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[Nobby]

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[Planck's constant]

Solve the following simultaneous equations for x and y:

(p-q)x+(p+q)y = (p+q)²
qx-py = q² -pq


Cheers

Nothing clever comes to mind immediately, so try brute force.

Multiply the first equation  by (q)
Multiply the second equation by (p-q)

You can now eliminate x and (reasonably) quickly arrive at,

y =2q       (I think, I did it very quickly)

etc

PS. Before you do anything rewrite the RHS of the second equation as -q(p-q)   ... you'll find out why

[rife168]

Would you be able to elaborate a bit, Argonaut? Particularly how you decided to do these steps:

Multiply the first equation  by (q)
Multiply the second equation by (p-q)

I haven't worked through it fully yet, but those steps don't seem intuitive..

[Planck's constant]

Would you be able to elaborate a bit, Argonaut? Particularly how you decided to do these steps:

Multiply the first equation  by (q)
Multiply the second equation by (p-q)

I haven't worked through it fully yet, but those steps don't seem intuitive..

At this juncture of your VCE career, there are 2 methods available to you for solving simultaneous equations of this type (ie 2 linear equations in 2 variables, x and y)

Method 1: Elimination
Method 2: Substitution

The method  proposed in my post is method 1 - Elimination

Multiplying the first equation by (q) and the second equation by (p-q) results in both equations having the same coefficient for variable x, namely p(p-q).

This means that you can now subtract one equation from the other which eliminates x, and results in an equation whose only variable is y
Therefore you can find y in terms of p's and q's
Use this result (y in terms of p's and q's) to find x by substitution into one of your original equations.

PS. The reason I said that nothing clever quickly comes to mind in my earlier post, is the fact that the section of AGM this problem comes (from memory, its been 2 years since I've seen this stuff) lends itself to some clever tricks.

[rife168]

Thanks, don't know why I didn't realise that.. I think I was too intimidated by how messy it was going to get if I multiplied the equations as such^^ and then expanded... Ouch.

[rife168]

Oh, by the way,

At this juncture of your VCE career, there are 2 methods available to you for solving simultaneous equations of this type

Are you implying there are more ways?

[Planck's constant]

Oh, by the way,

At this juncture of your VCE career, there are 2 methods available to you for solving simultaneous equations of this type

Are you implying there are more ways?

Matrices
If they have not come up yet, they soon will

[Nobby]

Oh, by the way,

At this juncture of your VCE career, there are 2 methods available to you for solving simultaneous equations of this type

Are you implying there are more ways?

The 'augmented' or something ones with row operations something or rather? And you try to get the little corner of 0's?


Matrices
If they have not come up yet, they soon will

[Nobby]

Oh, by the way,

At this juncture of your VCE career, there are 2 methods available to you for solving simultaneous equations of this type

Are you implying there are more ways?

The augmented ones? With row operations something or rather, and you try to get the little corner of 0's?


Matrices
If they have not come up yet, they soon will

[rife168]

yeah, iirc you have to convert an augmented matrix to reduced-row echelon form then you can work out the solutions, but that's another whole topic.

[Planck's constant]

The augmented ones? With row operations something or rather, and you try to get the little corner of 0's?

Yes and no.
At VCE level, even the Matrix method of solving simultaneous equations is restricted to 2 variables.
Because in order to solve simultaneous linear equations you need, more or less, to find the inverse of the coefficient matrix
Finding the inverse of a 2x2 matrix by rule, is simple enough (swap one diagonal .... etc etc)
But finding the inverse of higher order square matrices requires methods which are not taught at VCE.
You need to perform row operations which leads to the 'augmented matrix' method you alluded to.
But this is undergrad level Linear Algebra which I happened to study as part of UMEP maths last year.