how do i figure out 30-2x10+5

[regal sentinel]

i said that

[hazy pilot]

for the gradient yes

now what's our equation of the line?

[regal sentinel]

6,3 and one half

[regal sentinel]

oh

i have to write the equation

[sullen lotus]

$y=mx+b$

[midnight talon]

ImOakley

[hazy pilot]

well we identified our b as being -1

[sullen lotus]

$m=\frac{1}{2}$

[midnight talon]

ImOakley

[hazy pilot]

and our m being 1/2

[regal sentinel]

oh yeah

ohh

OHHHH

[sullen lotus]

that oh must be so satisfying

[regal sentinel]

so b = -1 m = 1/2

[sullen lotus]

no

[regal sentinel]

??

[sullen lotus]

b is the y intercept

m and c arent related they are separate properties of the line

[regal sentinel]

what does this mean

[sullen lotus]

oh wait

sorry

ur right

i didnt read ur message properly

go ahead

[muted cobalt]

yes, b=-1 and m=1/2 are true statements. you just need to compose these.

[hazy pilot]

yes and we get our equation of the line

[regal sentinel]

so

did i solve it

[sullen lotus]

this is where we started lol

[hazy pilot]

we actually started WAAAAAAAAY before that

[muted cobalt]

i was wondering where we started. i've been trying to backread but there's a lot.

[regal sentinel]

wdym

[hazy pilot]

from here

[muted cobalt]

is your goal to yield something of the form y = mx + b here?

[sullen lotus]

i mean the slope scenario

[hazy pilot]

oh okay lol

[hazy pilot]

wait it's been 1 hour 😭

[regal sentinel]

??

how do i compose

[muted cobalt]

allow me to backread to catch up properly on what your actual goal is here.

[sullen lotus]

put those into the equation y=mx+b

[regal sentinel]

y=21​x−1

[sullen lotus]

21?

[muted cobalt]

m is not 21.

[regal sentinel]

or y=1/2​x−1

[sullen lotus]

you got it

[regal sentinel]

BRO

[hazy pilot]

and there we go :D

this 1 hour journey :D

[muted cobalt]

indeed, if you plug in an x value to this, you will get the height of the line above the x axis at that point.

[hazy pilot]

seems like the equation of the line was the friends we made along the way :D

[sullen lotus]

an hour of work for this

[regal sentinel]

ok

ty guys

[hazy pilot]

but good job though

[regal sentinel]

bc u didnt explain the fundamentals

i just wanted to understand the fundamentals

[hazy pilot]

huh we did though

[regal sentinel]

poorly lol

[muted cobalt]

is there anything in particular here you feel like you have not internalized properly? i can try to help.

[sullen lotus]

which fundamentals

[regal sentinel]

i just wanted to understand the basics like why we put things in the order we do and what they mean and represent

and the letters

[hazy pilot]

the letters are just variables

remeber our original equation we worked with?

where we solved for x and got x = -5

[muted cobalt]

do you understand what y = mx + b is actually a statement about?

[regal sentinel]

no'

[muted cobalt]

alright. i can try to explain.

have you had coordinates explained to you?

[regal sentinel]

yes

x and y axis

[sullen lotus]

there isnt really a reason for putting operations in order its just that when it gets to algebra we use notation like 5x and stuff so if by convention we added before multiplication it would be very confusing to do 3+5x

[muted cobalt]

okay, good. so, this equation associates, with any given x value we plug in, a y value.

[sullen lotus]

which is why we have brackets in case we need to add before multiplication

[regal sentinel]

for what purpose

[muted cobalt]

right, i'm getting there.

consider our example derived earlier. y = x/2 - 1.

plugging in, say, x = 1, we will get out y = 1/2 - 1 = -1/2.

we can plot this like this.

[regal sentinel]

y = x?

so its equal

[muted cobalt]

y is not equal to x.

[regal sentinel]

they r equal

oh

[muted cobalt]

y is equal to x divided by two minus one.

which is not x.

