help-42

[marsh valley]

,w extrema of x^2+4y^2-12xy where x>=1 and x^2 + 4y^2 <=4

[potent lotus]

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[marsh valley]

🎊

[simple musk]

I appreciate it

[marsh valley]

You don't seem bad at it.
You didn't really need much help at all doing the work in this. I feel like you might just seek help too quickly if anything. Not that seeking help is a bad thing, just you might benefit by trying stuff a little bit more by yourself before asking for help so you avoid needing external help.

Try also to understand the big lines of problems. There are only so many things you can be asked in multivar calculus or other classes, and they are usually the same "main" types of problems, give or take some extra steps to word it in a more familiar setting.
For instance like I said earlier, optimization problems on compact regions like this are breakable into 3 steps : find the candidates on the boundary, find the candidates in the interior, pick the biggest and smallest of them.

[simple musk]

I will try to follow your advice mate, yeah I ask a lot of questions, but regarding this topic I think I got the gist of it , but now I would benefit from doing it with parametrization instead of lagrange multipliers, but let me try doing this same exercise first by myself and if I get stuck I post until where I left, thanks for the advice and the help really mate 👍

[marsh valley]

It was my pleasure. Feel free to ping if you get stuck! I'll try to help if I'm around.

[simple musk]

.solved

[calm coral]

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[normal frigate]

can I use a test series to test if the original series is decreasing for the alternating series test?
I used limit comparison test to test for absolute convergence
I don't want to differentiate the original series and I'd rather just differentiate 1/sqrt(n)

[leaden thunder]

There's no reason to differentiate the series

Find a series to compare it to

[normal frigate]

I have to find if it's conditionally convergent

I already used 1/sqrt(n) for LCT and found that it's not absolutely convergent

I got L=1/2

and 1/sqrt(n) diverges by the p-series test so my original function must also diverge
but since it's an alternating series, I must test for conditional convergence

[unkempt drift]

by similar reasoning, $\frac{1}{\sqrt{4n + \frac{1}{n^2}}}$ is decreasing

[potent lotus]

south

[unkempt drift]

well for n >= 1

just differentiate 4x + 1/x^2 to find the only turning point (which is a minimum considering the behaviour as n -> 0 and n-> infinity)

[normal frigate]

why did you split it up into two terms?

and I can't really see how you did so ;-;

[unkempt drift]

from n/sqrt(4n^3 + 1)

divide the numerator and denominator by sqrt(n^2)

[normal frigate]

to find the interval of conditional convergence?

[unkempt drift]

yes, the alternating series test is the right idea

if the conditions of the test are satisfied, you know the series is conditionally convergent

[normal frigate]

I'm not sure if I understand what you meant by this
the intervals in which the denominator is increasing is when it's conditionally convergent and when the denominator is decreasing, it's divergent?

[unkempt drift]

not at all

you want to show $b_{n + 1} \le b_n$ for all $n \ge 1$

[potent lotus]

south

[unkempt drift]

so we just need to check if n/(4n^3 + 1) is always decreasing for n >= 1

[unkempt drift]

then we know this for sure

[normal frigate]

so you'd just find the derivative

but I don't want to find the derivative of the original function

[unkempt drift]

yes, but there's a smarter way to take the derivative

[normal frigate]

okie

[unkempt drift]

so do you understand how this is the same as the original function

[normal frigate]

yes

completely identical

[unkempt drift]

right then, we need some function properties

the square root of an increasing function is always increasing

the reciprocal of an increasing function is decreasing

[normal frigate]

yup

[unkempt drift]

so that means

if we can show that 4n + 1/n^2 is always increasing for n >= 1

[unkempt drift]

that will satisfy this condition

[normal frigate]

so if the denominator, aka "4x + 1/x^2" is decreasing, the function is increasing (meaning ast wouldn't be satisfied) and it's divergent in that interval

if the denominator is increasing, the entire function is decreasing which means that ast is satisfied and it's conditionally convergent

idk if this is supposed to be decreasing or increasing

it's increasing

so the entire function is decreasing and it's conditionally convergent

ohh😭

since x<=1

[unkempt drift]

well actually even simpler, if 4x + 1/x^2 isn't decreasing, then 4x + 1/x^2 can't go to 0

so that's just the divergence test

[unkempt drift]

yep!

[unkempt drift]

this is actually the sign diagram method I used here to show it's a local minimum

[normal frigate]

were you saying that I can use 1/sqrt(n) to test if the original series is decreasing or not?
I got 1/sqrt(n) from the limit comparison test and L=1/2 so they both converge/diverge. I'm not sure if I can use said test series for AST.
(-1)^n-1 * 1/sqrt(n)

[unkempt drift]

no, you can't use 1/sqrt(n)

using 1/sqrt(n) is only considering the end behaviour as n -> infinity

[normal frigate]

my test series only shares convergence/divergence with my original series and notihng else?

ah

[unkempt drift]

so actually you can use 1/sqrt(n) to show the limit is 0

but not that it's decreasing

[normal frigate]

how come?

I know that nth term test and behavior of a series are related in that lim n-> 0 ≠ 0 means that it diverges
but lim n->0 = 0 is just inconclusive

[unkempt drift]

if you take the limit to infinity, you get $\frac{1}{\sqrt{4n + 0}} = \frac{1}{\sqrt{4}} \frac{1}{\sqrt n} \to 0$

[potent lotus]

south

[unkempt drift]

it's inconclusive for the nth term test

but this is a different condition, this is the one for the alternating series test

[normal frigate]

okay, that makes sense

[normal frigate]

tysm, have a great day

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[normal frigate]

disclaimer: I have no experience with proofs

how do I prove that this function is f(x)=0 only when x=0

[normal frigate]

the only thing I have so far is:
for x>0, as x approaches infinity, f(x) approaches infinity.
I'm not sure if that helps at all.
maybe I also need to prove that it's always increasing for x>0

[unkempt drift]

yes, you need to show it's always increasing

well, before you take the derivative of f(x)

note that f(x) is bounded below by 2 * 25^x - 2 * 16^x

ah actually the binomial theorem will do it

if you try using the derivative you'll get something in terms of the Lambert W function

[unkempt drift]

hmm, okay you can't generalise from when x is an integer to when x is a real number

[calm coral]

@normal frigate Has your question been resolved?

