My Math Forum System of equation

 Algebra Pre-Algebra and Basic Algebra Math Forum

 March 26th, 2014, 04:12 PM #1 Senior Member   Joined: Nov 2013 Posts: 137 Thanks: 1 System of equation I want to solve for u and v the following system: $\begin{cases} 0=(3a(u+v)+b)\\ 0=(3a(u+v)^2 + 2b(u+v)+c)\\ \end{cases}$ I thought in to resolve by sum of product, so: $\begin{cases} u+v=-\frac{b}{3a}\\ uv=\frac{b^2}{3^2a^2}-\frac{c}{6a}-\frac{u^2+v^2}{2}\\ \end{cases}$ Now appears another complicated term: uČ+vČ Anyway, I don't know how to solve this...
March 26th, 2014, 05:22 PM   #2
Math Team

Joined: Oct 2011

Posts: 14,597
Thanks: 1038

Re: System of equation

Quote:
 Originally Posted by Jhenrique $\begin{cases} 0=(3a(u+v)+b)\\ 0=(3a(u+v)^2 + 2b(u+v)+c)\\ \end{cases}$
k = u+v

3ak + b = 0 [1]
3ak^2+ 2bk + c = 0 [2]
-3ak^2 - bk = 0 : [1] * -k
=============
bk + c = 0 ; substitute back in:
b(u + v) + c = 0
bu + bv = -c
bu = -bv - c
u = (-bv - c) / b

Hmmm....dunno....

 March 26th, 2014, 06:16 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2665 Math Focus: Mainly analysis and algebra Re: System of equation u and v cannot be uniquely determined by that system. Since both equations are (as Denis noticed) in (u + v) we will find that if $u= p, v = q$ is a solution, then so are $u= p - 1, v = q + 1$ and $u= q, v = p$. In other words, the equations must be either linearly dependent or inconsistent.
March 26th, 2014, 07:45 PM   #4
Math Team

Joined: Oct 2011

Posts: 14,597
Thanks: 1038

Re: System of equation

Quote:
 Originally Posted by Jhenrique $\begin{cases} 0=(3a(u+v)+b)\\ 0=(3a(u+v)^2 + 2b(u+v)+c)\\ \end{cases}$ $\begin{cases} u+v=-\frac{b}{3a}\\ uv=\frac{b^2}{3^2a^2}-\frac{c}{6a}-\frac{u^2+v^2}{2}\\ \end{cases}$
How did you get denominator 3^2 a^2? The 3a is not squared:
I think your unnecessary brackets at start and end confused you; yes?

I make that:
-uv = (u^2 + v^2)/2 + b(u + v) / (3a) + c / (6a)

March 26th, 2014, 10:51 PM   #5
Senior Member

Joined: Apr 2013

Posts: 425
Thanks: 24

Re: System of equation

Quote:
 Originally Posted by Jhenrique I want to solve for u and v the following system:$\begin{cases} 0=(3a(u+v)+b)\\ 0=(3a(u+v)^2 + 2b(u+v)+c)\\ \end{cases}$
$u+v=-\frac{b}{3a}=\frac{-b\pm \sqrt{b^2-3ac}}{3a}$ and so $b=\pm \sqrt{3ac}$ and $u=\pm \sqrt{\frac{c}{3a}}-v$ where $v$ is any number.

March 27th, 2014, 03:12 AM   #6
Math Team

Joined: Oct 2011

Posts: 14,597
Thanks: 1038

Re: System of equation

Quote:
 Originally Posted by Dacu ..and $u=\pm \sqrt{\frac{c}{3a}}-v$ where $v$ is any number.
or v = +-SQRT[c/(3a)] - u where u is any number

 Tags equation, system

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post baltoxionokota Differential Equations 5 July 17th, 2012 03:42 AM Rekooo Algebra 1 December 3rd, 2010 12:55 AM mariorossi Algebra 3 September 18th, 2010 02:51 PM Sara so Algebra 4 September 5th, 2010 01:36 PM gabor7896 Linear Algebra 2 September 11th, 2009 05:49 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top