[regal sentinel]

oh wait they r not equal i see

[muted cobalt]

except for one value. but we can get to that later.

anyway, yes. this is a procedure which can get us a y value from an x value.

if we do this for every possible x value, we get this.

[muted cobalt]

i actually drew the y in the wrong direction here, my mistake.

[regal sentinel]

so where they meet?

[muted cobalt]

where what meets? i do not follow.

[regal sentinel]

we got y value from an value

so where the two meets

[muted cobalt]

oh, sure. there is certainly a sense in which this is the part where the x=1 line meets with the red line.

anyway. you can certainly verify that, for any x value we put in, we get out the height of the red line above the x axis. you can try it.

for instance, for x=2, we see that y = x/2 - 1 = 1 - 1 = 0.

so, y = x/2 - 1 uniquely describes this red line.

the reason "y = mx + b" is so important is that, by choosing the correct m and b, you can describe any line this way unless it is vertical.

[regal sentinel]

yes its flat

so it would be 0

[muted cobalt]

flat? moreso that it's touching the x axis here.

so this point is at y = 0.

[regal sentinel]

yeah

cuz its flat

x is flat

its on the ground

flat line

[muted cobalt]

alright, sure.

would you like an explanation for what m and b come from?

[regal sentinel]

yes

try to be concise

or compendious rather

[muted cobalt]

okay. so, let's consider this line for now:

we want a procedure to turn any x value we want into the y value of the line at that x value. for instance, if we plug in 3, we want how high up this point is.

so, consider this: for the case of x = 0, what should our procedure give us?

[muted cobalt]

well. i've got to go for about an hour, i'm afraid.

[regal sentinel]

what do you mean turn into?

[muted cobalt]

take this, for instance.

we would like to know how high up our line is above x = 3. we want a procedure that takes 3, and lets us get how high the line is. this is what i mean by turning the x value into the y value.

(i do really have to go now. i'll be back later.)

[muted cobalt]

if you understand, try to answer this.

(for the particular line shown in the image.)

[wide mango]

pemdas

[muted cobalt]

we're over that now, i think.

[regal sentinel]

if we had to go up to 3 it would be 3?

[muted cobalt]

well. we want to see how high the line is, at x = 3. if you look at the graph, that is not 3.

[regal sentinel]

oh its like 1 and a half

@muted cobalt

or like 1.3 or 1.4

however you wanna put it

[muted cobalt]

correct. yes.

so, what we're looking for is an equation which will take any x value like this to the correct y value. in this case, 3 to 1.5.

so, let's start with x = 0. what should the y value be there?

[regal sentinel]

0?

[muted cobalt]

not true. that would imply that at x = 0, the line goes through y = 0. you can see, however, that the line does not pass through that point.

[regal sentinel]

oh

[regal sentinel]

wait am i supposed to look at the number zero or still at 3

oh

yes

[muted cobalt]

simply; at x=0, what is the y value of the line?

x=3 was an example.

[regal sentinel]

i dont get the question, if its at 0 would the other also not be zero

or ru asking me when it meets the line

[muted cobalt]

this blue line is x = 0.

how high up is the line when it meets x = 0?

[regal sentinel]

u mean when it meets the red line?

[muted cobalt]

sure.

[regal sentinel]

3

[muted cobalt]

yes.

so, if we had our equation as "y = 3", this would be true at x = 0.

[regal sentinel]

how is the red line x = 0

i dont get that

[muted cobalt]

the red line is not itself x = 0, no.

the red line is what we are working towards describing. i am simply asking, when it passes through x = 0, how high is it?

[regal sentinel]

oh

u mean when it passes

through 0

[muted cobalt]

yes.

[regal sentinel]

through the line

oh ok

[muted cobalt]

this is what we are trying to get.

[regal sentinel]

i get it

[muted cobalt]

we are trying to get an equation which will tell us the answer for any number, not just 0.

so, we have that, at x=0, the line is at y=3. understand so far?

