#help-10
1 messages · Page 325 of 1
I have a question that asks to find linear combination for w=-4,2 with u=3,6 and v=3,-1
$$\begin{cases}
& 3\lambda_{1}+3\lambda_{2}=-4 \
& 6\lambda_{1}-\lambda_{2}=2
\end{cases}$$
Totalani
When I solve this im not getting the correct answer and im not sure what im doing wrong
yea
show work
sure sec
well what I did was I multiplied first one with 2
and subtracted the one below
so I got $7\lambda_{2}=-10$
Totalani
yeah
hold on a gosh darn minute
aight we good
thanks a lot ❤️
I was being too fast I guess
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What is $\frac{dy'}{dy}$ and $\frac{\partial y'}{\partial y}$? I am doing ODE right now and I amconfused about uniqueness and existance theorem
Kintiru
@strange summit Has your question been resolved?
<@&286206848099549185>
I am doing second order ode
would this not apply to second order ode?
@strange summit Has your question been resolved?
@strange summit Has your question been resolved?
first off, $\partial$ and $\mathrm{d}$ are just different notation for the same thing
bruhh
$\partial$ usually indicates that the function being differentiated has multiple parameters
bruhh
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excuse me
then you can transpose after plugging numberes
excuse me
why is it C + B - A?
i dont understand
wait
so this is my thinking
I got -5i + 3j for vector AB
and then i have vector OC which is -1i + 0j
now im thinking
vector AB must be equal to vector CD
so i have to find what to add to -1i + 0j to get -5i + 3j
so i get
-4i + 3j
for vector OD
???
yuh that should be right
think of it like solving for x
ok let me try again
why isnt it this then
the answer is -6,3
when im getting -4,3
oh cause u didnt distribute the - to the OC
oh
it would be 1j not -1j
ye
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✅
yuh
when the coords have zero in them its easier to just not include them in vector format imo
it just leaves u with the ones that u need
also dont forget the tilde under i,j,k
fair
one is in the complex plane the other is in vector plane
ohh yes
oh ive heard about that
i cant WAIT TO LEARN THAT... 😭 fake numbers
are u serious
its really easy
im assuming that ur at a little bit lower then me considering i just did that
2(OC) - 2(OG)
and complex is really easy
ohh what grade u in?
12
first time i see such thing
L
really??
is this right
2 am math sessions
ye
ok thanks
but u should keep the 2 outside
what u mean outside
like 2(x+y)
ye
ok thanks
keeps it clean
im not the right person to ask
i slept in every class pretty much
is this speshialist maths or called something diff for u
perhaps
holidays
plus kdrama
its 2:31 am
u vic?
mayhaps
bruh
what
kew is horrendous
nothing there
im moving there casue all my family live there
thats like a c tier suburb
vic is the opposite
pace of life is really slow in perth
yeh only 30% of students go through the atar pathway
in WA
- english is not counted if its not in ur top 4
and im bad at english
even if theyre not ur top 4
i think ur just not getting the right idea about atar
^ onto ur aggregate
how so
top 4 are obvi ur main ones
and 10% of the outer subjects are taken as extra
but all subjects are still scaled
oh its not like dat in WA
oh
they ignore them
nvm then
what career path u thinking of?
do u think toorak or kew is better
neither💀
bruh
damn
- the atar cut offs for uni are higher in vic
im gonna die
if u dont get the req atar u can always get in through other means
true that
but i dont wanna waste 2 years of my life suffering
just to do more work to get into the course
get the atar then🤷♂️
true
ok thanks for the help
BTW WHAT WAS THE HARDEST TOPIC IN YEAR 11 SPEC?
last question
proofs💀
what like proving congruency and similarity
ight
its only hard cause its a newly learned topic
what is that upside down U
thats not in the study design
but its good to know
cause there have been a few questions in the year 12 book that req u to know that for some reason
bruh
can laways drop in year 12 to further
so ur a sheep?
for uni?
yuh
spesh is like very useless for those engineerings
??????????????????
ou the WA one
oh its not req its either or
no they make u do a course
but ur doing all of them
oh thats dumb
u perma moving to vic?
when i can just do over 2 years
yeah unless my dad gets another job change
they wanted us to move to paris
oui oui
alright then youll have a bit more space to breathe
do u know st catherines school
u can add me if u want so we dont keep taking up a channel
yeah
ye
is it godo
good
im going st catherines brothers going st kevins
bc hes already going to an edmund rice school in perth
its okay from what ive heard
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just checking if there is a better explanation
<@&286206848099549185>
A better explanation than what?
