Simultaneous equations: Can you tell me how to solve the simultaneous equations X^-xy+3y^=15 X^+xy-y^=5

[reggiwilliams]

Can you tell me how to solve the simultaneous equations
X^-xy+3y^=15
X^+xy-y^=5

 

[stapel]

Can you tell me how to solve the simultaneous equations
X^-xy+3y^=15
X^+xy-y^=5

We'll first need you to explain what the equations are.

For instance, normally I would assume that "X^-xy" means "X^(-xy)", which typesets as [imath]X^{-xy}[/imath]. But then you have "3y^" with nothing after the carat, so you are using the carat notation in a non-standard way. We'll need you to tell us how you're defining the carat symbol.

Also, you have two equations in three unknowns (namely, X, x, and y), which generally means that no unique numerical solution can be found. So please include corrections of the equations, or else please specify that you really are working with three variables.

When you reply, please include a clear listing of your thoughts and efforts so far, so we can see where you're getting stuck. Thank you!

 

[reggiwilliams]

I attach typed version
I trust this clears up any discrepancies and await your solution along with workings.

 

[Dr.Peterson]

Can you tell me how to solve the simultaneous equations
X^-xy+3y^=15
X^+xy-y^=5

Possibly you are using "^" as if it were the exponent 2 (which is very nonstandard). And quite likely you didn't mean X to be different from x.

If so, the equations are [math]x^2-xy+3y^2=15\\x^2+xy-y^2=5[/math]
Nonlinear systems like this can be tricky; you could try solving one for y and substituting in the other, or you might first do something like adding the two equations to make that easier.

Is this for a class you are taking?

 

[reggiwilliams]

Stumbled across question when revising for exam in June. Have been doing a correspondence course.
Can’t find in any textbook the method to solve. Only way I can see is by trial or error which can’t be right.
Even when you isolate y still get a long fourth degree polynomial equation which I can’t factorise

 

[Dr.Peterson]

Stumbled across question when revising for exam in June. Have been doing a correspondence course.
Can’t find in any textbook the method to solve. Only way I can see is by trial or error which can’t be right.
Even when you isolate y still get a long fourth degree polynomial equation which I can’t factorise

I added the equations together, as I suggested, and solved the resulting equation for y in terms of x, then plugged that into the second equation. That results in a quartic equation, but one that is really a quadratic in x^2, and is easily solved.

There may well be another way that makes it easier; one possibility may be to rotate these two conics to standard positions, which would result in more symmetry.

​

Possibly if we knew what topics are covered by the exam for which this was written, we could think of other tools to use.

 

[reggiwilliams]

Here are my workings which show y= -or + sqr of 5 and similarly x = - or + sqr of 5
Is that correct
From trial or error also discovered x=-3 y=1 and when x=3 y =-1 but have no idea the method to get to this answer

 

[Dr.Peterson]

Here are my workings which show y= -or + sqr of 5 and similarly x = - or + sqr of 5
Is that correct
From trial or error also discovered x=-3 y=1 and when x=3 y =-1 but have no idea the method to get to this answer

If you really squared both sides here, you would still have a radical:

​

You need to isolate the radical before squaring.

 

[reggiwilliams]

See where I went wrong. squaring both sides properly left with equation 5y^4 -30y^ + 25 which when factorised gives me (5y^-5)(y^-5)
y = +or - 1
y=+or -Sqr of 5
When y=1 x=-3
When y=-1x=3
When y =sqrt of 5 x=sqrt of 5
When y =-sqrt of 5 x=-sqrt of 5
Graph tells you there has to be four possible solutions
Think I have grasped it know.

 

[lookagain]

See where I went wrong. squaring both sides properly left with equation 5y^4 -30y^ + 25 which when factorised gives me (5y^-5)(y^-5)

reggiwilliams,

you have grasped it (as far as I can tell), but you are not writing all of the parts correctly. You need punctuation to
separate sections so that they do not merge together, you need to show all exponents, and you need to show an
equals sign with the zero on the opposite side so that you are actually writing equations. It will benefit you to
factor out the greatest common factor of a quadratic at one point to ease its factoring into the product of two
binomial factors:

5y^4 - 30y^2 + 25 = 0
5(y^4 - 6y^2 + 5) = 0
5(y^2 - 1)(y^2 - 5) = 0

y^2 - 1 = 0
y^2 = 1
y = + or -1

or

y^2 - 5 = 0
y^2 = 5
y = + or -sqrt(5)

When y = 1, x = -3.
When y = -1, x = 3.
When y = sqrt(5), x = sqrt(5).
When y = -sqrt(5), x = -sqrt(5).

 

[reggiwilliams]

Thanks for all your help
If I come across a question like this again I won’t forget how to approach it