[regal sentinel]

can u explain y=3 just so im sure again

i get the x-0

[muted cobalt]

you can see that, if we just consider the slice of the grid where the x coordinate is 0, the point the line passes through is at the y coordinate 3.

[regal sentinel]

oh

so its just the result we got from it

[muted cobalt]

yes.

[regal sentinel]

ok

[muted cobalt]

for every such slice of the grid, there is only one point that touches the line. we want this point for every possible x coordinate.

at 0, the point is at the y coordinate 3.

so, if we're clear on this.

[regal sentinel]

yes

[muted cobalt]

notice that if you move the x coordinate over by 1, the line's height changes by the same amount every time.

what is this amount?

[regal sentinel]

which one

[muted cobalt]

if we instead changed from looking at x=0 to looking at x=1, the height we measure would change.

for instance, what's the height of the line at x = 1?

[regal sentinel]

2 and a half

[muted cobalt]

yes. and then x = 2?

[regal sentinel]

2

[muted cobalt]

so, you can see that every time we change our x by 1, we change our y by -1/2.

does this make sense?

[regal sentinel]

yes

[muted cobalt]

very good. so, we know that at x = 0, y = 3.

and we know that adding 1 to x adds -1/2 to y.

so, from this we get the final equation: y = 3 - x/2.

does this make sense so far?

[regal sentinel]

from which

[muted cobalt]

well, we know that when we plug in x = 0, we get y = 3.

and we know that adding 1 to x will subtract 1/2 from our answer.

hence "- x/2".

[regal sentinel]

i get the concept

[muted cobalt]

this is the equation that can tell us how high up the line is for any x value we plug in.

[regal sentinel]

composing is a bit weird for me tho

yeah

[muted cobalt]

that's fair. but just imagine it this way.

[regal sentinel]

i get y = 3

[muted cobalt]

if we looked at, say, x = 5, we know that every time we increase x by 1, we need to add -1/2 to the answer.

if we do this 5 times, we add -5/2 to the answer.

x = 5, and we have -5/2. i hope you see how this is -x/2.

[regal sentinel]

yeah

but cant u just start there

instead of that

[muted cobalt]

hm?

[regal sentinel]

cant u just start at 0 and say y is 3

thus 3

why do u have to subtract if u start there

[muted cobalt]

well, the point is that yes, at 0, y is 3.

[regal sentinel]

yes

[muted cobalt]

y is something else at other values.

[regal sentinel]

is that just to clarify

that u need to subtract

so u write that

[muted cobalt]

yes.

[regal sentinel]

ok

[muted cobalt]

if you wanted to find the value at x=5, you would start with 3, and then subtract -5/2.

you can see how if we take y = 3 - x/2, and plug in 5, we get y = 3 - 5/2.

understand?

[regal sentinel]

whats x/2 again

a half?

[muted cobalt]

x divided by 2, yes.

[regal sentinel]

ok

i kinda get it

[muted cobalt]

or "-1/2x" if you prefer to phrase it that way.

[regal sentinel]

the subtracting thing and composing

is the main issue

i get if we start at +

its 3

0*

then we go down to 1 and its 3.5

[muted cobalt]

2.5.

[regal sentinel]

yeah

2.5*

[muted cobalt]

well, consider this:

[regal sentinel]

so ur removing a half every time

[muted cobalt]

yes.

when x is equal to 0, x/2 is also 0.

so y = 3 - x/2 just evaluates to y = 3.

[regal sentinel]

ok but when we areat 5 and get 0.5 how would we write the equation

or know how to write it

[muted cobalt]

y = 3 - 5/2 = 3 - 2.5 = 0.5.

[regal sentinel]

y is not 3 tho

or ru writing from the start

like the whole thing

or just the last one

oh wait u are writing the whole thing

[timber bobcat]

how is this sill going

[regal sentinel]

thats why

Oh

i see

u are writing the equation of all the ones we solved

[muted cobalt]

it is at x = 0, and then for every x after that there's a contribution of -1/2 to the answer.

hence we just add -1/2 * x to 3.

