A system of equations.

**iammax27** · March 9th 2010, 06:28 PM

Dear Mathematical community,

(x^2)yz+3y-z = 31
(x^2)+(y^2)-z = 11
3xy+zy = 24

I've been working with this problem for the last week or so and it seems every way I approach the problem it becomes over complicated. I've tried every last elimination and substitution method I know along with using inverse matrices. So any help, or constructive comments would be much appreciated.

Thanks,
Max Thompson

P.S. I know the real integer solutions to this problem, the real issue is I don't know how to algebraically arrive at said solutions.

**Prove It** · March 9th 2010, 06:36 PM

Originally Posted by iammax27

Dear Mathematical community,

(x^2)yz+3y-z = 31
(x^2)+(y^2)-z = 11
3xy+zy = 24

I've been working with this problem for the last week or so and it seems every way I approach the problem it becomes over complicated. I've tried every last elimination and substitution method I know along with using inverse matrices. So any help, or constructive comments would be much appreciated.

Thanks,
Max Thompson

P.S. I know the real integer solutions to this problem, the real issue is I don't know how to algebraically arrive at said solutions.

The substitution method might be easier than the elimination method in this case because you can easily write equation 2 in terms of $z$ , whereas making the variables have the same coefficients could be tiresome.

**Prove It** · March 9th 2010, 06:48 PM

OK, seeing as this was quite difficult, I've had a go of it.

$x^2yz + 3y - z = 31$
$x^2 + y^2 - z = 11$
$3xy + yz = 24$ .

From equation 2 we can see that

$z = x^2 + y^2 - 11$ .

Substituting into the other equations:

$x^2y(x^2 + y^2 - 11) + 3y - (x^2 + y^2- 11) = 31$
$3xy + y(x^2 + y^2 - 11) = 24$ .

Working on the second of these equations:

$y(3x + x^2 + y^2 - 11) = 24$

$x^2 + 3x + y^2 - 11 = \frac{24}{y}$

$x^2 + 3x - 11 = \frac{24}{y} - y^2$

$x^2 + 3x + \left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 - 11 = \frac{24}{y} - y^2$

$\left(x + \frac{3}{2}\right)^2 - \frac{53}{4} = \frac{24}{y} - y^2$

$\left(x + \frac{3}{2}\right)^2 = \frac{53}{4} + \frac{24}{y} - y^2$

$x + \frac{3}{2} = \pm \sqrt{\frac{53}{4} + \frac{24}{y} - y^2}$

$x = -\frac{3}{2} \pm \sqrt{\frac{53}{4} + \frac{24}{y} - y^2}$ .

You will now have to substitute this into

$x^2y(x^2 + y^2 - 11) + 3y - (x^2 + y^2- 11) = 31$

and solve for $y$ .

**iammax27** · March 9th 2010, 06:54 PM

(@ Prove it) I also took substitution as my first approach due to the enormity of work I would have to do to simplify using elimination, however, with the substitution I found myself arriving at two distinctive end results. One, I almost got everything in terms of x but found myself with a polynomial expression to the degree of 7 with 2 extra z's. Then the other way I found myself dealing with a giant complex fraction with x and z still present in the equation (same as before, 2 z's). I showed this to my physics professor and he's still a work in progress. Along with my Math professor, whom I sat with for an upmost of 2 hours trying to solve this problem.

So if anyone can actually work through it and show me their work I would love to see, and discuss said work, however, I know most people don't have lots of free time to spend doing arbitrary mathematics.

Still any thoughts or comments are much appreciated.

(Sorry, I wrote this response before I saw that you had worked it out, but thanks I'll take A look at your answer and get back to you)

**iammax27** · March 9th 2010, 07:02 PM

@ prove it

Thanks for the help, that was fantastic. I think the problem was that I sometimes become too tunnel visioned that I miss to do things like completing the square. Still thanks for the help.

Thread: A system of equations.

LinkBack

Thread Tools

Display

A system of equations.

Similar Math Help Forum Discussions

A system of 4 equations help

Linear system of equations. 3 equations, 3 variables

system of equations

system of equations

system of equations

Search Tags