And what do you already have?
i said according to the sum to infinity formula, for increasing values of n, r^n will converge to 0 making the formula a/1-r in which this will only occur when |r| < 1
.
tryna see if theres a better explanation for it
bruddah this is taken up
Oh sorry I’m new
np
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did you use ratio test?
So you are looking at the formula for the partial sums to n and arguing that they don't converge
That's a fine explanation
ye basically that r^n for increasing values of n will converge to 0 when |r| < 1
taking place only when |r| < 1
alr thanks
Yes
appreciate it
That's already pretty good. Faster would be to notice that the sequence itself cannot converge to 0
could you explain what you mean by that. thanks
The infinite sum can only converge if we have a null sequence
Which we don't have here
when |r| < 1 ?
Yes
srry again but what exactly is a null sequence
A sequence converging to 0
you can also find the radius of convergence
You're welcome. If you don't need this channel anymore you can close it
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hello
what is the integer on the RHS?
i didnt learn integer
is the last number on the right a 9?
its 2.9
gotcha so youre trying to find x such that it equals 2.9 right?
yeah
in your opinion where do you think we should start?
remove the sqrt(5^x) of the denominator ?
how would you like to do that
negative exponents or multiplication or another way?
perf
what can i do with this
lemme double check smthg before we move on
@lucid pendant Has your question been resolved?
@lucid pendant has your teacher taught any substitution?
sry for how long that took
wdym by substitution ?
the only way i found to do it was by u substitution
i substitue what ?
u=5^x
i cant solve it with log or idk
youll use log in the second half
but theres apparently no way to bruteforce log for the whole thing
not that i found
@lucid pendant
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help on the underlined parts
not rly sure how to explain it
@timid silo Has your question been resolved?
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hi, can someone recommend me one single resource for studying precalc and calc. level being maybe AP level or a little above
multiple resources to study the same topics end up overwhelming me when i'm new to a topic
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P(x) is a polynomial with real coefficients. P(x+a) divided by P(x-b) has a remainder of x+a
P(x-b) divided by P(x+a) has a remainder of bx+a+2
Find a+b
<@&286206848099549185>
ok so
have you considered what the quotient would be?
like turn P(x-b) into P(x+a)Q(x) + bx+a+2
and same for P(x+a)
and try to figure out what Q(x) should be
so what did you get as the quotient?
For one p(x+a)=p(x-b)R(x) + x+a and the other was p(x-b)=p(x+a)Q(x) +bx+a+2
what do you think R(x) and Q(x) should be?
I have this gut feeling where p is a first degree polynomial
||the degree of P(x+a)= degree of P(x-b)||
Since the inside (x+a) is the remainder
if it was a first degree polynomial, P(x+a) = P(x-b)(nx+m) + ..., but then the degree of P(x+a) would be higher than P(x-b), which doesnt make sense
Indeed
so the remainder for both of them should be constant
and theres a very specific constant
Indeed tried using quotient theorem
There is some value K where p(k)=0
But there wasn't any thing to eliminate k
(we actually dont need that here, and i'll explain how to get Q(x) and R(x), and we'll see why we dont need it)
Okay
so P(x) would be in the form of jx^n+kx^(n-1)+... right
Yes
Same degree but different coefficient
the leading coefficient doesnt actually change
I know that
nice
So the qx will have a 1
Yes
znsn
$P(x-b)=P(x+a)+bx+a+2$
znsn
now it would be really useful to get rid of those P(x+a) and P(x-b)
how would we do that?
