MHB Solve the equation 8x^2−2xy^2=6y=3x^2+3x^3y^2

[lfdahl]

Find all the real numbers $x$ and $y$, that satisfy the following equations:

\[8x^2-2xy^2 = 6y = 3x^2+3x^3y^2\]

 

[Albert1]

Find all the real numbers $x$ and $y$, that satisfy the following equations:

\[8x^2-2xy^2 = 6y = 3x^2+3x^3y^2\]

my solution:

Spoiler

let $A=8x^2-2xy^2 $
$B= 6y$
$C= 3x^2+3x^3y^2$
from $A,B$ we have $xy^2+3y-4x^2=0-----(1)$
from $B,C$ we have $x^3y^2-2y+x^2=0-----(2)$
$(1)+(2)\times 4$ we get:
$y^2(4x^3+x)=5y$
so $y=0,x=0 $ or $y=\dfrac {5}{4x^3+x}---(3)$
put (3) to (1) or (2) we obtain all the solutions
but a simpler way :
for (3) is symmetric to the line : $y=x$
the solutions of A,B,C must lie on the line $x-y=0$
we may set $y=x$ and from (A)(B) we get:
$8x^2-2x^3=6x$
$x(x-1)(x-3)=0,$
that is $x=0,1,3$
but $x=3$ does not satisfy (2)
so all the real numbers $x$ and $y$, that satisfy the given equations:
are $(x,y)=(0,0)$ or $(x,y)=(1,1)$

 

[lfdahl]

my solution:

Spoiler

let $A=8x^2-2xy^2 $
$B= 6y$
$C= 3x^2+3x^3y^2$
from $A,B$ we have $xy^2+3y-4x^2=0-----(1)$
from $B,C$ we have $x^3y^2-2y+x^2=0-----(2)$
$(1)+(2)\times 4$ we get:
$y^2(4x^3+x)=5y$
so $y=0,x=0 $ or $y=\dfrac {5}{4x^3+x}---(3)$
put (3) to (1) or (2) we obtain all the solutions
but a simpler way :
for (3) is symmetric to the line : $y=x$
the solutions of A,B,C must lie on the line $x-y=0$
we may set $y=x$ and from (A)(B) we get:
$8x^2-2x^3=6x$
$x(x-1)(x-3)=0,$
that is $x=0,1,3$
but $x=3$ does not satisfy (2)
so all the real numbers $x$ and $y$, that satisfy the given equations:
are $(x,y)=(0,0)$ or $(x,y)=(1,1)$

Bravo, Albert! A very nice solution! Thankyou for your participation.

 

[solakis1]

my solution:

Spoiler

let $A=8x^2-2xy^2 $
$B= 6y$
$C= 3x^2+3x^3y^2$
from $A,B$ we have $xy^2+3y-4x^2=0-----(1)$
from $B,C$ we have $x^3y^2-2y+x^2=0-----(2)$
$(1)+(2)\times 4$ we get:
$y^2(4x^3+x)=5y$
so $y=0,x=0 $ or $y=\dfrac {5}{4x^3+x}---(3)$
put (3) to (1) or (2) we obtain all the solutions
but a simpler way :
for (3) is symmetric to the line : $y=x$
the solutions of A,B,C must lie on the line $x-y=0$
we may set $y=x$ and from (A)(B) we get:
$8x^2-2x^3=6x$
$x(x-1)(x-3)=0,$
that is $x=0,1,3$
but $x=3$ does not satisfy (2)
so all the real numbers $x$ and $y$, that satisfy the given equations:
are $(x,y)=(0,0)$ or $(x,y)=(1,1)$

Spoiler

how do you know that this the only real solution??

For e.g the equation: ax+b=c ,has a solution : x =c-b/a ,a different to zero.

But we can prove that this solution is unique

The same we can prove for the equation :$$ax^2+bx+c=0$$,$$a\neq 0$$

The OP wants all the real solutions

 

[Albert1]

Spoiler

how do you know that this the only real solution??

For e.g the equation: ax+b=c ,has a solution : x =c-b/a ,a different to zero.

But we can prove that this solution is unique

The same we can prove for the equation :$$ax^2+bx+c=0$$,$$a\neq 0$$

The OP wants all the real solutions

Spoiler

$A,B\,\,and \,C $ are not symmetric to $y=x$
but their solutions $y=\dfrac {5}{4x^3+x}$ do
so they must situate on $x-y=0$, or the line :$x=y$
with this in mind and searching back we can find all the solutions