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August 8th, 2014, 05:26 AM  #1 
Member Joined: Aug 2014 From: Lithuania Posts: 62 Thanks: 3  Systems of Linear Equations 
August 8th, 2014, 05:35 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,691 Thanks: 2670 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: 62 Thanks: 3 
Thanks for helping! I understood your point, but the numbers I've got are just ridiculous (the equation I've got: $\displaystyle 2197169b^213b^4+b^8144b^5=13b$. 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: 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: 7,691 Thanks: 2670 Math Focus: Mainly analysis and algebra  Quote:
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: 7,691 Thanks: 2670 Math Focus: Mainly analysis and algebra  
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: Which is a cubic in 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: Adding these equations gives: And subtracting these equations gives: So: Which simplifies to: By inspection x = 1 is a root of the cubic , so: The other roots are: 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 

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