[hazy pilot]

heav is explaining to them what an equation of a line is now

[muted cobalt]

i'm working on it. we're getting somewhere.

[regal sentinel]

yeah

[muted cobalt]

alright. is it intuitive why we have y = 3 - x/2 now?

[regal sentinel]

me?

[muted cobalt]

yes.

[timber bobcat]

if this continues for long enough then maybe ironvril will be doing quantum mechanics tomorrow

[muted cobalt]

don't worry, now that we understand lines we can do the next obvious step, multivariable calculus.

okay, really though.

[muted cobalt]

now some terminology:

[regal sentinel]

idk

[muted cobalt]

okay, look at it this way.

what's x/2 at x = 0?

[regal sentinel]

3

why ru saying x/2

[muted cobalt]

no, i mean just, what is x divided by 2 if x is 0?

[regal sentinel]

half of 3 would be 1.5

[muted cobalt]

i'm not talking about y right now.

[regal sentinel]

oh

0

[muted cobalt]

yes. and if x = 1?

[regal sentinel]

ru asking me the same

0.5

or what

[muted cobalt]

yes.

do you see how this is what we need to subtract from 3 to get our y position, at any given x value?

since it decreases by 0.5 every time x increases by 1.

[regal sentinel]

yes bc it goes down

yes

and it also goes the other way?

around

[hushed badger]

@regal sentinel dedication is real bro

[muted cobalt]

sure thing.

at x = -1, y = 3.5.

[regal sentinel]

I have a admission test next week bro

[regal sentinel]

is this a question

[muted cobalt]

it's a statement.

about the line.

[regal sentinel]

yes?

[muted cobalt]

so anyway.

now you can see, that for a given x value, y = 3 - x/2. right?

[regal sentinel]

yes i understand

it goes down a half or up depending

[muted cobalt]

okay. good. so, this describes our line.

[regal sentinel]

yes

[muted cobalt]

that value, 3, where x = 0. we call this the 'y intercept'.

it is where the line crosses the y axis, at x = 0.

[regal sentinel]

yeah where the lines meet

[muted cobalt]

and the value -1/2, we call this the 'slope' or 'gradient'. it is how fast the line is changing.

as we increase our x value.

[regal sentinel]

yes

its a slope goes up or down depending which way

[muted cobalt]

often, you will see the general form of this as 'y = mx + b', where b is our y intercept and m is our gradient.

here, b was 3, and m was -1/2.

if we set these values differently, we can describe any line like this, unless it is vertical.

understand so far?

[regal sentinel]

m is how steep it is

right

[muted cobalt]

yes.

[regal sentinel]

so in our case a half

[muted cobalt]

indeed. negative one half, since our line went down as x went up.

[regal sentinel]

and b is where they meet

yes i understand

[muted cobalt]

okay, good. now, as an exercise:

can you determine the same thing for this line?

[sullen lotus]

bro

1332 messages

[regal sentinel]

4

[timber bobcat]

[muted cobalt]

that is indeed the y intercept. what is the slope?

[regal sentinel]

1?

[muted cobalt]

not quite. when we increase x by 1, y does not increase by 1.

for instance, this would imply that at x = 1, y = 5. but that is not what we see on the graph.

[regal sentinel]

how do i determine

[muted cobalt]

you saw last time that when we increased x by 1, we increased y by -1/2.

what happens with this line?

[regal sentinel]

wait

it was easier cuz there were lines on the last ones

isnt it the same like 0.5 how can u tell

[muted cobalt]

there are on this one too. it's a bit more zoomed out.

just look at x = 1. what's the y coordinate of the line there?

you correctly identified that at x = 0, y was 4.

what is it at x = 1?

[regal sentinel]

oh

4

why didnt i check that

[muted cobalt]

i can zoom in a little for you if you need. it is not 4 at x = 1.

[regal sentinel]

how can u tell

[muted cobalt]

make sure to click on the image to expand it fully.

[regal sentinel]

i did lol

[muted cobalt]

do you know where x = 1 is?

