Solve the given simultaneous equations

[chwala]

Homework Statement

See attached.

Relevant Equations

simultaneous equations

In my approach,

$(x+y)^2=4a^2$

$x^2+y^2=4a^2-2xy$

and

...

$x^2+y^2=xy+a^2$

then,

$4a^2-2xy=xy+a^2$

$3a^2=3xy$

$a^2=xy$

$⇒x=a, y=a$

Bingo!!

Any other approach is welcome ...

 

[phinds]

If you have a question, ask it, else why are you posting your work here?

 

[topsquark]

Homework Statement: See attached.
Relevant Equations: simultaneous equations

View attachment 331233

In my approach,

$(x+y)^2=4a^2$

$x^2+y^2=4a^2-2xy$

$x^2+y^2=xy+a^2$

$4a^2-2xy=xy+a^2$

$3a^2=3xy$

$a^2=xy$

$⇒x=a, y=a$

Bingo!!

You need to show more steps and where they come from.

Where did you get $x^2+y^2=xy+a^2$ from? It's correct, but it comes out of nowhere in your solution method.

Your solution to $a^2=xy$ is not the most general. x = a, y = a is certainly a solution, but any $y = a^2/x$ will also do. You need to plug this back into one of the original equations to get an expression for x and y in terms of a.

If your purpose is to check your solution method, a couple of points to consider:
1. Make it clear where each line is coming from. Be specific.

2. Explicitly work out each line step by step. This makes it much easier for anyone (especially yourself) to check for errors.

3. Review your work as if you didn't write it. That way you will be able to see more easily where you may have skipped steps when writing out the solution. (This is an acquired skill.)

-Dan

 

[chwala]

You need to show more steps and where they come from.

Where did you get $x^2+y^2=xy+a^2$ from? It's correct, but it comes out of nowhere in your solution method.

Your solution to $a^2=xy$ is not the most general. x = a, y = a is certainly a solution, but any $y = a^2/x$ will also do. You need to plug this back into one of the original equations to get an expression for x and y in terms of a.

If your purpose is to check your solution method, a couple of points to consider:
1. Make it clear where each line is coming from. Be specific.

2. Explicitly work out each line step by step. This makes it much easier for anyone (especially yourself) to check for errors.

3. Review your work as if you didn't write it. That way you will be able to see more easily where you may have skipped steps when writing out the solution. (This is an acquired skill.)

-Dan

Note that,

$x+y=2a$

from

$\dfrac{x^2+ax+y^2+ay}{xy+3a^2}=1$

we shall have,

$\dfrac{x^2+2a^2+y^2}{xy+3a^2}=1$

on cross -multiplication we get,$x^2+2a^2+y^2=xy+3a^2$

...

Cheers.

 

[topsquark]

Note that,

$x+y=2a$

then,

$\dfrac{x^2+2a^2+y^2}{xy+3a^2}=1$

then,

$x^2+2a^2+y^2=xy+3a^2$

...

Cheers.

First, I know how to solve this one. I know where the equation in line 3 comes from. I was giving you advice about just writing down an equation with no explanation where it came from.

Second, you just repeated the mistake. Where does the second equation come from? There is no equation on this page that says $xy + 3a^2 =$ something, much less how you applied it to the first equation in this post to get there.

What I was saying is that you need to start with
$\dfrac{x}{y+a} + \dfrac{y}{x+a} = 1$

and derive $x^2 + y^2 = xy + a^2$, or at least say that it comes from there.

For the most part your posts are clear enough. This one simply isn't.

-Dan

 

[SammyS]

First, I know how to solve this one. I know where the equation in line 3 comes from. I was giving you advice about just writing down an equation with no explanation where it came from.

Second, you just repeated the mistake. Where does the second equation come from? There is no equation on this page that says $xy + 3a^2 =$ something, much less how you applied it to the first equation in this post to get there.

What I was saying is that you need to start with
$\dfrac{x}{y+a} + \dfrac{y}{x+a} = 1$

and derive $x^2 + y^2 = xy + a^2$, or at least say that it comes from there.

For the most part your posts are clear enough. This one simply isn't.

-Dan

@chwala :

Dan, a.k.a. @topsquark , makes some good points above.

I only disagree with his final statement. You do similarly confusing things in many of your threads.

 

[pasmith]

Homework Statement: See attached.
Relevant Equations: simultaneous equations

View attachment 331233

In my approach,

$(x+y)^2=4a^2$

$x^2+y^2=4a^2-2xy$

and

...

$x^2+y^2=xy+a^2$

then,

$4a^2-2xy=xy+a^2$

$3a^2=3xy$

$a^2=xy$

$⇒x=a, y=a$

Bingo!!

Any other approach is welcome ...

If $x + y = 2a$ and $xy = a^2$ then $x$ and $y$ are the roots of $$\begin{split}<br /> 0 &= (z - x)(z - y) \\<br /> &= z^2 - (x + y)z + xy \\<br /> &= z^2 - 2az + a^2 \\<br /> &= (z - a)^2.\end{split}$$ Thus $(x,y) = (a,a)$ is the only possibility.

 

[PeroK]

Why not simply substitute $y = 2a - x$ into the first equation? That seems an obvious way to generate a quadratic in $x$.

 

[chwala]

Why not simply substitute $y = 2a - x$ into the first equation? That seems an obvious way to generate a quadratic in $x$.

This looks interesting...nice one you end up with

...
$(x-a)(x-a)=0$

$x=a$.

Cheers!