Algebra Pre-Algebra and Basic Algebra Math Forum

 August 9th, 2018, 03:15 PM #1 Newbie   Joined: Mar 2017 From: Norway Posts: 12 Thanks: 0 Please help I am struggling to solve these simultaneous equations: $a^2+ab+b^2=7$ $b^2+bc+c^2=28$ $c^2+ac+a^2=21$ Any help would be highly appreciated.
 August 9th, 2018, 04:51 PM #2 Newbie   Joined: Mar 2017 From: Norway Posts: 12 Thanks: 0 I couldn't go any further than this: $(a+b)^2=7+ab$ $(b+c)^2=28+bc$ $(a+c)^2=21+ac$
 August 9th, 2018, 05:14 PM #3 Senior Member   Joined: Sep 2016 From: USA Posts: 559 Thanks: 324 Math Focus: Dynamical systems, analytic function theory, numerics Sum the first and third equation. Notice the LHS of each must sum to the LHS of the 2nd equation. Thanks from topsquark
 August 9th, 2018, 05:24 PM #4 Newbie   Joined: Mar 2017 From: Norway Posts: 12 Thanks: 0 $a^2+ab+b^2+c^2+ac+a^2=28$ $a^2+ab+b^2+c^2+ac+a^2=b^2+bc+c^2$ $2a^2+ab+ac-bc=0$ and now what
August 9th, 2018, 05:34 PM   #5
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,042
Thanks: 815

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by Farzin $a^2+ab+b^2+c^2+ac+a^2=28$ $a^2+ab+b^2+c^2+ac+a^2=b^2+bc+c^2$ $2a^2+ab+ac-bc=0$ and now what
This may or may not be what SDK is moving you toward but notice that you can solve for c in terms of a and b in your last equation.

-Dan

 August 9th, 2018, 05:39 PM #6 Senior Member   Joined: May 2016 From: USA Posts: 1,306 Thanks: 549 This problem has basically been answered over at FMH. https://www.freemathhelp.com/forum/t...73-Please-help It can be turned into a quartic. if the assumption is further made that the answers are integers, it can be shown that $|b| < 4.$ I gave one real answer over there, so there is at least one other real answer. Last edited by JeffM1; August 9th, 2018 at 05:41 PM.
 August 10th, 2018, 02:20 AM #7 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,139 Thanks: 721 Math Focus: Physics, mathematical modelling, numerical and computational solutions a=1, b=2, c=4 I got this answer by making the assumption that the numbers were all positive and then guessing some numbers to feel out the problem though, not by implementing a clever method or anything. Last edited by Benit13; August 10th, 2018 at 03:09 AM.
 August 10th, 2018, 06:36 AM #8 Global Moderator   Joined: Dec 2006 Posts: 20,302 Thanks: 1974 7(a - b) + 28(b - c) + 21(c - a) = a³ - b³ + b³ - c³ + c³ - a³ = 0 Hence c = 3b - 2a. Substituting for c in the 3rd equation gives 9b² - 12ab + 4a² + 3ab - 2a² + a² = 21, so 3b² - 3ab + a² = 7. Subtracting that from the first equation gives 4ab - 2b² = 0. Hence 2b(2a - b) = 0. Substituting b = 2a into the first equation gives a² = 1. Hence (a, b, c) = (1, 2, 4) or (-1, -2, -4) or (√7, 0, -2√7) or (-√7, 0, 2√7). Thanks from topsquark and JeffM1
August 10th, 2018, 09:25 AM   #9
Senior Member

Joined: May 2016
From: USA

Posts: 1,306
Thanks: 549

Quote:
 Originally Posted by skipjack 7(a - b) + 28(b - c) + 21(c - a) = a³ - b³ + b³ - c³ + c³ - a³ = 0 Hence c = 3b - 2a. Substituting for c in the 3rd equation gives 9b² - 12ab + 4a² + 3ab - 2a² + a² = 21, so 3b² - 3ab + a² = 7. Subtracting that from the first equation gives 4ab - 2b² = 0. Hence 2b(2a - b) = 0. Substituting b = 2a into the first equation gives a² = 1. Hence (a, b, c) = (1, 2, 4) or (-1, -2, -4) or (√7, 0, -2√7) or (-√7, 0, 2√7).
Very pretty. Once you "see" the difference of cubes, you can avoid the quartic altogether.

August 10th, 2018, 10:54 AM   #10
Senior Member

Joined: Sep 2016
From: USA

Posts: 559
Thanks: 324

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
 Originally Posted by topsquark This may or may not be what SDK is moving you toward but notice that you can solve for c in terms of a and b in your last equation. -Dan
Nope I was in a hurry and made a mistake. I thought the sum had a factor of $a$. I strongly dislike questions like this at a lower level since it gives the impression that math is just a bunch of clever tricks to be remembered.

In reality, despite the large amount of structure in this problem, it is still extremely difficult or impossible to solve most nonlinear systems exactly. Even in the case you know ahead of time that there are integer solutions, trying to eliminate variables is still the worst possible way to proceed. In this case you would just do Newton and round to the nearest integer.

There is no case where playing around with the equations is the "right" thing to do yet at low level this is the only thing students are equipped to do. These problems are toxic.

 Thread Tools Display Modes Linear Mode

 Contact - Home - Forums - Cryptocurrency Forum - Top