[normal frigate]

this is what symbolab gave me but I can't see how the derivative relates to the lambert w function

[unkempt drift]

if you try solving for g(x) = 0 you'll need the Lambert W function

which is a pain

[ancient thistle]

rewrite as u^3(2u^log2(25/8) - 1 - u), u = 2^x and prove that 2u^log2(25/8) - 1 - u passes through 0 once

[normal frigate]

I heard that you could use rolle's theorem

[normal frigate]

idk where to go from here tbh

I've only taken up to calc 2 ;-;

[keen flower]

i'm just thinking... let's say we know e^x and ln(x) are strictly increasing
then we can prove a lemma: for all positive reals a,b,x : a^x > b^x <=> a > b

the proof would use: a^x = e^(x ln a) > e^(x ln b) = b^x

this lemma can be used to show 2*25^x - 8^x - 16^x > 0 for x > 0

i think you can prove another lemma: for all positive reals a,b and all negative reals x : a^x > b^x <=> a < b
then 2*25^x - 8^x - 16^x < 0 for x < 0

[calm coral]

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[cerulean sigil]

With these partial derivatives, is this sufficient to show that f(x,y) is not differentiable at (0,0)?

[worthy juniper]

I think not. While it is true that the partial derivatives are not differentiable at (0,0). Remember what is stated in Criterion for differentiability

If partial derivatives exist and are continuous at a point, then the function is differentiable at that point.

However, the same cannot be said in reverse. So, you simply proven that the function may not be continuously differentiable but not the fact it is non-differentiable

To prove this definitively, you can try using the limit definition of differentiability

[cerulean sigil]

I showed that the total derivative is not continuous though. Is that not different?

[worthy juniper]

It is in a way.

When we say differentiable, it mostly means a tangent plane exists at that point

But for continuously differentiable, it means the derivative exists and the derivative behaves smoothly at that certain point

You proven it is not continuously differentiable

But take note it is still mathematically possible for a function to be differentiable even if the partial derivatives are discontinuous

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[winter elbow]

looks like some binomial expansion stuff

$(x^2-\frac1x)^{100} = \binom{100}{0} (x^2)^{100} - \binom{100}{1} (x^2)^{99}\frac1x + \binom{100}{2}(x^2)^{98}{\frac1x}^2...$

So here's what I've observed

[torpid canopy]

your binomial coefficients are off

should start from 100C0

Also, every other sign should be negative

[winter elbow]

mb, tysm

[potent lotus]

Acknowledged Scumbag

[winter elbow]

Alright so
x^200 > 100 pick 0
x^197 > -100 pick 1
x^194 > 100 pick 2

[torpid canopy]

I think this would be easier if you wrote the expansion as a summation

using the binomial theorem

[winter elbow]

but how would that help if I just want to find the coefficient of certain variable

[torpid canopy]

Well generally we have

$(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^{k} y^{n-k}$

[potent lotus]

Ari

[winter elbow]

I see, that makes sense

Alright, I'll do the rest myself

thanks for pointing out the mistake

appreciate it

.close

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[fading sparrow]

is my method mathematically sound, and are there perhaps quicker ways to show this?

[fading sparrow]

prove that it is impossible to have a cuboid for which teh volume, the surface area and the perimeter are numerically equal (the perimeter of a cuboid is the sum of the legnths of all its twelve edges)

[vagrant dune]

this seems okay. proof by contradiction.

[fading sparrow]

okok ty

.close

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[spare chasm]

I simplified and reached here

[calm coral]

@spare chasm Has your question been resolved?

[spare chasm]

<@&286206848099549185>

[calm coral]

@spare chasm Has your question been resolved?

[vagrant dune]

hi do you know the value of trig(pi/18) as a standard result?

id use that if i were you. cant think of any other way lol sorry

[upbeat venture]

the only thing I can think of that isn't hand wavy is to bound it between 2 integers and use some logic to find which it's closest to

[calm coral]

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[wet venture]

Some dude in my class gave me this and idk how to solve it

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[wet venture]

Ig we have 0<an+1/an<bn+1/bn<1

[wet venture]

But I'm stuck

[jaunty leaf]

@fierce pebble

[west cosmos]

why stricly under 1

[wet venture]

Not sure about strictly

[west cosmos]

an+1/an=?

[wet venture]

Idk

[west cosmos]

1+(1/an)

[wet venture]

No

Idk how to say it in english

It's like a function with n€N

Like a0,a1,a2...

A sequence

That's the word I think

<@&286206848099549185> ?

[calm coral]

@wet venture Has your question been resolved?

[wet venture]

Nope my boy, still wondering

[calm coral]

@wet venture Has your question been resolved?

[true falcon]

Hello, what do you need?

[wet venture]

So i have this and i don't really know how to solve it

[wet venture]

.

a and b are sequences

[wet venture]

Well ig i'll never have the answer

.close

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[quiet fractal]

where should i start?

[worthy juniper]

Right. So, you're supposed to make a mathematical expression out of this

[quiet fractal]

ohh 😅 didn’t catch that part of the problem

[old falcon]

I don't think you need to do that

[quiet fractal]

?

[worthy juniper]

Soz think of it like this, aren't you making 5 similar terms, then what do you do to 5 similar terms?

[old falcon]

do you know what like terms means here?

"like terms"

[quiet fractal]

well the lesson is combining like terms

(sorry i should’ve said that)

[old falcon]

no it makes sense

[worthy juniper]

Well, it's for visualization but it's not necessary

[old falcon]

yea maybe its a good idea

can you create an expression with 2 like terms?

[old falcon]

its not totally necessary I'm just curious because most of the problem is understanding what the definition is telling you

[zealous mulch]

I have no idea what they mean by this

[old falcon]

It's just a definition question, i think

[quiet fractal]

it’s alright guys, i’ll do this. i’m sorry for confusing y’all. i’m not great with wording what i want to think

have a good day

.close

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[old falcon]

No no wait up

[old falcon]

We can do an example, if its helpful?

