instead of i, ii and iii i will use a,b, and c. so our 3 equations become:

7a + 8b + 7c = 2

6a + 7b + 5c = 8

6a + 9b + 6c = 2.

subtract equation 2 from equation 3 to obtain:

2b + c = -6 (*)

now, we need another equation with just b and c in it, to see if we can eliminate another variable. since we want the "a" terms to cancel, multiply equation 1 by 6, and equation 2 by 7:

42a + 48b + 42c = 12 (equation 1a)

42a + 49b + 35c = 56 (equation 2a)

then subtract equation 1a from equation 2a to get:

b - 7c = 44 (**), and now we multiply (**) by 2 to get:

2b - 14c = 88. subtract (*) from this, and we have:

-15c = 94, so c = -94/15.

using (*) we have:

2b - 94/15 = -6

b - 47/15 = -3

b = -3 + 47/15 = -45/15 + 47/15 = 2/15

and finally, using equation 1, we have:

7a + 8(2/15) + 7(-94/15) = 2, so:

a + 16/105 - 94/15 = 2/7 thus:

a = 2/7 - 16/105 + 94/15 = 30/105 - 16/105 + 658/105 = 672/105 = 96/15 = 32/5.

now let's see if our solution is correct (the first two times i did this, i made errors, so don't feel bad if you did, too):

7(32/5) + 8(2/15) + 7(-94/15) = 224/5 + 16/15 - 658/15 = 4704/105 + 112/105 - 4606/105 = 210/105 = 2 (so equation 1 checks out).

6(32/5) + 7(2/15) + 5(-94/15) = 192/5 + 14/15 - 94/3 = 576/15 + 14/15 - 470/15 = 120/15 = 8 (so equation 2 checks out).

6(32/5) + 9(2/15) + 6(-94/15) = 192/5 + 6/5 - 188/5 = 10/5 = 2 (all three equations check out).

so a = 32/5, b = 2/15, c = -94/15 is indeed the correct solution (with some rather ugly arithmetic attached).