Solve these simultaneous equations

[chwala]

Homework Statement

See attached.

Relevant Equations

Simultaneous equations

In my approach,
$\dfrac{x^2}{y}+ y =2.5x$

and we know that

$2y^2=9-x^2$

$y^2=\dfrac{9-x^2}{2}$

$(\dfrac{x^2}{y}+ y)(\dfrac{x^2}{y}+ y)=(2.5x)^2$

$\dfrac{x^4}{y^2} + x^2+x^2+y^2=6.25x^2$

$\dfrac{x^4}{y^2}+2x^2+\dfrac{9-x^2}{2}=6.25x^2$

$\dfrac{2x^4}{9-x^2} +2x^2+\dfrac{9-x^2}{2}=6.25x^2$

$4x^4+2x^2(18-2x^2)+(9-x^2)^2=6.25x^2(18-2x^2)$

...

$36x^2+81-18x^2+x^4=112x^2-12.5x^4$

$12.5x^4+x^4-94.5x^2+81=0$

$13.5x^4-94.5x^2+81=0$

$x_1=6$ We cannot have $y$ values for $x_1=6$

and for $x_2=1 ⇒y= ±2$

There should be a better approach...will explore later.

 

[topsquark]

Homework Statement: See attached.
Relevant Equations: Simultaneous equations

View attachment 331231In my approach,
$\dfrac{x^2}{y}+ y =2.5x$

and we know that

$2y^2=9-x^2$

$y^2=\dfrac{9-x^2}{2}$

$(\dfrac{x^2}{y}+ y)(\dfrac{x^2}{y}+ y)=(2.5x)^2$

$\dfrac{x^4}{y^2} + x^2+x^2+y^2=6.25x^2$

$\dfrac{x^4}{y^2}+2x^2+\dfrac{9-x^2}{2}=6.25x^2$

$\dfrac{2x^4}{9-x^2} +2x^2+\dfrac{9-x^2}{2}=6.25x^2$

$4x^4+2x^2(18-2x^2)+(9-x^2)^2=6.25x^2(18-2x^2)$

...

$36x^2+81-18x^2+x^4=112x^2-12.5x^4$

$12.5x^4+x^4-94.5x^2+81=0$

$13.5x^4-94.5x^2+81=0$

$x_1=6$ We cannot have $y$ values for $x_1=6$

and for $x_2=1 ⇒y= ±2$

There should be a better approach...will explore later.

I didn't look over the whole thing, but one suggestion: leave fractions as fractions (that 5/2 thing at the start.) Decimals are usually approximations... better to maintain the exact expressions. And who knows? Something might cancel later.

And your solution method is probably as good as it's going to get for this one.

-Dan

Addendum: I just looked at the graph. You are missing two solutions.

 

[chwala]

I didn't look over the whole thing, but one suggestion: leave fractions as fractions (that 5/2 thing at the start.) Decimals are usually approximations... better to maintain the exact expressions. And who knows? Something might cancel later.

And your solution method is probably as good as it's going to get for this one.

-Dan

Addendum: I just looked at the graph. You are missing two solutions.

Yes, i am aware that there are other solutions...which can easily be determined when solving for $x$... i am seeking other better ways of solving this particular problem...

 

[SammyS]

Homework Statement: See attached.
Relevant Equations: Simultaneous equations

There should be a better approach...will explore later.

Multiply both sides of the first equation, $\displaystyle \frac x y +\frac y x = \frac 5 2 \$ by $\displaystyle 2xy$ .

Rewrite as $\displaystyle 2x^2 - 5xy + 2y^2 =0$ .

Factor and solve for $x$ or for $y$.

Plug the result into the other given equation and solve.

 

[chwala]

Multiply both sides of the first equation, $\displaystyle \frac x y +\frac y x = \frac 5 2 \$ by $\displaystyle 2xy$ .

Rewrite as $\displaystyle 2x^2 - 5xy + 2y^2 =0$ .

Factor and solve for $x$ or for $y$.

Plug the result into the other given equation and solve.

@SammyS thanks let me check this approach...

 

[SammyS]

Homework Statement: See attached.
Relevant Equations: Simultaneous equations

In my approach,

. . .

$13.5x^4-94.5x^2+81=0$

$x_1=6$ We cannot have $y$ values for $x_1=6$

and for $x_2=1 ⇒y= ±2$

There should be a better approach...will explore later.

Your workings are correct down to and including the degree 4 polynomial equation I have quoted.

Note that multiplying that equation by 2 gives:

$\displaystyle 27x^4-189x^2+162=0$

After factoring out 27 and discarding it, you get:

$\displaystyle x^4-7x^2+6=0$ .

This factors easily into $\displaystyle (x^4-1)(x^2-6)=0$ , so that either $\displaystyle x^2=1$ or $\displaystyle x^2=6$ , rather than $\displaystyle x=1$ or $\displaystyle x=6$ as you had.

That still leaves you with some details to finish up to arrive at the overall solution.

 

[chwala]

@SammyS yeah I will look at this again... will be back in a few days...