I'm not confused I just wanna help

[quiet fractal]

maybe yeah, that might help a bit

[old falcon]

.reopen

[calm coral]

✅ Original question: #help-42 message

[old falcon]

so I can make an expression with 2 terms here

$5x+4x$

[potent lotus]

jan Niku

[old falcon]

all of the terms in this expression are like terms

[quiet fractal]

ohhh

i see what you mean

[old falcon]

by that i mean, they all contain the same variable

and its to the same power

we can create others

like $5+4$ is an expression with 2 like terms

[potent lotus]

jan Niku

[old falcon]

so is $9x^2 + 17x^2$

[potent lotus]

jan Niku

[quiet fractal]

i’m not sure if this is what my teacher wants me to do

[old falcon]

the important thing about the terms being 'like terms' is that they can be combined

[old falcon]

no, this is where i mean i think its just a definition

if you can identify a group of terms are all like terms, then you can combine them

it doesnt matter how many there are

[quiet fractal]

i know from other problems that it can’t be one number or something like that

[quiet fractal]

ohh

[old falcon]

so whats important is that you know that 'like terms' can be combined

as long as they are all 'like' to eachother

you might have $5x + 2x + 9y + 17y + 2y$

[potent lotus]

jan Niku

[old falcon]

these cant all be combined together, even though they can all be combined into some other terms

but i think the question is just implying they are all like to each other

idk if that helps or gives away the answer too much

[quiet fractal]

i can type out the question if it helps

[old falcon]

isnt it just in the picture?

[quiet fractal]

yeah but sometimes people ask for me to type out the question 😅

[old falcon]

thats crazy

[quiet fractal]

😅

[old falcon]

do you get what the answer should be then?

[old falcon]

I'm sorry if someone here made you do that

[quiet fractal]

yeah i gót ít. what you gave was a good starting point

that was my question from the beginning 😅

what a good starting point would be

[old falcon]

well hopefully it explains the whole problem

but if you want to think more thats fine

[quiet fractal]

yup, it definitely did. math wording sure sucks sometimes

[old falcon]

https://www.mathsisfun.com/algebra/like-terms.html

these definition questions can be really weird sometimes

[quiet fractal]

yeah 🥹

[old falcon]

feel free to ping if you run into more issues

[quiet fractal]

alright, thanks

i’ll close rq

[old falcon]

[quiet fractal]

.close

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[potent harbor]

hi

[potent harbor]

i literally do not understand this, i missed class and i cant get this to click..

its apparently Linear Programming: The Graphical Method

but

it looks like ancient texts to me

i know i have to find the 4 corner points..

i just dont know how.

[stark breach]

see the conditions in "subject to"?

they're all (basically) lines, right?

try drawing them

[potent harbor]

wym twin

[stark breach]

yk how you can describe a line as $ax + by = c$ for constants $a, b$ and $c$?

[potent harbor]

im a bot , like this ?

[potent lotus]

that_one_gal9

[stark breach]

seems right, but which condition is that?

[potent harbor]

2x+7y

i think..

[stark breach]

oo dont do that one yet! thats the last step

[potent harbor]

OH

[stark breach]

yeah!

[potent harbor]

ok i know i js have to find corner points

but like

idk where to start

[stark breach]

the conditions!

[potent harbor]

what is the conditions

i missed this whole lecture

and dont know anythign

[stark breach]

"subject to ..."

[potent harbor]

oh..

bye

[stark breach]

huh?

[potent harbor]

what about the x > 0 and y > 0

[stark breach]

also conditions!

[potent harbor]

what do i do w those

[stark breach]

draw them out too?

basically each of those will give you an area in the xy-plane

[potent harbor]

oh wait its clicking

[stark breach]

and for all of the conditions to be true (at a given point), all of those areas need to overlap (at that point)

sooo draw out the conditions first

[potent harbor]

okok

[calm coral]

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[blissful storm]

I haven't been able to come up with anything

.close

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[blazing hinge]

when doing gradient descent method and you get a quadratic for h'(t). which t* do you pick if there's 2 solutions to h'(t) = 0?

[calm coral]

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[keen surge]

A rectangle is inscribed in an isosceles triangle whose sides have length 5, 5, and 6. One side of the rectangle lies along the base (the unequal side) of the triangle. What is the greatest area that such a rectangle can enclose?

[keen surge]

!status

[calm coral]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[keen surge]

2

i'm clueless about how to give variables to the edges of the rectangle

i doubt that would even take me to somewhere

[fiery dove]

I would do this

so the side lengths are y and 2x

[finite oasis]

Hint: try to relate the smaller isosceles triangle formed at the top of the original triangle by the rectangle with the original triangle

Using similarity

[keen surge]

copy

i'll try that

and let you know

[finite oasis]

Oki

[keen surge]

h = 4k

well 😄

guess i can solve it know

thanks

is there a point i'm missing

or should i just go ahead and start

[finite oasis]

What even is k

[keen surge]

5a = 3b

a = 3k

b = 5k

[finite oasis]

Ok the similarity coefficient

[keen surge]

ah yes

that one

alright

thanks!

.close

[calm coral]

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[keen surge]

!status

[calm coral]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[keen surge]

4

u = ln (secx + tanx)
answer: 2 sqrt(u) + c

how i'd do?

[short pumice]

,w differentiate 2sqrt(ln(secx+tanx)) wrt x

[potent lotus]

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Requested by leftover5270

[short pumice]

good job

[keen surge]

thanks!

.close

[calm coral]

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[calm coral]

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[thick sigil]

i currently feel like i have no idea on where to begin
i know i have to do something involving synthetic divsion maybe?

[glad parrot]

Try some divisors of 8 as root

[warm warren]

Try 1

[thick sigil]

so things like 1,2 and 4?

[warm warren]

Or -1

[thick sigil]

• or - right

[warm warren]

Yes you can try the factors of 8 but try 1 or -1 first because they are fast and easy to do

[thick sigil]

alright

so using synthetic divsion, that brings me to remainders of

3 10 4 -8 and 0

was that the right way to try 1?

[warm warren]

What are you dividing by

Oh

No no no

You just plug in 1

[thick sigil]

oooh

[warm warren]

3 + 7 - 6 - 12 + 8

You immediately see this is a root

[thick sigil]

so i didnt have to use synthetic divsion to find that it would give me a 0

plugging it in simply can work

[glad parrot]

Yes

[warm warren]

Plugging always works

But you can clearly see that trying x = 8 is not the easiest thing

But for x = 1 it is super fast

[thick sigil]

would i just be plugging in factors of 8 and 3 until something doesnt work, or is there an efficient way to know?

that might be a dumb question, i dunno if that made sense

[glad parrot]

-2 is also a root by the same check

[warm warren]

I’d try -1 next to see if I can get another easy root without doing factorisation first

Yeah I’d try -2 -1 1 2 first

After that it’s too hard

If I find something then you can do polynomials long division to get the resultant cubic or quadratic

Those are far easier to deal with than quartics

[thick sigil]

alright yeah -2 is a root

so im confused where i go from here, there isnt another easy root right?

