<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>