I already got that bx+a+2 =-x-a
oh 💀
alright
ok put everything on one side
and you get bx+x+2a+2=0
now since we want a+b, theres a value we want to plug in for x
2
exactly
And I experienced a stroke
it's actually so bad at math it's amazing
i asked it to give me a combination of even numbers that add up to 20
once
it gave me 5+5+5+5
so 💀
fr
why u checking using gpt lmao
if ur uncertain just try to just look through your processes and check if you messed anything up
gpt is ass
Felt reliable for a moment
Found like three lapace integrals correct
it can do higher level math using info online and all but it doesnt have a "foundation" of understanding math
Ah i see
it could solve the riemann conjecture and still fail at adding 1 and 1
So it would be bad to ask him for my bio hw right?
idk how reliable gpt is in other subjects but it isnt in maths
I mean i just downloaded it so i wouldn't know
💀
ahh yes 3 + 3 = 5
what is it
P(x) is a 4th degree polynomial with real coefficient where
P(x)>=x
P(1)=1
P(2)=4
P(3)=3
P(4)=?
It was 2002 olympiad round 2 I think
Such a lovely question
interesting
idk if this is optimal but i would try solving the series of equations
I graphed the thing
a + b + c + d + e = 1
16a + 8b + 4c + 2d + e = 4
81a + 27b + 9c + 3d + e = 3
256a + 64b + 16c + 4d + e = ?
that would be way too hard lmao
The coefficient method will take so long, also youll need to know cramers rule
cramers rule 😭
whats that?
using determinant of matrices to find solutions
finding the determinants also seem long though
I used like one substitution and a graph, solved it in 5 minutes
how did u use the information that P(x)>=x
yeah that makes sense
Id plug the points to find qx now
15a + 7b + 3c + d = 3
80a + 26b + 8c + 2d = 2
255a + 63b + 15c + 3d = ? - 1
i dont think you can solve it this way lol
if we could use that we couldve just taken the difference multiple times
but we dont have enough points for that
there are no more roots
how
For x=1 and 3 qx=0
and those are double roots
k(x-1)^2(x-3)^2
Since qx is never negative it must have double roots
so P(x) = 2(x-1)^2(x-3)^2+x
Indeed
and P(4)=22
Congrats
nice problem
I used to do these like butter
ngl i fell off 💀
I was smart idk what happened
I suddenly am learning thermodynamics and losing wits
Yeah
dw i wont 💀
It was easy ngl
sorry im in hs rn and have never heard of double roots. what do they do for an equation? how can you use that to get k(x-1)^2(x-3)^2
if it touches and goes up like this, it's a double root
why is it called a double root?
but be careful of things like this cuz it might look like a double root but it's a quadruple root
the graph i sent was (x-1)^2
yeah
Yes
but
oh i just never heard it called a double root before
💀 become windows ceo
Bro no shit even teachers don't know what they are saying
idk why i decided to take ap phys c
Ap is simple
yeah but my teacher 💀
Go electric or sum
im learning ahead of my year on yt and stuff so the order in which im learning stuff is messed up lol. ik how to do synthetic division but never heard of double roots before lmao
I do that too
i havent heard of synthetic division in such a long time 😭
i havent done it in such a long time either
i actually forgot so much of what i learned last year
No such thing as division in physics🔥
Wait till they teach you cal 1-3 in 3 months
me when $\frac {q_{enc}} {\epsilon_0} $
oops
wrong direction
i dont remember if calc 3 includes mv
multivar
i only partially know partial derivatives 😭
we were too focused on suprema and infima
I cease to know those stuff lol
sup is lowest upper bound of a set and inf is highest lower bound of a set
we had to use it in order to define integrals
I know those
spivak moment
But all i ever use integrals is dynamic and heat transfer
do you only do physics in college(or uni)?
