Solving a System of Equations for Real Solutions

This video is about solving a System of Equations
If you need to post a picture of your solution or idea:
#ChallengingMathProblems #Systems

EXPLORE:

[falknfurter]

The complex solutions look also really nice: If you substitute the first equation z=-(x+y) into the second one you'll get x^2+xy+y^2=0. This can be solved as y = x(-1/2+i*sqrt(3)/2). Note that the expression in the parenthesis is a complex third root of unity! It turns that out you can choose an arbitrary non zero x and get y = x*cis(2*pi/3) and z = x *cis(-2*pi/3) (+permutations). This seems like a nice parametrization of the solutions of the original equations.

[mannpambhar]

Hello SyberMath, Where Do You Find This Black Screen ?

[YahiaNebti]

You could put ab + ac + bc from the beginning instead so you'll get rid of restrictions, but I enjoyed it !

[joaquingutierrez]

Nice!!

I did it as follows

We have x + y = -z

Adding 1/x and 1/y, putting 1/z on the right and substituing z we get (x + y)/xy = 1/(x + y)

By doing some algebra we get

x^2 + xy + y^2 = 0

Then dividing by y^2 and putting u = x/y we get

u^2 + u + 1 = 0 (*)

Whose solutions are complex, then x = yu is complex, a contradiction.

If we want complex solution then we proceed as follows:

Let w be a solution of (*)

Then x = wy

Let y take any complex value =/= 0.

Then our general solution is

y = y
x = wy
z = -( x + y ) = -( 1 + w )y

This is easily seen to be a solution using (*) and knowing that w is a primitive cube root of unity

[pgseducation]

Starting from 1/x+1/y+1/z=0 and ends up with x=y=z=0 makes something fishy fishy, as 1/0 is NOT a number.

[-skydning-]

Hello, where can i suggest you a problem?

[namunamu]

جميل جداً
ماهو البرنامج الذي تستخدمه في الكتابة اثناء الشرح
What is the program that you use to write during the explanation

[dustinbachstein]

Nice little puzzle which I solved in my head :)

[satyapalsingh]

Very nice explanation . Thanks, dear professor.

[dneary]

I would go straight to Viéte's rules: x, y, z are roots of the equation a^3/xyz-1=0 - then you have a/(xyz)^(1/3) = 1, omega, omega^2 (where omega = 1/2(-1+sqrt(3)i)) (and you end up with x=1, y=omega, z=omega^2).

[timetraveller]

my solution:
notice that x=y=z=0 is not a solution since dividing by 0 is not possible
notice if we add the first 2 fractions
we get
(x+y/xy)+1/z
and from the the first equation we get x+y =-z
therefore
(x+y/xy)+1/z
=(-z/xy)+1/z
=(-z²+xy)/(xyz)
since x≠y≠z≠0
we can multiply both sides by xyz
and get -z²+xy=0
z²=xy
doing the same procedure
but with the addition of the 1st and 3rd fraction and 2nd and 3rd fraction
and we get
x²=zy
and y²=xz
now if we add all the fractions together
we get
(xy+xz+zy)/xyz=0
since x≠y≠z≠0
we can multiply both sides by xyz
and we get
xy+xz+zy=0
substituting
xy=z²
xz=y²
yz=x²
and we get
xy+yz+zy=x²+y²+z²=0
and now from x+y+z=0
we get x+y=-z
⇒-(x+y)=z
substitue into x²+y²+z²=0
and simply and get
x²+xy+y²=0
doing the same procedure for y and z
we get 2 other equations
y²+zy+z²=0
z²+xz+x²=0
now if we look at the discriminant
of these quadratic equations
we see these all are imaginary
hence there are no real solutions

[dontsetyourlimitsyt]

I did it differently:
Taking 2nd equation
xy+yz+zx=0
xy=-z(x+y)
xy=z²
Therefore they are in Geometric Progression.
Therefore
z=xr
y=xr²
Substituting this in first equation.
x(1+r+r²)=0
But 1+r+r² has no real roots.
Therefore x=0
Therefore xyz=0
Contradiction!

[tharagleb]

I looked at the equations and thought the solution would have to be symmetric in X, Y and Z, but if so they must all be equal. But at least one of X, Y, Z must be positive and at least one negative. (to satisfy 1/x + 1/y + 1/z = 0) Therefore there is no solution.

[arturovinassalazar]

Wait wait wait! Wasn't it supposed xyz ≠ 0? why x=y=z=0???

[flutepeace]

I got the answer from thumbnail
It really was very amazing question

[tatshatsarangi]

I guess this is a question from the book" elementary mathematics" classic text book by hall & knight😉

[volsyb]

You can just add these equations and realize that the sum of reciprocals is always greater or equal to 2. Therefore, there are no solutions.

[Qermaq]

I combined the denominators like you did, but then I just set the equations equal to each other. xyz(x+y+z)=xy+yz+xz. When I multiplied each side out, I got x^2yz +xy^2z + xyz^2 = 2(x^2yz +xy^2z + xyz^2), thus 1 = 2. That's when I knew a solution would only be possible if 1 = 2. If only I knew if that were true! :D

[sonalichakraborty]

Come on bruh....most of us would appreciate u more if u posted just two difficult Or moderately difficult problems a week instead of posting an easy one every day