3 variable system of equations

[obsidian ermine]

I have a system of equations (am doing vector spaces so i need to find a formula for the linear combination of a set of generators). Im not sure if the vector part is clear or even useful information but I cannot for the life of me solve the system of equations. Ive watched tutorials and its so simple but every time I do it (ive tried different ways, both substitution and elimination) it does not work and is wrong. Is there something that im missing? is there an operation that you can do in 2 variable systems of equations that you cant in 3 variable ones? thanks for your help, heres the question :

u vector (2 ; 1 ; -3), v vector (3 ; 1 ; -5), w vector (1 ; -1 ; 1)

with these we have to solve this system, where X1, X2 and X3 are known numbers

2a + 3b + c = X1
a + 2b - c = X2

• 3a - 5b + c = X3

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[spare fossil]

im not sure if iam getting what you mean, im sending an image could you confirm?

[obsidian ermine]

i can send you the question as well but its in french

[flint inlet]

yes please do

[spare fossil]

is this what you are trying to say?

we havent done vectors yet and this is matrices so i am probably wrong, but ive heard matrices are just vectors and thats how you solve a system of equations in matrices

[obsidian ermine]

no its not matrices

i dont think vectors is neccessarily important tbh

[spare fossil]

do you need to find a b and c in terms of x1, x2, x3?

[obsidian ermine]

yes exactly

[spare fossil]

i made a mistake here wait

[obsidian ermine]

its not matrices anyways tho

[obsidian ermine]

nothing to do with matrices tho, i dont even understand this x)

[spare fossil]

oh so with just plain algebra?

[obsidian ermine]

i think so? try it just plain algebra and then ill see if it works

[spare fossil]

add equation A and equation B, that would remove a "c" term

[obsidian ermine]

when i check if the vectors are "libre" (which means instead of x1 x2 x3 i equate the equations to 0 its fine i just cant do it like this idk)

[spare fossil]

and then add equation B and equation C, that would again remove a "c" term

[obsidian ermine]

c = X1 + X2 + X3

[spare fossil]

what?

[obsidian ermine]

wait idk how i got that let me check my notes

yeah

add all the equations

all 3

that works right?

[spare fossil]

that would still have a term in "c"

[obsidian ermine]

yeah

c = X1 + X2 + X3

[spare fossil]

instead im trying to remove that c term

[spare fossil]

ah i see

[obsidian ermine]

i tried it before i think let me check

maybe not

[spare fossil]

so you can then put that as the value of "c" in equations A and B to get 2 equations with two variables "a" and "b"

[obsidian ermine]

yes i will try it now but i thought i already did

3a + 5b = X1 + X2
-2a -3b = X2 + X3

you got the same?

[spare fossil]

which equations did you use?

[obsidian ermine]

A + B and B + C

[spare fossil]

ah ok

[obsidian ermine]

wasnt that what you were proposing before?

[spare fossil]

yep

[spare fossil]

do you know how you can solve for a and b?

[obsidian ermine]

yeah im quite slow tho lol ill try to hurry up a bit

[spare fossil]

no worries

[obsidian ermine]

I have b = (-5X2 - 3X3 - 2X1 )/7

we can check that rn if we plug in X1 = 6, X2 = 2, X3=-7

it should give 1

[spare fossil]

im, not sure if that is correct, i got b = 2X1 + 5X2 + 3X3 maybe you could look though your process once?

[obsidian ermine]

yours works

where is my mistake tho i dont get it

ill send idk if you can decipher what im doing but yeah

[spare fossil]

okay so what i prefer to do is to make the coefficients of one of the variables to match

[spare fossil]

for say a, i would
multiply the first equation with 2,
multiply the second one with 3

[obsidian ermine]

its easier than working with fractions thats true

[spare fossil]

it was -3b/2

when you multiply with 3 it becomes -9b/2

[obsidian ermine]

i multiplied by 2 tho

are we talking abt the same place?

[spare fossil]

last line

-3/2 b

that should be -9/2

[obsidian ermine]

oh

ok but i still have a sign error somewhere because i get -1 and not 1

[spare fossil]

ah so

[obsidian ermine]

yeah im stupid i switched sides and didnt put -

[spare fossil]

A + B = C say

then B = C - A

[spare fossil]

yea

[obsidian ermine]

so its solved thanks for your help

[spare fossil]

nws yw

[obsidian ermine]

ig i just have to hope not to make any stupid mistakes lol

i tried this like 5x and got it wrong every time idk if it was just random stupid stuff or if i was doing smthn wrong or what

but the strat is to eliminate 1 variable and then just solve the 2 variable system?

[spare fossil]

yea thats it

[obsidian ermine]

ok thanks a bunch

[opal frigate]

@obsidian ermine has given 1 rep to @spare fossil

[obsidian ermine]

how do i close?

[spare fossil]

+close

[obsidian ermine]

+close