(Which i wolframalpha)
💀
interesting
bio 😭
Its so bad bro
how long are your classes lol
How you mean
1.30 hour
how many do you take in a day
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this is just a super tedious question and its getting annoying so i wanna make sure that i didnt make a mistake
Ok
i divided the 6a^3-a^2-12a-5 by x+1 and just factored it, factored everything and i have
i really dont wanna expand this
<@&286206848099549185> is there a way to do tghis without expanding
@brittle tundra Has your question been resolved?
a^2 - a -2 isnt (a-2)(a-3)
@brittle tundra Has your question been resolved?
oh so what happened was I multiplied a^2-a-2*3a^2-5a, factored it which became a(3a-5)(a-2)(a+1)/2a(a+1)(2a-1)(3a-5) from factoring the 6a^3 stuff
why did you do that?
uhhh to simplify
first thing is cancelling
I’d say find the restrictions first
wouldn't i have to factor in the first place in order to find restrictions
This is what I got
correct now you can just cancel a lot on the left
Now restrictions then cancel
oh yeah you cant cancel 0's lol
for restrictions would i have to make same gcf for the subtraction part
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✅
In denominator 2a+1 **
Why aren't you cancelling 3a-5 too?
what happens to the 2a-1
ones minus ones plus
isnt it 6a^2+7a-5
which is (2a-1)(3a+5)
right if multiplied its 10a-3a
which is +7a
oh shit
i messed up the dividion
i divided it by x+1
like the 6a^3
Sure
.close
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A quadratic function, leading coefficient is positive. x=2, y=11
x=5, y= 12
x=11, y=30
what’s f(8)
possible answers are 18,20,21,24
<@&286206848099549185>
Hello I am here to do what I can to help you out, what's the question is it the one above?
yeah
Well there is no way for me to tell you what f(8) is when you only provided two data points, x=11 y=30 and x=5 and y=12. The options for how many functions could pass through those are infinite, are you sure there isn't more information that you are missing.
Yeah I guess it makes sense since the rest is scratched all out.
????/ The big red marker scratching the next problem in the picture
wait so how would i solve number 11
You know I thougth I knew this but this looks hella confusing, algebra is not my strongest point, but just to assure, I would call for help to reassure yourself.
this is precal
can you get someone else to help me pretty please
what i notice is that from 2-5 y=1/3x 5-11 y=3x
<@&286206848099549185>
If you're willing to do hard work
assume quadratic y=ax^2 + bx + c = f(x)
put 3 points
play with them
get a b c
and put in the equation and put x=8
Or rearrange them to result in 64a+8b+c = f(8)
Like for example
You were solving those 3 equations
and got 64a + 8b + c itself while solving it
rather finding a , b , c and then putting it in 64 a + 8 b + c
It doesn't work in all questions but just if your luck subsides
so is there any way I can do this without guesstimating
@timid vapor Has your question been resolved?
explain
assume quadratic y=ax^2 + bx + c = f(x)
input (5,12)(11,30)(2,11)
Do you see 3 eq 3 var?
yeah no no i get what you are saying
Thats very smart thank you sir
may i ask what level of math you are on?
.close
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High school?
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Wdym lvl of math
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Well I just know Ive studied calculus
Alright
thank you
.reopen
I'm still confused
I have 121a + 11b + c = 30, 25a + 5b + c= 12, so 4a + 2b+ c =6, 64a +8b +c=?
so where do i go from here
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Hello, I have my geometry midterm tomorrow and I need help on a couple questions.
rate of change
Wait 23 doesnt show any line st
Which one is that?
change in y/ change in x
Rise over run gradient equation
Is it y2-y1/x2-x1
Precisely
to show the equation for the line
but if it just asks for the rate of change you just show the rate of change nothing more
oh okay so yeah you would give it on there!
Any tips for turning it into slope intercept?
You just take the slope (4)
times it by the x either first or second it doesnt matter
so (4)(3) + c = 4
12+c=4
4-12
c=-8
Alright cool, thank you.
Could I have help on this one? This one is bugging me a lot.
just make sure to use the corresponding y to each x
whats your problem with it
I don’t understand the square rooting, and how to setup the equation.
okay so the elementary pythag theorm is just a^2+b^2=c^2
Yeah I understand that part
Are you trying to understand why the equation is the way it is?
Yeah, so I put the length/ width squared + l/w squared = d
And then I add them, square root and find the multiples?
