> No.9656192

>>9656189
that means hes retarded like
>>9655919

>> No.9656211
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9656211

>This thread
>The amount of autismo generated
Holy kek

>> No.9656220

>>9656134
/thread