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in other terms:
\[x \sqrt{y}=108, y \sqrt{z}=80\sqrt{3}, and, z \sqrt{x}=225\sqrt{3}\]

any ideas?

This is mind boggling o.o

I got 6 minutes then i have lunch.

Getting confused from just looking at it.

its a algebra 2 question i believe

You pretty much have to do them all at once...And the x(sqrt y) thing confuses me.

i got lunch, i was thinking to square the whole equations

any ideas?

\[x \sqrt{y}=108 \\ y \sqrt{z}=80 \sqrt{3} \\ z \sqrt{x}=225 \sqrt{3}\]
this is right?

ye

and is x and y and z integers are real?

I don't know :|

and is x and y and z integers or real?
*

I just want to learn how to do these questions, because I like to learn.

I understand that but I was asking if you knew anything about the variables

I guess we will assume they are real

ok

This is taking a bit of time, need 2 minutes.

@ganeshie8 do you see any kind of short cut here

x=27

y=16

cool I got x=27 too

Can't find Z though..

5 radical 3

\[z=\frac{80^2 (3)x^4}{108^4}\]

x+y+z = 42+\[5\sqrt{3}\]

o

75

cool stuff

x+y+z=27+16+75

42+75=127

so i guess they were integers after all

What if they weren't what would you do?

isn't 27+16 equal to 43?

and isn't 43+75=118

bad math confirmed.

well we just assumed they were real numbers in the beginning

So thats why I got a 145/150 on this Mu Alpha contest

that includes the integers and non-integers

I was thinking that there should be a short cut for some reason but I do not see one

Two awards, [Fan medal], https://www.youtube.com/watch?v=jH-miNEnIgE

One more question maybe?

ok I will try to answer

\[a^2_1+a^2_2+a^2_3=7 \\ \text{ find max of } 3a_1+2a_2+a_3\]

yeah

I'm thinking that we can try to use that one thing
it starts with an L
lebniz?

Leibniz, I am in 6th grade, I dont know this stuff well, explian more of it.

oh I was thinking of a calculus thing

I do pre calculus

i'm so dumb it was another l word

lagrange multiplers

What is that exactly?

http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers.aspx
this is what it is

K then?

we need to solve that bottom system for (x,y,z)

and we also have that first equation

\[3=\lambda 2x \\ 2=\lambda 2y \\ 1=\lambda 2z \\ x^2+y^2+z^2=7\]

3x+2y+z = what number?

I honestly don't know what that is :(, I read it but it doesnt make my brain into order

so maybe we need another way to solve besides lagrange multiplers then

any ideas out there

that is the only one I have unfortunately

It is hard to learn pre cal in 6th grade :(

yeah I think lagrange multipler is a cal 3 method

or cal 2

:_:

I have, one more >_<

Let x and y be real numbers satisfying 2/x = y/3 = x/y. Determine x^3

Im assuming 0, but that just seems wrong

\[x^3+y^3+z^3=(x+y+z)^2-3(x^2y+y^2x+y^2z+z^2y+z^2x+x^2z)-9xyz\]

is there any way to find\[x^3+y^3+z^3\]?

\[6=xy \\ 2y=x^2 \ \text{ note multiply both sides by } x \text{ and then use first equation }\]

:O ok :)

Or did you cross multiply 2 * 3 and x*y

that is what some people like to call it

that is what I did for each pair of fractions that were equal

I only needed the first two equations though

Im just going with my method, proceed your explanation

my method is done :p

you just replace xy with 6

2(6)=x^3

Huh? I dont remember 2xy=x^3?

did you read anything I said?

that isn't meant as a rude question by the way

I multiplied both sides by x
2y(x)=x^2(x)
2xy=x^3

the first equation says xy=6

replace xy with 6
so you have 2(6)=x^3

12 = x^3, x = 2 radical 3

you don't need to find x

ok

\[x^3=12 \text{ also this doesn't give a solution of } x=2 \sqrt{3}\]

ik that was x^2 :/

you do have to show work on these right?

i guess

lol what does that mean?

yes 3:

3;

what kind of face is that? an agreeing face? or a " I don't know" face?

its a 3 and a type of colon 3;

well I might go to sleep now then
goodnight

me too 1:24 am is hurting me head