Here’s the full equation
what do you mean leangth/width squared + l/w squared
its just length^2 + Width^2 = hyptoneuse^2
Like in the place for a^2 + b^2
yes
https://www.youtube.com/watch?v=yfGtbNgcrQ8&ab_channel=MathematicalVisualProofs
watch this 1 minute video btw @rotund raptor
This is a short, animated visual proof of the Pythagorean theorem (the right triangle theorem) using a dissection of a square in two different ways. This theorem states the square of the hypotenuse of a right triangle is equal to the sum of squares of the two other side lengths. #mathshorts #mathvideo #math #pythagoreantheorem #pythagorean #tria...
but it can be used to find both the length and width too!
but you have to be careful as to not break the rules of order of operations
Thank you.
so if you are ever concerned that you may have done something wrong just put your answers back into the equation and see if it gives you the same thing
yes it gives us the length of the hypotenuse
the square of the length and the width is just to account for the height gained and distance traveled by a diagonal line
Go ahead! 🙂
Thank you
Do you have any tips for proofs?
Just know your lingo, and pay attention to detail like the 90 degree angles and such
👍
you can also back track like if you dont know where to start
so like well if I knew blank was congruent to blank then that would make this and that equal
Okay thank you very much, I have one last question.
For these, what do I look at to identify the AAS, SSS etc
The curved lines indicate the angles and the straight lines indicate the sides?
We have done it in class and I’ve found sometimes it’s not the case
Im sorry for the wait a horrid song came on
Don’t worry about it, take your time.
Sorry, I am terrible at math, probably super easy for you guys lmao.
no actually believe it or not geometry was the hardest math for me
Yeah, just trying to get through it, I’ve always struggled with math.
besides my burning hatred for such simple 2d figures that make me extrememly frustrated on competition math tests that i always seem to get wrong
we arent a judgemental space, the fact that you are reaching out and seeking help shows us all we need to know and what we care to know, I can assure you everyone in here is glad that you are seeking help in areas you dont neccesairly need but are unsure in
but anyways back to the matter at hand
Thank you, and thank you for all of the help also.
so this is aas because there are two congruent angles and one side. How I see it if I were to flip the triangle upside down itd be exactly the same as the other
Ah, so it’s like that because of the congruent angles?
And the side dont forget!!
What do you mean HL
One of them is hypotenuse leg
Oh like the example on the right
Nah, that one is side side side.
I don’t know what an hypotenuse leg one would look like.
Ohhhhh
like this
well if you are looking for if the triangles line up with their right angles and congruent sides
if it shows the hypotenuse for both triangles are congruent and some other side than its most definitely true
or if both sides are congruent than the triangles are most def congruent in that case aswell
Okay, thank you so much.
is that all you need?
yep
.close
.CLOSE
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For what values of n is n!+24 a perfect square?
@severe basin Has your question been resolved?
Use mod 7
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If we know that dim(U1+U2)=dim(U1)+dim(U2)-dim(U1 intersect U2)
then isn't this statement equivalent to
prove U1 + U2 is a direct sum if and only if, dim (U1 intersect U2)=0
Which it immediately follows that the intersection is empty
?
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What do I have to do?
Some surd stuff
Multiply both num and denominator by denominator
So 5sqrt(5)/5
U can simplyify it further
Sqrt(5)?
Yes
So for the rest I just multiply num and denominator by denominator and simplify?
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Hey, I have a question: What exactly is the dot product in vector geometry? (Also known as the scalar product? And for what does one need it?)
needed for projection
Could you allaborate?
I need to draw a picture but man i have nothing to write with right now
itd be nice if you could project that image of yours onto the screen
Mine?
no thats a joke for what herels said
this is just one of our scripts
youre at 1.6.2 you ought to wait until you actually start using it extensively to then ask questions like this
thats like learning multiplication for the first time then asking "how is it useful"
but i don't really understand the definition..
the length in red should be the dot product of a and b
only if its an orthogonal projection tho
if its orthogonal the dot product is 0 isnt it
nah you dont understand
?
I mean if we do an orthogonal projection, we get the dot product
ah i understood it
but if the two vectors are orthogonals, we definitely cant do a projection
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@warm tide Has your question been resolved?
