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You can use substitution to solve it
wym
Wait lemme write it on paper
ook
So we have those two equations, we will use one of them to find x or y then substitute it in the other one to find one of them then substitue the one you found in on of the equations, this sounds complicated but lemme explain it on paper it will be easier
yes plz explain on paper
We will use the second one to find x with y in it, we divide both sides by y the ones on the left will cancel each other and we will be left with x = 47/y
Sorry it got tilted
why divide both sides by y
To get rid of the y thats next to the x, we only want the x for now, or you can divide both sides by x to only have y, like this
Either one of them is correct
Is everything clear so far ?
At this point You are just trying to get either x or y alone
You cant remove it because there is nothing to divide it with so it is kept there
In maths the two sides of the = are always equal to each other so if you wanna add something you add it to both sides and they will still be equal
Wanna vc so i can explain it while talking ?
im in class lmao
Oh
Wait is this an exam ?
Alright
that we're doing
the qu estion is behind
im asking here cz teachers speaking english
n i cant und erstand
x+y = 14
xy = 47 right
so im not gon ask my teacher
you want to make x the subject or y the subject in the second one? @frosty slate
make it x = or y =
you know simultaneous equation
whats th at
Because you said and
it means they are both true
oh find x = what
y = what
wait u r refering the picture?
im lost
xy = 47
yeah
yez
makes sese

you get that?
both sides = ?
.
,tex [x = \frac{47}{y} ]
!Ego
When you divide both sides by Y the x will be alone, which will make you closer to solving it, you take the x and substitute it in the other equation which will make it only have y
yes
Yeah but we divided both sides by y
After you have the x = 47/y you will take it into the first equation which eill become 47/y + y = 14
uhhhhhhhhh yk what ill wait till the times comes then only i attempt these questions again 🥹
si hard
now move 14 back, $33+y^2=0$
Wait, are you sure the question is right ?
y is an imaginary number
now b^2 -4ac < 0
wat
your question isn't valid
it wont give you a solution
The question is not your grade because y will be an imaginary number
!Ego
The age of 17/18 i think
you sure you didn't send us the wrong number or something
because thats easily done
if x +2y = 14 and xy = 47 whats x and y
thats now valid innit
because now you have 33 +2y+ y^2 = 0
im sure
a thousand million pe rrcent srue
Was it given by the teacher ?
its in the paper
Its invalid according to your grade
typo by teacher?
i learn a lot invalid stufs
plus minus innit
I hate that sh*
$y = +\sqrt{33} i or -\sqrt{33} i$
lmao trust me
I have been working with shapes for a while i forgot the - part
i dont even know what complex number system is bro
i just know if b^2 -4ac < 0 not valid
cus you can't root a negative in the normal number system
more invalid stuffs for my grade
x-y=3, find the value of x^3-2x^2y+xy^2-3xy+3y^2
a=b=4 find the value of a^3-b^3-12ab
i learne complex stuffs
🤷♀️
Pretty simple just $\sqrt{-1} = i$
do you write the i on the outside
of the root
or outside
Just ugnore it
Outside
okayigs
what grade you in @cursive hare
Last one before college
age?
19
ok im 16
wow
,tex [ \sqrt{-15} = \sqrt{15} i ]
!Ego
Also you can close this post by typing +close when you are done
Alright have a nice day
+close