[sleek vessel]

Now that you have two roots, you're left with a quadratic after you divide

You can factor that manually

[warm warren]

You can be clever and expand (x-1)(x+2)

Then divide the problem by that

So you only do long division once

[thick sigil]

what would that look like?

[warm warren]

You try it and show us

[thick sigil]

lemme screen shot what i had before at first

So i had this, im confused by the division im being told to do

[warm warren]

Did you do this

[thick sigil]

how do i expand it?

i might be overthinking it

[warm warren]

You are

It’s just distribution

(a + b)(c + d)

[thick sigil]

x^2 - x - 2?

[warm warren]

Hmm

• 2x - 1x is not -x

[thick sigil]

oh wait

thought it was otherway around mb

@warm warren

[warm warren]

Yeah

Now divide your problem by this thing

[thick sigil]

f(x) = 3x^4 + 7x^3 -6x^2 -12 + 8

x^2 + x -2
?

[warm warren]

Yup

[thick sigil]

if i am being honest, i do not know how to divide this by that

[warm warren]

Why not

[thick sigil]

drawing blank on it

[warm warren]

Can you divide it by x-1?

[thick sigil]

probably yeah

[warm warren]

What’s the difference

[thick sigil]

the divisor having more to it is probably throwing me off

might be able to check a resource rq

[warm warren]

It’s just 1 more term you have to think about

I would say, do it with x-1, then look at the steps you take and apply them exactly to the bigger divisor

The beginning shiuld look something like this:
x² times what is 3x⁴? Must be 3x².
Then you write 4x⁴ under
Then you go, okay what extra pieces did I get? I should have +x * 3x² and also -2 * 3x² as well, so I should write 3x⁴ + 3x³ - 6x² under the original problem and subtract them

[thick sigil]

alright i think i get it

brb

3x^2 - 4x - 4
@warm warren ?

so so far i have
found both roots 1 and -2 by using factors of 8
y intercept would be at 8
and i have used foil on (x-1)(x+2) to find what i can divide the problem by to get to the next step

after finding 2 roots, am i consistently allowed to then expand them and divide the quartic by said expansion?

[warm warren]

,w (3x^4 + 7x^3 -6x^2 -12)/(x^2 + x -2)

[potent lotus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by frosst

[warm warren]

Yuck

,w (3x^4 + 7x^3 -6x^2 -12x + 8)/(x^2 + x -2)

[potent lotus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by frosst

[warm warren]

+4x

[warm warren]

Look here

[thick sigil]

is there an alternative to having to divide the polynomial by this btw?

[calm coral]

@thick sigil Has your question been resolved?

[calm coral]

@thick sigil Has your question been resolved?

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[calm coral]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[remote mural]

1

[uneven shell]

What do you know about similar triangles?

[remote mural]

Nothing

[inner zealot]

Are schools not teaching students about similar triangles anymore? You're the second person I've seen today say they don't know about similar triang.es. 🤔

[calm coral]

@remote mural Has your question been resolved?

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Channel closed

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as this action is irreversible, and this channel may abruptly lock.

[calm coral]

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[tall bane]

Why does he put brackets in -1?

[leaden thunder]

/-1 is awkward looking because it could be read as "divided by then take away 1" which is nonsense

/(-1) is divided by negative 1 because of PEMDAS

[uneven shell]

It's like this in general

You don't write $7 \cdot -5$ but $7 \cdot (-5)$

[potent lotus]

USS-Enterprise

[eternal shard]

pemdas mfs disagree

[uneven shell]

Also, if you will, you can think it the other way. The first step, where we are multiplying by 2, the 2 is also in brackets. So it's $/ \cdot (2)$

[potent lotus]

USS-Enterprise

[uneven shell]

But we like to omit brackets whenever we can

Wouldn't be practical to write $(((7) \cdot (x^2)) + ((4) \cdot (x))) = (7)$

[potent lotus]

USS-Enterprise

[calm coral]

@tall bane Has your question been resolved?

[velvet osprey]

stylistic convention

[eternal shard]

like people out there use -1

[tall bane]

But what does it actually change

[tall bane]

It’s really not I’ve always used it

/ is not a division

But :

Just like ratios

[velvet osprey]

ok so you have never written a computer program

programming languages regularly use / to mean division jsyk

[velvet osprey]

if you are so insistent on having it your way, then honestly: it changes jack shit and fuck all

[eternal shard]

Me when the picture says $| \cdot (-1)$ and not $/ (-1)$ kek

[eternal shard]

where | denotes command line

[calm coral]

@tall bane Has your question been resolved?

[calm coral]

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[tall bane]

We’re not fucking on a computer but doing math.

[graceful dust]

what

[eternal shard]

what

[tall bane]

.close

[calm coral]

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[sharp ice]

Hi

[sharp ice]

How the hell do i solve this

Even chat gpt gives me a different answer everytime i ask it

[rustic osprey]

In the future, please make the first message you send the actual question since that's what the bot pins.

The question is using the rules found here: https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/03%3A_Determinants/3.02%3A_Properties_of_Determinants. What I'd advise is to identify the row/column operations that were performed, then use the rules to determine how the determinant changes after each operation.

[sharp ice]

Really? Wow

Super crazy

Doctor am sick how do i get better?

[eternal shard]

eat an apple

[sharp ice]

Then i throw some medical book in his face

And i say your answer is here

[sharp ice]

This link. It helps me so much

You gave me unlimited knowledge just now

[sharp ice]

Dude you should be some mayornof a country or a president or something

[eternal shard]

What is the deal with you right now?

[sharp ice]

Am inmpressed at his helping skills

[marsh valley]

It’s just easier for everyone if you to read them and point out what you do/don’t understand than doing everything from scratch.

[clear delta]

This is an incredibly rude way to speak to people. Be kind or leave.

[marsh valley]

They not only gave you a ressource exactly about your question, but was also nice enough to provide specifics as to what you should read in it.