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The vectors e1 and e2 form a base in the plane. Two other vectors f1 and f2 are chosen as a new basis. F1 = 7e1 + 8e2. F2 = e1 + e2. Express the vectors e1 and e2 as linear combination of the new basis vectors.
e1 = xF1 + yF2
e2 = vF1 + wF2
so
1 = 7x +1y
0 = 8x + 1y
0 = 7v + 8w
1 = 8v + 1w
You have to solve for x,y, v,w
Alternatively, you could either "solve" those equations for e1 and e2 (similar to a pair of linear simultaneous equations), or equivlaently convert that statement into one about matrices
Oh okay so x = -1 and y = 8?
you could also just think of the change in coordinate matrix from the first base to the 2nd (it would be the matrix with columns (7, 8) and (1, 1) and i guess going backwards from 2nd basis to 1st would be the inverse of that first matrix
I think that is too advanced for me but what do I do with x and y now?
if you saw a system like
3x + 4y = 2
4x +3y = 1
you would know how to solve it right?
and get x = ..., y = ... ?
@random olive ?
I think so
e1 = xF1 + yF2
e2 = vF1 + wF2
ok so this is exactly the same thing, you would be solving for F1 and F2 (analagous to x and y) but you don't have given numbers on the other side, you just leave the solution in terms of e1 and e2
and youll have some coefficients times the e1 e2 vectors which will just be the linear combination the problem is asking for
[other way round no: solving f1 = 7e1 + 8e2 and f2 = e1 + e2 for e1 and e2?]
hm?
Question is to express e1 and e2 in terms of f1 and f2
erm yeah i was just copy and pasting this message, guess it needs correcting
If I solve out e1 and e2 I get e1= (f1-8e2)/7 and e2= f2-e1?
so you have
F1 = 7e1 + 8e2
F2 = e1 + e2
now solve for e1 / e2 just like you would the simple example with x y, and leave F1 and F2 as unknowns, e1 e2 will just be expressed in terms of them
This algebra video tutorial explains how to solve systems of equations by elimination and how to solve systems of equations by substitution with 2 variables.
Access Full-Length Premium Videos: https://www.patreon.com/MathScienceTutor
Algebra For Beginners: https://www.youtube.com/watch?v=M...
you need e1 and e2 only in terms of f1 and f2, not in terms of each other also
refresh yourself on how to solve systems ^
8e1 and 9e2?
if you are doing a linear algebra class you better be clear on how to solve systems of linear equations 😄 go watch the video
i'm sure youve done it before and itll come back quickly if you watch it
I understand the video but I don’t know how to solve it when we have f1 and f2 unknown as well
Well when I solve for e1 I get f12 = -e1
It's similar to when there are numbers, but you leave them as they are, if that makes sense?
Okay but e1 = -1 then?
Say I give you a set of linear equations, $7x + 8y = 4$ and $x + y = 5$, you'd be able to solve that, right? (you don't have to actually solve it for me)
@unreal musk
Yes!
And the steps may be say either multiplying the second one by, say, 7 and subtracting that from the first one, or rearranging one of those equations for a variable and substituting that back in, right?
Yes right!
Cool, now, if I now change the question a tiny bit: solving
[
7x + 8y = u \text{ and } x + y = v
]
where $u$ and $v$ can be any numbers of my choosing, and I tell you to solve that for $x$ and $y$ (with those being in terms of $u$ and $v$), would you know how to do that?
@unreal musk
I don’t know, if I multiply the second equation with -7 do I multiply v with it as well?
Yep, you'd need to multiply v by -7 too, but that's the idea!
Oh okay!
So if I for example multiply my second equation with -8 to get rid of e2
I multiply f2 with it as well?
Yep, that's it
pretty much like that!
So I have this?
And I can cancel out 8e2 and -8e2
And then I combine them
Then I get this
Be very careful: f1 and f2 are distinct objects, so you can't combine them like that
Oh okay
You should get f1 - 8f2 though
Okay so I have f1 - 8f2 = -1e1
Then one small thing to get e1 by itself 
Moving over -1?
how do you mean "moving over"?
That's it, perfect 
Careful, there wasn't a negative so that -56f2 should be +56f2, but otherwise you get the idea 
Oh sorry you’re right
]