Equation for Finding Solutions of x and y

[Albert1]

find solutions of $x$ and $y$

$4x+3y-2x\begin{bmatrix}
\dfrac{x^2+y^2}{x^2}
\end{bmatrix}=0
$

 

[Prove It]

Re: find solutions of equation

find solutions of $x$ and $y$

$4x+3y-2x\begin{bmatrix}
\dfrac{x^2+y^2}{x^2}
\end{bmatrix}=0
$

Do you mean to find y in terms of x?

 

[Albert1]

Re: find solutions of equation

Do you mean to find y in terms of x?

if x,y not specialized value , then express y in terms of x

 

[Prove It]

Re: find solutions of equation

Spoiler

Expand, move everything with a y to one side, and then complete the square to solve for y.

 

[Evobeus]

Re: find solutions of equation

find solutions of $x$ and $y$

$4x+3y-2x\begin{bmatrix}
\dfrac{x^2+y^2}{x^2}
\end{bmatrix}=0
$

Is [ ] denoting the greatest integer function or is it just a bracket replacement of ( ) ?

 

[Albert1]

Re: find solutions of equation

Is [ ] denoting the greatest integer function or is it just a bracket replacement of ( ) ?

[ ] denotes the greatest integer function

 

[Evobeus]

find solutions of $x$ and $y$

$4x+3y-2x\begin{bmatrix}
\dfrac{x^2+y^2}{x^2}
\end{bmatrix}=0
$

Spoiler

First we easily see that $x \ne 0$
So, now divide both sides of the given equation by $x$
We get,
$4+3 \frac{y}{x} -2\begin{bmatrix}
\dfrac{x^2+y^2}{x^2}
\end{bmatrix} =0$
From there we see that $\frac{y}{x}$ must be an integer.
Thus,
$y=k*x$ where $k$ is an integer.
Now replace $y$ with $k*x$ in the given equation and divide it by $x$.
This gives us
$2k^2-3k-2=0$
Solving for $k$ gives us $k=2$ or $k=-\frac{1}{2}$
Therefore,
$x$ is any non zero real number and $y=k*x$ where $k=2$ or $k=-\frac{1}{2}$

 

[Albert1]

Spoiler

First we easily see that $x \ne 0$
So, now divide both sides of the given equation by $x$
We get,
$4+3 \frac{y}{x} -2\begin{bmatrix}
\dfrac{x^2+y^2}{x^2}
\end{bmatrix} =0$
From there we see that $\frac{y}{x}$ must be an integer.
Thus,
$y=k*x$ where $k$ is an integer.
Now replace $y$ with $k*x$ in the given equation and divide it by $x$.
This gives us
$2k^2-3k-2=0$
Solving for $k$ gives us $k=2$ or $k=-\frac{1}{2}$
Therefore,
$x$ is any non zero real number and $y=k*x$ where $k=2$ or $k=-\frac{1}{2}$

Spoiler

$y=2x , \,\,\,\,\,\,\,\,\,\,\,\bigcirc \\
y=-\dfrac{x}{2}, \,\,\,\,\,\,\,\,\times$
$y=kx $
(where $k$ may or may not be an integer)

 

[Evobeus]

Spoiler

Spoiler

First we easily see that $x \ne 0$
So, now divide both sides of the given equation by $x$
We get,
$4+3 \frac{y}{x} -2\begin{bmatrix}
\dfrac{x^2+y^2}{x^2}
\end{bmatrix} =0$
From there we see that $\frac{y}{x}$ must be an integer.
Thus,
$y=k*x$ where $k$ is an integer.
Now replace $y$ with $k*x$ in the given equation and divide it by $x$.
This gives us
$2k^2-3k-2=0$
Solving for $k$ gives us $k=2$ or $k=-\frac{1}{2}$
Therefore,
$x$ is any non zero real number and $y=k*x$ where $k=2$ or $k=-\frac{1}{2}$

My solution (the one quoted above {^^} [ in this post ] ) is correct in terms of the idea or methodical procedure or algorithm (algorithm is a good word) but is wrong in terms of the answer

Although Albert's post very effectively helped me (and I am sure many who read my solution), but I cannot understand Albert's claim clearly (sorry for that --it is my fault )

Spoiler

$y=2x , \,\,\,\,\,\,\,\,\,\,\,\bigcirc \\
y=-\dfrac{x}{2}, \,\,\,\,\,\,\,\,\times$
$y=kx $
(where $k$ may or may not be an integer)

So, after spending so many words on something that could be described in a single line I describe the error and the solution again :
First : See that $x \ne 0$ [why ? ] [Reason : $x^2$ is in the denomianator ]
Second : Divide both sides of the given equation by $x^2$ [It is possible because $x \ne 0$ and $x,y \in \mathbb{R}$ ]
Third: we get an expression
$4+3*\frac{y}{x} - 2 ( \lfloor \frac{x^2+y^2}{x^2} \rfloor ) = 0$
Fourth:We get an expression
$3* \frac{y}{x}$ = integer
So ,
$3y=k*x$ where $k$ is integer.
earlier I wrote :
$y=k*x$ where $k$ is an integer but actually $\boxed{3}y=k*x$ where $k$ is an integer.

Now proceed as done in the solution and get the solution.

 

[Albert1]

Spoiler

My solution (the one quoted above {^^} [ in this post ] ) is correct in terms of the idea or methodical procedure or algorithm (algorithm is a good word) but is wrong in terms of the answer

Although Albert's post very effectively helped me (and I am sure many who read my solution), but I cannot understand Albert's claim clearly (sorry for that --it is my fault )

So, after spending so many words on something that could be described in a single line I describe the error and the solution again :
First : See that $x \ne 0$ [why ? ] [Reason : $x^2$ is in the denomianator ]
Second : Divide both sides of the given equation by $x^2$ [It is possible because $x \ne 0$ and $x,y \in \mathbb{R}$ ]
Third: we get an expression
$4+3*\frac{y}{x} - 2 ( \lfloor \frac{x^2+y^2}{x^2} \rfloor ) = 0$
Fourth:We get an expression
$3* \frac{y}{x}$ = integer
So ,
$3y=k*x$ where $k$ is integer.
earlier I wrote :
$y=k*x$ where $k$ is an integer but actually $\boxed{3}y=k*x$ where $k$ is an integer.

Now proceed as done in the solution and get the solution.

Spoiler

so the solutions
(1)y=2x correct
(2)y=-0.5x wrong
check it again