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August 8th, 2014, 05:26 AM   #1
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Systems of Linear Equations

I need to solve these equations for a, b, c, and d.
Really important!

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August 8th, 2014, 05:35 AM   #2
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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.
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August 8th, 2014, 06:01 AM   #3
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Sidenote: this is a system of quadratic equations, not linear. (The first one is linear, the others are nonlinear.)
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August 8th, 2014, 06:17 AM   #4
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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.
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August 8th, 2014, 07:32 AM   #5
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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.
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August 8th, 2014, 08:06 AM   #6
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Quote:
Originally Posted by kaspis245 View Post
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.
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August 8th, 2014, 08:09 AM   #7
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Originally Posted by Pero View Post
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.
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August 8th, 2014, 08:21 AM   #8
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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.
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August 8th, 2014, 08:41 AM   #9
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... the other roots of the cubic are negative, so only solutions for a^2 = 1, i.e. a = +/-1.
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August 8th, 2014, 10:07 AM   #10
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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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