[regal sentinel]

do u count the small boxes

[muted cobalt]

you do not need to. just look at the y axis, where the numbers are.

[regal sentinel]

theres not a 1

well yes there is

[muted cobalt]

there is at the bottom.

[regal sentinel]

but we cant see it

wait

3?

[muted cobalt]

nope.

[regal sentinel]

well 6

[muted cobalt]

yes. 6.

[regal sentinel]

oh

yes

oh

[muted cobalt]

right. we started at 4, and got to 6.

[regal sentinel]

https://tenor.com/view/ahoklollmao-angry-cat-angry-black-cat-angry-black-cat-gif-8431628872897666919

[muted cobalt]

what is the slope here?

[regal sentinel]

in total

6?

well from 4 to 6

it would be 2

[muted cobalt]

yes.

every time we increase x by 1 here, we increase y by 2.

so, what is our full equation for y? we know it starts at 4.

[regal sentinel]

b = 4 y = 2

[muted cobalt]

m = 2.

so. what's our full equation?

recall the form of this is y = mx + b.

[regal sentinel]

yeah m

y = 2x + 4

[muted cobalt]

yes. very good.

this fully characterizes this line, you can plug in an x value to this and get the correct height.

here's another one:

[regal sentinel]

b = -1 m = one and a half

[muted cobalt]

indeed. can you do the equation?

[regal sentinel]

do i have to specfiy m and b

y = 4x + 2

or can i say it like that

[muted cobalt]

you should say it like that.

as in, y = (whatever).

can you do that for this line?

[regal sentinel]

new one?

[muted cobalt]

the blue one.

[regal sentinel]

can i do what for it

[muted cobalt]

make an equation for it of the form y = mx + b.

you did it for the one before that, if you recall.

[regal sentinel]

i did?

[muted cobalt]

this one is not 4x + 2. the last one was.

[regal sentinel]

this one?

[muted cobalt]

no.

[muted cobalt]

this one.

[regal sentinel]

i did?

y = 4x + 2

[muted cobalt]

you've plugged in m = 4 and b = 2 here.

[muted cobalt]

but this is not true for this line.

[regal sentinel]

wait

??

b = -1 m = 0.5

[muted cobalt]

1.5.

[regal sentinel]

wait

oh yeah

1.5

[muted cobalt]

you have those right. you just need to put them into y = mx + b.

[regal sentinel]

y = 1.5x - 1

?

[muted cobalt]

yes. good.

this is correct.

how about this?

[regal sentinel]

y = 1x - 0.5

[muted cobalt]

not quite.

if you plug x=1 into this, you'll get 0.5. but y is not 0.5 at x=1.

[regal sentinel]

oh

[muted cobalt]

also, b is not -0.5 either.

[regal sentinel]

y = 1x - -1

[muted cobalt]

not quite either.

what is y at x=0?

[regal sentinel]

0.2?

if u count boxes

[muted cobalt]

it is directly between 0 and 1.

[regal sentinel]

so 0.5

[muted cobalt]

yes.

and what is y at x = 1?

[regal sentinel]

i thought -1 bc they meet there before

[regal sentinel]

0

[muted cobalt]

also no. that is at x = -1.

[regal sentinel]

oh

1

its at 1

[muted cobalt]

yes.

so, what are your values for m and b?

[regal sentinel]

y = 0.5 - 1

[muted cobalt]

you're missing x.

[regal sentinel]

oh

[muted cobalt]

and at least 1 number there is wrong.

[regal sentinel]

y = 0.5x - 1

[muted cobalt]

0.5 is correct for m.

[regal sentinel]

https://tenor.com/view/cat-wiggle-crazy-cat-zoomies-gif-9877358298643393466

[muted cobalt]

-1 is not correct for b.

do you see why?

[regal sentinel]

yes bc it meets at 0.5

[muted cobalt]

yes. so what is the correct equation?

[regal sentinel]

y = 0.5 + 1

[muted cobalt]

hm?

you are missing x again and no.

please, list out for me. what is m and what is b, here?