[sharp ice]

This message that you sent me. Incredible meaning and amazing advice.

[drifting seal]

lol

[marsh valley]

It’s just basics mate

Anyways

[sharp ice]

A resource exactly about my question. So if i ask you a rocket science theoretical physics question. You gonna hand me a highschool physics book

Cause it seems i asked an advanced question and then was handed with a link for a row operations tutorial

[eternal shard]

Do you want a book on Topology instead or what

[clear delta]

this is not an advanced question girlie

[sharp ice]

U got one?

[marsh valley]

It’s a specific page about exactly how row and column operations help compute determinants.

Which is exactly what your question is about.

[graceful dust]

[sharp ice]

Do i have to repeat myself? He sent me a tutorial on operations. I obv know them and am stuck in this question

[drifting seal]

chatgpt has really conditioned you to thinking you're entitled to answers without doing any cognitive labor yourself

[clear delta]

show work

[drifting seal]

he literally gave you a resource

😭

[marsh valley]

You can say that you don’t really understand and would like more help… without being rude?

Like ffs hahaha

[sharp ice]

Yea i throw some random link at your face with no further clarification

[eternal shard]

blud wrote more than a clarification

[sharp ice]

Chat gpt gave me an answer. I said are you sure? Then it said oh wait its wrong heres the real answer

So for the mod saying this question is basic

Js trying to make fun of me

[drifting seal]

chatgpt isn't that advanced in math

[sharp ice]

Yea your more advanced ik

[drifting seal]

*you're

[sharp ice]

You got your ielts 9 score too

[graceful dust]

we could all just put this behind us and stop typing unless you want to help

[drifting seal]

nah im a native speaker

[clear delta]

sigh.

okay i'm going to give you a chance to restart this interaction. Please follow the steps in the bot message above: ask your question in a clear manner, show your work and explain where you are stuck, and demonstrate some effort beyond asking a machine.

[sharp ice]

No clue how to procceed with this since the colums are switched. And even transposed the letterd arent in their right position

What now

[clear delta]

the columns are switched? what happens to the determinant of a matrix if you switch two columns?

[sharp ice]

What hapoens

Nothing happens?

Something happens?

[marsh valley]

Swapping two columns swaps the sign of the determinant.

[sharp ice]

Yea and how would that help here

[clear delta]

from the link we posted earlier

[clear delta]

can you swap two columns of A to make it look more like B?

not exactly like it, but more reminiscent of it?

[sharp ice]

Yep, then whe gotta do alot of swapping

Row swapping column swapping

Right?

[graceful dust]

yea

and some other transformations

[sharp ice]

Does anybody actually know the answer

Or is everyone here just throwing whatwver info they think they remember

One and done advice

[graceful dust]

lol

[sharp ice]

Drive by tips

[clear delta]

i wouldn't be able to write it down offhand, but i know some of the numbers in it

[graceful dust]

i know how to do it in principle. you don't need to already have the final answer to possibly be helpful

[sleek vessel]

Even if we do, we are not supposed to give you the direct answer

People are giving you tips on how to solve it

[sleek vessel]

There are more transformations than just swapping

Scaling is one that'll be helpful here

Scaling and adding is another

[graceful dust]

if someone hands you a function involving a bunch of standard high school functions and asks you to take its derivative, you don't need to compute it before you can help

[sharp ice]

Ive asked for olympic race advice, and i recieved a walking tutorial

[graceful dust]

this problem is really not that hard lol

[sleek vessel]

Lmao you've got to be trolling at this point

[vague trench]

lilbludyapping moment

[graceful dust]

do you understand the things in the link civil sent? do you understand you want to apply some transformations from it to A to make it look more like B?

[vague trench]

There’s no clever trick here. It’s just tedious.

[graceful dust]

keeping track of what each transformation does to the determinant

[sleek vessel]

This one isn't even that tedious tbh

[vague trench]

All of them are to me

[sharp ice]

Yea makes sense this helps

[sleek vessel]

Fair enough, anything computational is tedious

[sharp ice]

Great. Ok. Everyone joking around suddenly summoned the info i need to solve this

[graceful dust]

at this point i can't tell if you are serious

[sleek vessel]

You were given that info in the first response message

[graceful dust]

or just trying to make fun of us

[clear delta]

right okay
before you insult our helpers again and throw their help back in their face while proclaiming that you already know everything, i'm going to need you to demonstrate that you have made an effort on this problem -- this will look like a picture of your paper with some pen/cil writing on it.

if your next message does not have that, then this thread will be closed and you will be muted.

[sharp ice]

Other than the introductory to row operations link. Everything has been insulting

[graceful dust]

i just gave some instruction

[sleek vessel]

Point out what messages have been insulting please

[graceful dust]

people have been mildly insulting over how ungrateful/entitled you are coming off as, sure

[sharp ice]

As soon as i saw it and was confused, and even gpt didnt know it. I came here

[sleek vessel]

Anyway, please read what @clear delta said. Thank you

Well we have given you some tools

Now give it at least a try

[sharp ice]

The tutorial?

[sleek vessel]

See where that gets you

If you're stuck come back and ask

[graceful dust]

wait so were you kidding when you said this

[sleek vessel]

Holy shit

[sharp ice]

All you said is to keep track. As if i dont keep track in every single question.

[sleek vessel]

...

[graceful dust]

bruh

[clear delta]

as i mentioned, you have not demonstrated any effort on this problem (copying and pasting it into chatgpt does not constitute effort). we have given you the tools; over the course of the next hour please give it the ol' college try.
toodles

.close

[calm coral]

Channel closed

Closed by @clear delta

Use .reopen if this was a mistake.

[graceful dust]

sorry i couldn't help myself

[calm coral]

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[tall bane]

When doing system of equations when doing quadric formula, what does it give us?

[cerulean nebula]

??

[spark stratus]

quadratic formula you mean?

quadric usually means "quartic" instead as in the degree-4 quartic formula

also, this question doesnt really say much of anything, you could just show a homework problem youre stuck on and we can direct where you go

[tall bane]

I just want to know what it does when solving these

[uneven shell]

Your question does not make sense.

Therefore we can not tell you what "it gives you".

[cerulean nebula]

Try finding a question related to what you want to understand.

[calm coral]

@tall bane Has your question been resolved?

