HARD Math question~math people, prove yourselves

<p>No no no no no no no.... I would NOT try using matrices on this one... Don't use substitution either... Both work, but they are not the easiest way to do it.... I started using matrices and then was like "ehhh..." I tried it a different and MUCH easier way.</p>

<p>With the equations:
25 = a + b + c + d
40 = 8a + 4b + 2c + d
60 = 27a + 9b + 3c + d
95 = 64a + 16b + 4c + d</p>

<p>Subtract the first equation from the second equation, the second equation from the third equation, and the third equation from the fourth equation.
This gives you:
15 = 7a + 3b + c
20 = 19a + 5b + c
35 = 37a + 7b + c</p>

<p>Now subtract the first equation from the second and the second equation from the third.</p>

<p>5 = 12a + 2b
15 = 18a + 2b</p>

<p>Now subtract the first equation from the second equation:
10 = 6a, a=5/3</p>

<p>Put 5/3 in for a:
5 = 12(5/3) + 2b
5 = 20 + 2b
2b = -15, or b = -15/2.</p>

<p>Now plug a and b into one of the 3 variable equations:
15 = 7a + 3b + c
15 = 7(5/3) + 3(-15/2) + c
15 = 35/3 -45/2 + c (multiply by 6)
90 = 70 -135 + 6c
155 = 6c, c = 155/6 (or 25.83333)</p>

<p>Now plug a, b, and c in the original equation
25 = a + b + c + d
25 = 5/3 -15/2 + 155/6 + d (multiply by 6)
150 = 10 -45 + 155 + 6d
30 = 6d, d = 5</p>

<p>Therefore the solution is a = 5/3, b = -15/2, c = 155/6, d = 5</p>

<p>I worked with matrices for 20 minutes before I realized there must be an easier way.... I did it this way in less than 10 minutes.</p>

<p>wow... if u can do that with a ti89 i should go buy one. the 84 comes nowhere near the 89</p>

<p>errr, you can do this in a ti-83 in under 2 minutes. Just punch the numbers in.</p>

<p>[X][Y]=[Z]</p>

<p>[1 , 1, 1, 1 [a [25
8 , 4, 2, 1 * b = 40
27, 9, 3, 1 c 60
64,16,4, 1] d] 90] </p>

<p>[Y]=[X]^(-1)[Z]</p>

<p>[a [5/3
b = -7.5
c 155/6
d] 5]</p>

<p>uhhh, anyone mind explaining that?</p>

<p>I think it's interesting how the coefficients are all powers of 1, 2, 3, and 4:</p>

<p>25 = (1^3)a + (1^2)b + (1^1)c + (1^0)d
40 = (2^3)a + (2^2)b + (2^1)c + (2^0)d
etc.</p>

<p>It's far too coincidental not to be ignored.
Just putting that out there in case there's a really elegant solution all of us are missing.</p>

<p>fizix: I noticed that as well. </p>

<p>I think my solution is pretty "elegant" though, much easier than substitution or matrices... The only easier method that has been recommended so far is plugging it into your calculator... (I didn't have my graphing calculator with me so I did it by hand.) Mine only looks really long and drawn-out because the OP wanted an explanation as to how we arrived at our answers.</p>

<p>That being said, there probably is an easier solution...</p>