Question

hey,I have system of equation to solve, can anyone have a look?

\[\left(\begin{matrix}\frac{ x+2x+y }{ 2 }=2(\sqrt{6} +4\sqrt{3}) \\ x ^{2}+y ^{2}=(2x)^{2}\end{matrix}\right)\]

Answer 1

You can solve second equation a bit more.. :)

Answer 2

\(y^2 = 3x^2\)

Answer 3

so firstly:

\[\frac{ 3x+y }{ 2 }=2(\sqrt{6}+4\sqrt{3})|*2\]
\[3x+y=4(\sqrt{6}+4\sqrt{3})\]
\[3x+y=4\sqrt{6}+16\sqrt{3}\]

Answer 4

then I can say that y=..
\[y=4\sqrt{6}+16\sqrt{3}-3x\]

Answer 5

"Squaring it" is not solving your problem?

Answer 6

I dont know how to 'square' when there is three numbers:/

Answer 7

I only know the rule (a+b)^2=a^2+2ab+c, but I dunno what to do,when it is (a+b+c)^2

Answer 8

@waterineyes

Answer 9

@hartnn can you have a look?

Answer 10

is that x+2x+y ??

thats just 3x+y

Answer 11

ok, i see u got that

Answer 12

yeah, the original problem was a geometric one,I just need to solve this equation in oder to find the answer

I found that (a+b+c)²=a²+b²+c²+2ab+2ac+2bc

Answer 13

so,
(a+b+c)^2 = a^2+b^2 +c^2 + 2ab+2bc+2ac

Answer 14

oh gosh, lots of work

Answer 15

you sure, what you posted is original question ?

Answer 16

well, the orignal question is hard to translate, but I will try... It is the outcome of the original question

Answer 17

thought so, you sure this outcome is correct ?

Answer 18

The half perimeter (p) of a right triangle is
\[p=2(\sqrt{6}+4\sqrt{3)}\]
One of the angle is 60 degrees (so another one is 30 degrees, because it is a right triangle)). I need to find an area of this triangle.

So what I did: since I know, that one of the angle is 30 degrees,it means that the side in front of this angle is 1/2 the lenght of hypotenuse.

If we say that the side in front of 30 degrees is x, then hypotenuse is 2x and another side is y (because we don'tknow anything about it)

so we get the equation:

\[\frac{ x+2x+y }{ 2 }=2(\sqrt{6}+4\sqrt{3})\] (it is because p is half of Perimeter)
another equation is from pythagoras theorem:
\[x ^{2}+y ^{2}=(2x)^{2}\]

Answer 19

I dunno,maybe there is an easier way @hartnn

Answer 20

so you know side opposite to 30 degree angle is hypo/2
but you don't know side opposite to 60 degree angle is sqrt 3/ 2 times hypo ?

Answer 21

no,I didnt know this

Answer 22

I dont think they teach this at school, where I am from tho:/

Answer 23

ok, so now you know y

Answer 24

so basically I dont even need the system of equation?

Answer 25

not actually....you get the same using pythagoras theorem

Answer 26

I am confused, can you show me what I need to do?

Answer 27

area is xy/2
perimeter is x+y+2x
let me think of a shorter way if any

Answer 28

instead of squaring 1st equation

take square root from 2nd equation

Answer 29

y = sqrt 3 x

Answer 30

plug this in 1st equation

much easier than squaring

Answer 31

oh, so y=\[y=x \sqrt{3}\]

Answer 32

yes

Answer 33

and now put that in the first equation?

Answer 34

yup

Answer 35

\[x \sqrt{3}=4\sqrt{16}+16\sqrt{3}+3x\]

no idea how to solve this, tho

Answer 36

its a linear equation!

Answer 37

but I cant sum up any of those together, can I?

Answer 38

4 sqrt 16 or 4 sqrt 6 ??

Answer 39

6,sorry

Answer 40

ok I think I nailed it

Answer 41

ok
\(x \sqrt 3 +3x = 4 \sqrt 6 + 16 \sqrt 3 \)
yes ?

\(x (3+\sqrt 3) = 4 \sqrt 6 + 16 \sqrt 3\)

Answer 42

\[x \sqrt{3}-3x=4\sqrt{6}+16\sqrt{3}\]

Answer 43

yes and now you need to divide by (3+sqert 3),right?

Answer 44

\(x \sqrt 3 (1+\sqrt 3) = 4 \sqrt 3 (\sqrt 2 + 4 \sqrt 3)\)

Answer 45

oh wait how did you get it

Answer 46

why is there a -3x ?

Answer 47

I would have done somethinglike that:

\[x(\sqrt{3}+3=4(\sqrt{6}+4\sqrt{3})|:(\sqrt{3}+3)\]