[regal sentinel]

m is slope

which is 1

b is where they meet

0.5

[muted cobalt]

m is not 1.

you are correct that at x = 0, y = 0.5.

at x = 1, what is y?

[regal sentinel]

0.5

0

[muted cobalt]

no.

here is the line again.

[regal sentinel]

ok so

b is where they meet

[muted cobalt]

yes.

[regal sentinel]

0.5

correct?

[muted cobalt]

yes.

[regal sentinel]

m is slope

it is 1

[muted cobalt]

it is not 1.

a slope of 1 means that when we increase x by 1, we increase y by 1.

increase x by 1 here. what does y increase by?

[regal sentinel]

1?

= 2

?

[muted cobalt]

no.

look.

here is x = 1.

what is y here, where it meets the line?

[regal sentinel]

1??

[muted cobalt]

yes. we start at 0.5, and go to 1.

what is this difference?

[regal sentinel]

0.5

[muted cobalt]

yes.

so, what are m and b?

[regal sentinel]

we said that already

so they r both 0.5?+

[muted cobalt]

yes.

[regal sentinel]

y = 0.5x - 0.5

[muted cobalt]

not - 0.5.

[regal sentinel]

y = 0.5x + 0.5

[muted cobalt]

yes.

correct.

[regal sentinel]

does it mean anything

if its - or +

[muted cobalt]

- 0.5 would mean that b = -0.5.

[regal sentinel]

oh ok

so negative

excellent

[muted cobalt]

this is 0.5x - 0.5.

[regal sentinel]

yes4

[muted cobalt]

this is 0.5x + 0.5 (in purple).

[regal sentinel]

i understand

yes

[muted cobalt]

alright. let's try this, then.

[regal sentinel]

y = 2x + 2

[muted cobalt]

yes. good.

you did not need my help this time.

okay. let's try something else:

[regal sentinel]

y = 6?

[muted cobalt]

not 6.

[regal sentinel]

0

y = 0?

[muted cobalt]

it is not 0.

[regal sentinel]

??

[muted cobalt]

i know the line is going through it, but that number is 5.

[regal sentinel]

oh

y = 6

[muted cobalt]

it is still not 6.

[regal sentinel]

5

y = 5

[muted cobalt]

yes.

the slope, in this case, is 0.

now here's a bit of a tricky case.

[regal sentinel]

yes

x = 2?

[muted cobalt]

you got it.

[regal sentinel]

can i say dat

[muted cobalt]

you can.

and it's the correct thing to say.

x = 2 does fully characterize this line. good job.

well. i suppose i'll leave it there, for now. i hope you learned something, maybe.

[regal sentinel]

yes

https://tenor.com/view/nerd-plink-glasses-side-eye-smart-gif-11422187893182432522

these were the ones i actually wanted help with @muted cobalt

[muted cobalt]

oh, sure. do you know order of operations?

[regal sentinel]

yes

not all of them

but yes

[muted cobalt]

well. things in parentheses come first.

for instance, (10*9) + 1 = (90) + 1 = 91.

and 10 * (9 + 1) = 10 * (10) = 100.

then, exponentials come before other things. 2^3 + 1 = 8 + 1 = 9.

then, generally, multiplication, division, addition, subtraction.

[regal sentinel]

some are easy like 7.1 is 7

[regal sentinel]

ik that

[muted cobalt]

okay. so you can do 6.1 fine.

which ones are you struggling with?

[regal sentinel]

no i cant lol

[muted cobalt]

okay. well. you have subtraction and multiplication. which one do you do first?

[regal sentinel]

30 -2 x 10 +5

2x10 = 20 + 5 = 25 30-2 28 25+28 = 53

[muted cobalt]

that sure is a string of things you did.

you do 2 * 10 first, yes.

to get 30 - 20 + 5.

you do this, as i understand it, from left to right. 30 - 20 = 10. 10 + 5 = 15.

do you understand this?

[regal sentinel]

2x10 = 20

then u do plus

20+5

25

45

[muted cobalt]

you should do strings of subtraction and addition from left to right i believe.

