# Solve these simultaneous equations

• chwala
chwala
Gold Member
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.

chwala said:
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.

SammyS and chwala
topsquark said:
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...

chwala said:
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
SammyS said:
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...

chwala said:
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 and MatinSAR
@SammyS yeah I will look at this again... will be back in a few days...

## Related to Solve these simultaneous equations

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## What are simultaneous equations?

Simultaneous equations are a set of equations containing multiple variables. The solution to these equations is the set of values that satisfy all equations simultaneously.

## What methods can be used to solve simultaneous equations?

Common methods to solve simultaneous equations include substitution, elimination, and using matrices (such as the Gaussian elimination method). Graphical methods can also be used for systems with two variables.

## How do you solve simultaneous equations using the substitution method?

To solve using substitution, solve one of the equations for one variable in terms of the other. Then, substitute this expression into the other equation. This will result in a single equation with one variable, which can be solved. Finally, substitute back to find the other variable.

## How does the elimination method work for solving simultaneous equations?

In the elimination method, you manipulate the equations to eliminate one of the variables. This is typically done by adding or subtracting the equations after multiplying them by suitable coefficients. This reduces the system to a single equation with one variable, which can be solved. Then, substitute back to find the other variable.

## Can simultaneous equations have no solution or infinitely many solutions?

Yes, simultaneous equations can have no solution if the lines or planes they represent are parallel and do not intersect. They can have infinitely many solutions if the equations represent the same line or plane, meaning they are dependent.



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