My Math Forum Systems of Linear Equations

 Algebra Pre-Algebra and Basic Algebra Math Forum

 August 8th, 2014, 05:26 AM #1 Member   Joined: Aug 2014 From: Lithuania Posts: 60 Thanks: 3 Systems of Linear Equations I need to solve these equations for a, b, c, and d. Really important!
 August 8th, 2014, 05:35 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,640 Thanks: 2624 Math Focus: Mainly analysis and algebra The first equation allows you easily to elimate $c$, and the last allows you to eliminate $d$. You will then have two equations (the second and third) in two variables (a and b) to solve. Post what you get from that.
 August 8th, 2014, 06:01 AM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Sidenote: this is a system of quadratic equations, not linear. (The first one is linear, the others are nonlinear.)
 August 8th, 2014, 06:17 AM #4 Member   Joined: Aug 2014 From: Lithuania Posts: 60 Thanks: 3 Thanks for helping! I understood your point, but the numbers I've got are just ridiculous (the equation I've got: $\displaystyle 2197-169b^2-13b^4+b^8-144b^5=13b$. I can't solve it . I've been trying to solve this problem for over a week now, but it's to hard for me. If you could solve it step by step, I would be very grateful.
 August 8th, 2014, 07:32 AM #5 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 The equation I get is: $b= \frac{26}{a^2 + 13 + \frac{12}{a}}$ I tried a = 1 and got the solution: a = 1, b = 1, c = -1, d = 13. Last edited by Pero; August 8th, 2014 at 08:04 AM.
August 8th, 2014, 08:06 AM   #6
Math Team

Joined: Dec 2013
From: Colombia

Posts: 7,640
Thanks: 2624

Math Focus: Mainly analysis and algebra
Quote:
 Originally Posted by kaspis245 Thanks for helping! I understood your point, but the numbers I've got are just ridiculous...
Yes, I agree. However, you now have a single solution, so you can start factorising your equation for $b$.

I think it's a badly set question to be honest. You can use equations like Pero produced to find some other possible values too. I'd be very tempted to cheat and use something like Wolfram Alpha to factorise the equation for me, because it would be a lot of work manually. Once you have the (integer) roots you can always show that they are roots (and factorise) by hand.

August 8th, 2014, 08:09 AM   #7
Math Team

Joined: Dec 2013
From: Colombia

Posts: 7,640
Thanks: 2624

Math Focus: Mainly analysis and algebra
Quote:
 Originally Posted by Pero I tried a = 1 and got the solution: a = 1, b = 1, c = -1, d = 13.
$a = 3$ also gives $b=1$, and then $c=-3$ and $d=13$, but that doesn't work with the middle two equations.

 August 8th, 2014, 08:21 AM #8 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 There is also a = -1, b = 13, c = 1 and d = 1. I also get: $a^6 + 26a^4 + 117a^2 - 144= 0$ Which is a cubic in $a^2$ No time to look at this further now, but that looks better than the equation in b.
 August 8th, 2014, 08:41 AM #9 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 ... the other roots of the cubic are negative, so only solutions for a^2 = 1, i.e. a = +/-1.
 August 8th, 2014, 10:07 AM #10 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Putting it all together: Replacing c and d gives: $\frac{13}{b} + b= a^2 + 13 \ and \ \frac{13a}{b} - ab = 12$ $\frac{13}{b} + b= a^2 + 13 \ and \ \frac{13}{b} - b = \frac{12}{a}$ Adding these equations gives: $\frac{26}{b}= a^2 + 13 + \frac{12}{a} \ hence \ b = \frac{26}{a^2 + 13 + \frac{12}{a}$ And subtracting these equations gives: $2b= a^2 + 13 - \frac{12}{a}$ So: $2b= \frac{52}{a^2 + 13 + \frac{12}{a}} = a^2 + 13 - \frac{12}{a}$ Which simplifies to: $a^6 + 26a^4 + 117a^2 - 144= 0$ $Let \ x= a^2$ By inspection x = 1 is a root of the cubic $x^3 + 26x^2 + 117x -144$, so: $x^3 + 26x^2 + 117x -144= (x - 1)(x^2 + 27x + 144)$ The other roots are: $x= \frac{-27 \pm \sqrt{153}}{2}$ and are both negative. So, there is only one positive solution for x, hence a = 1 or -1. These lead to the solutions: a = 1, b = 1, c = -1, d = 13 a = -1, b = 13, c = 1, d = 1

 Tags equations, linear, systems

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post david940 Linear Algebra 2 July 25th, 2014 06:12 PM Akcope Calculus 3 March 14th, 2014 01:49 AM Jake_88 Linear Algebra 1 June 15th, 2010 05:59 AM maxpalme Algebra 1 April 2nd, 2009 10:33 AM maxpalme Abstract Algebra 1 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top