[regal sentinel]

oh

[muted cobalt]

just do the first step, 2 * 10 = 20. you get 30 - 20 + 5.

[regal sentinel]

30 -2 x 10 +5

30 -2 x 10 +5 2x10 = 20 30-2 = 28?

or do u solve them first

before adding

in that case

[muted cobalt]

just do the multiplication first.

see what it becomes.

[regal sentinel]

30 -2 x 10 +5

-2x10?

20?

[muted cobalt]

30 - 2 x 10 + 5.

not 30 -2. 30 - 2.

[regal sentinel]

30 - 2 x 10 + 5.

[muted cobalt]

yes.

do the multiplication first to get 30 - 20 + 5.

then solve that.

[regal sentinel]

30 - 2 x 10 + 5.

2x10 = 20 30 - 20 = 10 + 5 = 15

[muted cobalt]

yes.

[regal sentinel]

lets goo

[muted cobalt]

do you need help on any of the other ones?

[regal sentinel]

6.2 is roots right

[muted cobalt]

it is.

[regal sentinel]

so how many times you have to time somethinb by itself

[muted cobalt]

not quite.

these are square roots, so they're all about multiplying things with themselves once.

sqrt(16) is asking, what multiplied with itself is 16?

[regal sentinel]

4 - 3?

[muted cobalt]

yes.

[regal sentinel]

= 1

[muted cobalt]

it is.

[regal sentinel]

6.3

[muted cobalt]

this is multiplication.

[regal sentinel]

and 7.2 and 7.3

(-2) x (-5)

two negatives make a pluis

no

(-2) x (-5) = 7

[muted cobalt]

no. you're doing addition there.

the correct thing to do is multiplication.

[regal sentinel]

oh

10

[muted cobalt]

yes.

[regal sentinel]

3x + 7 = 5x -1

did we do this one

already

[muted cobalt]

i haven't done it with you, at least.

do you need help with it?

[regal sentinel]

8

[muted cobalt]

hm?

[regal sentinel]

3x + 7 = 5x -1

[muted cobalt]

do you know how to do things like this?

[regal sentinel]

5x + 7 = 7 -1

[muted cobalt]

i'm not quite sure what you did there.

[regal sentinel]

12 = 6

[muted cobalt]

that is not a true statement. you seem a little confused also.

[regal sentinel]

i think i solved it

but i confused myself

[muted cobalt]

i can walk you through how to solve it if you want?

[regal sentinel]

yes

[muted cobalt]

okay. so, the general principles are:

you can add or subtract something from both sides of the =, and it will still be true.

[regal sentinel]

with any number u choose?

[muted cobalt]

yes.

you want your answer to look like x = (some number). you want to get all the numbers on one side, and all the variables on the other side.

[regal sentinel]

even if its not in the equation

[muted cobalt]

yes. it won't be false, so long as you're doing the same things to both sides.

[regal sentinel]

yes

[muted cobalt]

so, 3x + 7 = 5x - 1.

you can eliminate "3x" by subtracting it from both sides.

try it.

[regal sentinel]

-3 3x + 7 = 5x - 1 -3

[muted cobalt]

hm? no, you've done something strange here.

specifically, you are subtracting 3*x from both sides.

3x + 7 - 3x = 5x - 1 - 3x.

understand?

[regal sentinel]

but its already on one side

do i just add it to the other

[muted cobalt]

you subtract it from both sides.

[muted cobalt]

to get this.

[regal sentinel]

3x + 7 - 3x = 5x - 1 - 3x.

ok

[muted cobalt]

yes. now.

3x and -3x cancel out to 0.

so you get 7 = 5x - 1 - 3x.

do you understand this?

[regal sentinel]

how does it cancel out 0

[muted cobalt]

anything minus itself is 0.

3x - 3x = 0.

you can do addition in any order, so this is the same as "7 + 3x - 3x" if it helps.

