- #1
lfdahl
Gold Member
MHB
- 749
- 0
Find all the real numbers $x$ and $y$, that satisfy the following equations:
\[8x^2-2xy^2 = 6y = 3x^2+3x^3y^2\]
\[8x^2-2xy^2 = 6y = 3x^2+3x^3y^2\]
my solution:lfdahl said:Find all the real numbers $x$ and $y$, that satisfy the following equations:
\[8x^2-2xy^2 = 6y = 3x^2+3x^3y^2\]
Albert said:my solution:
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 $(x,y)=(1,1)$
Albert said:my solution:
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 $(x,y)=(1,1)$
solakis said: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 :\(\displaystyle ax^2+bx+c=0\),\(\displaystyle a\neq 0\)
The OP wants all the real solutions
The equation is asking you to find the values of x and y that make the equation true.
The degree of an equation is the highest exponent of any variable. In this case, the highest exponent is 3, so the equation is a third-degree equation.
Yes, this equation can be solved algebraically by using techniques such as factoring or the quadratic formula.
There is no specific method that must be used to solve this equation. You can choose the method that you are most comfortable with or that seems most appropriate for the equation.
Yes, there are restrictions on the values of x and y in this equation. For example, since there are no square roots in this equation, the values of x and y must be real numbers. Additionally, the equation may have unique solutions or multiple solutions, depending on the values of the coefficients.