[calm coral]

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[tulip plover]

i am hilariously failing to understand how definite integration and anti derivatives are related. why is the difference between two y values of the anti derivative equal to the area under the curve of the integrand at those points

[thorny stump]

Do you understand that $\int_0^x f(t) \dd t$ is an antiderivative of $f(x)$?

[potent lotus]

Nel

[tulip plover]

ive been told by some that trying to conceptually understand integrals is very time consuming and possibly real analysis

[thorny stump]

(if f is continuous)

[tulip plover]

no. my understanding is that if you differentiate the area function (integral) you just get the orignal function

that image doesnt differentiate the integral so i am confused

[thorny stump]

Start there I suppose: https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Intuitive_understanding

[tulip plover]

ok i am used to seeing the capital F of x to denote its an anti derivative

[tulip plover]

so yes i understand this now @thorny stump

[thorny stump]

Well, any continuous function f has multiple antiderivatives

An infinite amount, since you can just add an arbitrary constant term

So you can define $F(x) = \int_0^x f(t) \dd t$ for example, that's one antiderivative

[potent lotus]

Nel

[thorny stump]

Now, $F(b) - F(a) = \int_0^b f(t) \dd t - \int_0^a f(t) \dd t = \int_a^b f(t) \dd t$

[potent lotus]

Nel

[thorny stump]

That should answer your original question

[tulip plover]

why does the change in the anti derivative over two points correspond to change in area of the original function

maybe you can explain that with position vs time and velocity vs time

[thorny stump]

Are you asking why this holds?

Or why an integral measures the area under the curve

[tulip plover]

essentially

both...?

like i know an integral is a riemann sum

and im also confused why that area corresponds to change in an anti derivative

i know that (for example) area under the curve of a velocty graph gives change in displacement

[thorny stump]

I'm not really sure what to answer

Do you want a more physics related answer?

[tulip plover]

im not really sure what to ask either

all i know is that i'm hilariously lost

[topaz raft]

i think they are asking why plugging in two points into the anti derivative and subtracting them gives the area under f

[tulip plover]

no but I'm just trying to understand at a calculus 1 level

[tulip plover]

basically

[spark stratus]

theres a way you can view this for sums that can give some intuition

lets say you have the sequence
1 3 5 7 9 11 13 15...
if you do "the sum of the first n numbers", you get
1 4 9 16 25 36 49 64...

now visually, your 1 3 5 7 9 ... would be these lines here

and then the corresponding integral would be the area between the x-axis and that above

you can count the squares (sort of) and see that this is the case

now if you for example wanted to add: 5 + 7 + 9 + 11 + 13

you can do: (1 + 3 + 5 + 7 + 9 + 11 + 13) - (1 + 3)

you start the sum at a higher number, then you subtract the lower portion

this isnt ideally how you should view FTC but itll work

do you understand the sum version so far

[tulip plover]

yes

[spark stratus]

now for the integral version

[thorny stump]

(this is basically a Riemann sum)

[spark stratus]

yep this here would be a riemann sum, but a riemann sum isnt exactly necessary

[spark stratus]

first, lets say all the integral does is find the area under a curve

knowing the sum example, does the formula work

[tulip plover]

I don't know

[spark stratus]

here, we had: (sum from 5 to 13) = (sum from 1 to 13) - (sum from 1 to 3)

now the formula says (area from a to b) = (area from 0 to b) - (area from 0 to a)

by itself, would that formula make sense

[tulip plover]

yes

[spark stratus]

so that way, what we really have is why an area would represent an antiderivative

from there its a straight shot for why we subtract

did you follow the wikipedia link?

[tulip plover]

yes I did

[spark stratus]

did you believe it?

it sort of looks like you didnt in fact read the wikipedia link

@tulip plover hello?

[calm coral]

@tulip plover Has your question been resolved?

[tulip plover]

yeah

[calm coral]

If you are done with this channel, please mark your problem as solved by typing .close

[tulip plover]

.close

[calm coral]

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[cosmic garnet]

hi i need help with number four, i’ve been using tje quotient rule since we could use either one and still get an answer, but i’m confused on applying it to that question

[cosmic garnet]

i have the answer key and everything i’m just so lost on how to get there

[thorny stump]

!show

[calm coral]

Show your work, and if possible, explain where you are stuck.

[marsh canyon]

you dont have to use quotient rule you could rewrite the power

[cosmic garnet]

what do u mean

the denominator?

[marsh canyon]

yeah you can write that to be on the top with a differnet power

[kindred estuary]

Can you show your attempt?

[cosmic garnet]

yes sorry

i didn’t get far but on this part i’m unsure whether its zero or one since the numerator is -7

[kindred estuary]

Do you recall quotient rule?

[cosmic garnet]

yea i know it as vdu - udv / v^2

[kindred estuary]

Good, so what you should do, is identity v, u, dv, and du then just plug them all into the formula

[kindred estuary]

But this, so far, the work is not right

[cosmic garnet]

oh i think i accidentally dropped the squared on v

[kindred estuary]

Yes, that was the mistake

[cosmic garnet]

oh okok

the derivative of u would be zero right?

[kindred estuary]

Correct

Because derivative of a constant is zero

[cosmic garnet]

oooh ok

[carmine lotus]

haii

[cosmic garnet]

i plugged everything in, does this look right so far?

[cosmic garnet]

hi

[kindred estuary]

Not quite

Your derivative of v is wrong

Chain rule

[sonic rose]

And the denominator

tbh indicating the power rule rewrite was prob easier.

[cosmic garnet]

i used quotient rule cuz it was already in u/v and i couldn’t figure it out if i were to use product rule

[cosmic garnet]

its f’(g(x)) • g’(x) right?

[kindred estuary]

It depends on how the course is taught. Yes, power rule would be easier but if the class is working on practicing power/quotient, then they should apply those instead of rewriting the fraction

[kindred estuary]

Yes

[cosmic garnet]

we could’ve used either but he emphasized quotient would be easier since we wouldn’t have to rewrite the fraction , he gave us both answers i’m pretty sure as well

[potent smelt]

Can you derive the quotient rule from the product rule and chain rule? (Hint, let u/v = u * (1/v)

/me reads up. Nevermind, I had the wrong impression of what the issue was

[cosmic garnet]

ohh no worries😭

[potent smelt]

v(x) = (2x-3) ²

Can you tell me what v' is?