[regal sentinel]

wouldnt it be -6

[muted cobalt]

this is not 3. this is 3*x.

also, 3 - 3 = 0.

not -6.

[regal sentinel]

ok

[regal sentinel]

what is 3*x

[muted cobalt]

x multiplied by 3. we don't know what x is yet.

[regal sentinel]

ok

ok

[muted cobalt]

but we do know that x*3 - x*3 is 0, no matter what it is.

[regal sentinel]

i understand

yes

yesyes

[muted cobalt]

okay. so, on the right, we have 5x - 1 - 3x.

5x - 3x = 2x. understand why?

[regal sentinel]

bc 5-3 = 2

[muted cobalt]

yes, and this holds true even if you multiply all of those numbers by something.

[regal sentinel]

yes

[muted cobalt]

so. 5x - 1 - 3x simplifies to just 2x - 1.

so we have: 7 = 2x - 1.

understand so far?

[regal sentinel]

yes

[muted cobalt]

okay. we can get rid of the "- 1" by adding 1 to both sides.

can you do that?

[regal sentinel]

1x 7 = 2x - 1. 1x

[muted cobalt]

what?

we are not adding 1x to both sides.

we are just adding 1 to both sides.

[regal sentinel]

1 = 2x 1

[muted cobalt]

hm?

no. we have 7 = 2x - 1.

we want 7 + 1 = 2x - 1 + 1.

[regal sentinel]

7 = 2x - 1

yes

[muted cobalt]

well, not 2x - 1 anymore.

it's 2x - 1 + 1. which simplifies to 2x.

[regal sentinel]

oh

1 + 7 = 2x - 1 + 1

[muted cobalt]

yes.

can you simplify both sides?

[regal sentinel]

yes

8= 2x - 2

[muted cobalt]

not 2x - 2.

remember, we added 1. we did not subtract 1.

[regal sentinel]

8= 2x 2

[muted cobalt]

hm?

[regal sentinel]

8= 2x + 2

[muted cobalt]

also no.

we started with 2x - 1. we added 1.

what is the result?

[regal sentinel]

2x - 2

[muted cobalt]

no. that is what happens if you subtract 1.

[regal sentinel]

2x - 0

[muted cobalt]

yes. which is?

[regal sentinel]

2x?

[muted cobalt]

yes.

[regal sentinel]

ok

[muted cobalt]

so, we have 8 = 2x.

[regal sentinel]

yes

[muted cobalt]

right. now, we can remove the 2 from 2x by dividing both sides by 2.

[regal sentinel]

8/2 = 4 = 2x/2 = 1

[muted cobalt]

2x/2 is not 1.

[regal sentinel]

2 divided by 2 is 1

[muted cobalt]

yes. this is 2*x, however.

2*x / 2 = 1*x.

[regal sentinel]

so 0

what

we dont know what x is

[muted cobalt]

2/2 is 1, yes.

we don't. but consider this.

what happens if you multiply something by 2, and then divide it by 2?

[regal sentinel]

u get half

[muted cobalt]

not quite.

[regal sentinel]

if u

ok

2x2

= 4

4 / 2 = 2

[muted cobalt]

yes.

you can see that nothing happens.

[regal sentinel]

that is half

[muted cobalt]

you started with 2. you got 2.

[regal sentinel]

yes

[muted cobalt]

so nothing happens.

2x / 2 = x.

[regal sentinel]

yes?

[muted cobalt]

right. so we have 4 = x.

[regal sentinel]

what now

[muted cobalt]

x = 4 is your answer.

[regal sentinel]

oh

ok

so 4

Ty for your help

[muted cobalt]

right. you seemed quite confused on the steps here.

i would suggest trying some more of these.

[regal sentinel]

yes

[regal sentinel]

i dont know them 100%

not familiar

[muted cobalt]

it seems you struggle a bit with just arithmetic that involves variables in general.

[regal sentinel]

7.3?

[muted cobalt]

same thing. you just need to add, subtract, multiply and divide the correct things to put x on one side and a number on the other side.

i think someone should take over that is not me.