[sullen mist]

mb

[cosmic garnet]

i thought it would (2)^2 but it makes no sense😭

[kindred estuary]

Don't forget, chain rule

[cosmic garnet]

oh yeah

is it 8x-12?

[potent smelt]

Show work?

I think that is correct, but I don't want to rely on my mental math

[cosmic garnet]

2(2x-3) * 2
4(2x-3)
8x-12

thats what i did

[potent smelt]

Lgtm

[cosmic garnet]

so that would be v’ ?

since i used chain rule

[potent smelt]

Yup

[cosmic garnet]

oohh ok that makes sense

then the denominator would be 8x-12 too?

[potent smelt]

No...

(u/v)' = (u'v - v'u)/v^2

[cosmic garnet]

oh so its its just v then

oh yeah

[potent smelt]

What?

[cosmic garnet]

squared

[potent smelt]

Yes

Be careful though, v = (stuff)^2 so v^2 isn't (stuff)^2 but (stuff)^4

[cosmic garnet]

oh yeah

i’ll finish it now but thank you so much i was completely stuck on this part

.close

[calm coral]

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[calm coral]

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[cedar urchin]

!help

[calm coral]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[cedar urchin]

Hi i wanted to check if i got this right

<@&286206848099549185>

.close

[calm coral]

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[cosmic frost]

Calculus 1, can anyone tell me how tanswer this

[daring heart]

!status

[calm coral]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[cosmic frost]

1

[daring heart]

Do you know how to figure out when the domain of a function is restricted

[cosmic frost]

no, what does domain of function being restricted mean

[daring heart]

That just means inputs that aren't in the domain

Specifically, we restrict inputs to functions to numbers we can get outputs from

[cosmic frost]

is the domain equal to f(x) or x

[daring heart]

Almost

Think about when a fraction is undefined

We can write any number, but can we divide by any number?

[cosmic frost]

fraction is undefined when denominator is 0 right?

[daring heart]

Exactly

So it's not just the domain of f(x)

But domain of f(x), g(x) and specifically when g(x) is not 0

Because if f(x) or g(x) is undefined, we also can't figure out the value of f/g

A simple example is f(x) = x and g(x) = x^2, which are both defined everywhere

But the domain of f/g does not include 0, because g(0) = 0

[cosmic frost]

i see, but what does intersection mean in the question

[daring heart]

Intersection is another word for 'must be in both'

For x to be in the domain of f/g, it must be in the domain of f(x) and in the domain of g(x), so it is in their intersection

But the reason the answer is False is because that's not enough

[cosmic frost]

i see, yeah that was what i was also thinking

[daring heart]

Like we saw here, 0 is in the domain of g, but not in the domain of f/g 👍

[cosmic frost]

but why does g(0) = 0 like what if g(x) = 1 +x

[daring heart]

In that case g(-1) = 0

So -1 is not in the domain of f/g

[cosmic frost]

oh ok, basically -1 should be in both to be an intersection

[daring heart]

Yes 👍

[cosmic frost]

alright well thank you for your help

how do we close the channel?

[daring heart]

You use the .close command

[daring heart]

No problem!

[cosmic frost]

alright well see ya later

.close

[calm coral]

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[calm coral]

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[flat cosmos]

could anyone explain to me what this means, i dont know what any of this means

[uneven shell]

For starters

You do know what "the function g" means right

[flat cosmos]

means g(x) right

i dont really understand what the mcq part is asking

[mossy gulch]

Then, do you know the term "equal-length input-value intervals"?

[flat cosmos]

not really

[mossy gulch]

Ok, it literally means the inputs are equally spaced.

A visual way of interpreting it would be...

I've drawn here a random graph.

You can see the inputs are marked in red.

They are equally spaced.

[flat cosmos]

so equal-length input-value intervals is refering to the x value?

[mossy gulch]

Yes.

Because it has the key word "input-value".

and the function is of x.

it is solely depended on its value.

I'll remark that the input might not always be x, but for the context of most questions and your question, it is x.

You can ignore that as it's not very relevant.

[flat cosmos]

would the answer be C? wont the +4 from f(x) ruin the equal length?

[mossy gulch]

Yes, that's correct.

[mossy gulch]

But it would ruin the proportion.

[vague vigil]

hi po

[mossy gulch]

What was your thought process?

[flat cosmos]

well since g(x) is exponential wouldn't the ratio be constant

[mossy gulch]

Yes.

What about f(x)?

[flat cosmos]

f(x) throws off the pattern or the constant because it adds 4?

[mossy gulch]

Yes, but how would you answer if I asked you why the 4 ruins the pattern?

[flat cosmos]

hmmm im kinda stuck now

[mossy gulch]

Ok, I'll stop with the barrage of questions.

It's intuitive to say f would ruin it, but pondering why would give you better comprehension.

You won't need to think that deep though.

Would you like a quick explanation or keep it to yourself to think about?

[flat cosmos]

sure

[mossy gulch]

sure about what?

I'll assume the former.

In the general sense,
\begin{align*}
\dfrac{f(x + c)}{f(x)} &= \dfrac{g(x + c) + 4}{g(x) + 4} \
& = \dfrac{ab^{x + c} + 4}{ab^{x} + 4} \
& = b^{c} \cdot \dfrac{ab^x + \frac{4}{b^c}}{ab^{x} + 4}
\end{align*}

[potent lotus]

Erebus

[mossy gulch]

So here, b^c is a constant, and if we want the proportion of f(x) to be constant...

This term must be constant.

If b is 1, the pattern won't be ruined. But we can't choose what b is so C is the most logical choice.

[flat cosmos]

I see

Thanks a lot for the help

[mossy gulch]

No problem.

Stay curious.

[dim plover]

Nice slogan ☺️

[calm coral]

@flat cosmos Has your question been resolved?

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[calm coral]

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[rough blade]

\begin{cases}
\alpha_1 p^{a_1} + \alpha_2 q^{a_2} = \gamma_1 \
\alpha_3 p^{a_3} + \alpha_4 q^{a_4} = \gamma_2
\end{cases}

[rough blade]

p,q are primes

can we solve it ?

with all the other symbols being positive integers

oh wait

i made a mistake

the second one is not p , its q

[lyric ravine]

This definitely needs more context, you can arbitrarily choose alpha_i, p,q, a_i and gamma_is will then be forced if everything in the equation can vary.

[potent lotus]

Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lone flax]

This is not what you originally asked

[rough blade]

not the same but this is what i deduced

we get 2 equations

for p,q

[lone flax]

It isn’t the same as what we deduced I mean

[lyric ravine]

So it does need more context

If you have a question go to #❓how-to-get-help . If you just want to talk to people go to #discussion

[lone flax]

\begin{cases}
(\alpha_1 p)^{e_1} + (\alpha_2 q)^{e_1} = \gamma_1 \
(\alpha_3 p)^{e_2} + (\alpha_4 q)^{e_2} = \gamma_2
\end{cases}
$\text{mod} pq$

[potent lotus]

Cogwheels of the mind
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lone flax]

This was

[rough blade]

the only difference is that i didnt add mod pq

[lone flax]

And a_1=a_2, a_3=a_4

[rough blade]

oh absolutely

[lyric ravine]

Ok, so we are given a pair (gamma_1,gamma_2) and we need to find all integer tuples (alpha_i,p,q,e_i) satisfying that?

[rough blade]

no we're given everything besides p,q

we know there multiplication value though

we are given p*q

[lone flax]

If you are solving this more general question, I suppose it further can reduce to

Though I still can’t see how to derive p, q

[lyric ravine]

I mean from a practical standpoint you have only finitely many p,q to check because pq is fixed

[rough blade]

idk why im making it generic lol , we have values more values here

[lyric ravine]

p,q are primes, so in particular there will be only two ordered pairs? This is what's confusing me

[lone flax]

Oh

Forgot he said pq is known

Then yeah why bother, only two possibilities

[rough blade]

@lyric ravine the start of the problem that we got the equation from

[rough blade]

how

[lone flax]

(p,q), and (q,p). Though I guess the point is to factorize N

[lyric ravine]

Ok this makes a lot more sense now

[lone flax]

$$c_{1}=(2p)^{e_{1}}+(3q)^{e_{1}}$$
$$c_{2}=(5p)^{e_{2}}+(7q)^{e_{2}}$$
$\text{mod}, N$

[potent lotus]

Cogwheels of the mind

[rough blade]

And ?

[lone flax]

Idk, just simplified it a little bit

[calm coral]

@rough blade Has your question been resolved?

[lyric ravine]

You might want to read this:
https://www.ctfrecipes.com/cryptography/general-knowledge/maths/modular-arithmetic/modular-binomial

[lyric ravine]

Seems like the general case mentioned here is also covered

[calm coral]

Channel closed

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[lone flax]

👍

[calm coral]

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[candid island]

Does anyone know where I went wrong in my calculations? x should be 3, but I can't find my mistake. I'm using the addition method.

[gritty sequoia]

in 1 you multiply numerator and denominmator of first fraction by 2 not the whole equation by 2

[autumn zenith]

In (2) it should be 128×16 not just 128 and (1) 42×2

[candid island]

But I have to multiply the whole equation by 2 to get rid of the triple fraction, right?

[candid island]

I multiplied 8 / 2y-4 by 16, which is why I got 128.

[rustic robin]

you need to multiply all the denominators to get rid of the triple frac

[gritty sequoia]

not the whole equation if you consider thing below 21 as denominator multiplying 21 by 2 and denominator by 2 works

essentially saying $2\left(\frac{21}{\frac{3x+5}{2}}\right) \neq \frac{42}{3x+5}$

[potent lotus]

oppenheimer

[gritty sequoia]

rather = $\frac{42}{\frac{3x+5}{2}}$

[potent lotus]

oppenheimer

[gritty sequoia]

you can use this $\frac{a}{\frac{b}{c}} = \frac{ac}{b}$

[potent lotus]

oppenheimer

[gritty sequoia]

converting $\frac{42}{\frac{3x+5}{2}}$
into $\frac{84}{3x+5}$

[potent lotus]

oppenheimer

[gritty sequoia]

you could apply this at the start tho, simplifying $\frac{21}{\frac{3x+5}{2}}$ as $\frac{42} {3x+5} $

[potent lotus]

oppenheimer

[gritty sequoia]

@candid island do you understand it?

[candid island]

Not sure, let me try it real quick

[autumn zenith]

You multiply by $\frac{c}{c}$ here

[potent lotus]

ch3rry

[candid island]

like this?

[autumn zenith]

Isnt this the same as you did before?

[candid island]

Why is 2 is still under the fraction? If I multiply by 2, doesn't that cancel out and 21 becomes 42?

[candid island]

Oh you're right

[autumn zenith]

Multiplying by $\frac{2}{2}$ would give tht result

[potent lotus]

ch3rry

[candid island]

So $\frac{21}{\frac{3x+5}{2}} times 2 = \frac{42}{6x+10}$

[potent lotus]

para

[autumn zenith]

No multiplying by 2 would only change the numerator

[candid island]

But doesn't that solve the 2 under the fraction?

[autumn zenith]

https://youtu.be/omxHc5E0TCw?si=1jXnxr-h5IzBKhlz

This might help

[calm coral]

@candid island Has your question been resolved?

Channel closed

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[calm coral]

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[tawdry ruin]

should i start by solving for y or z?

[leaden thunder]

It really doesn't matter

[tawdry ruin]

it doesnt?

i started with y but a little confused

im doing 2x - (-2x-z+7) + 3z = 13 but how would i go about solving it

would i do -2x - 2x? or -2x + 2x

[ocean cypress]

The simplest way to think about this is to just take the transpose of the cofactor matrix multiplied by the reciprocal of the determinant and use that to left multiply the constant vector.

[tawdry ruin]

need help here

[half trench]

were you taught to use substitution to simplify like this or is it just easier to you btw

[leaden thunder]

-(-2x) = 2x

[tawdry ruin]

uh idk im self teaching myself so thats why idk what to do here

[half trench]

ohh ok

[tawdry ruin]

so they cancel eachother out?

what if i have a positve number, like +7, would it be -7?

[half trench]

what riemann is saying is that the negative sign flips the -2x again

so its 2x+2x

[tawdry ruin]

oh wait do you mean how im plugging in y into the other problem? if so i was taught this way and its easier for me

[tawdry ruin]

okay i see

but would it flip positives into negatives?

[half trench]

yeah in my opinion its easier to just get two terms with the same coefficient and cancel them with adding/subtracting the equations the whole way but this still works fine