Question 257803
Here we have a 3x3 system.
Let the cables be A, B, and C.
(i) 3A + 1B + 2C = 95
(ii) 3A + 2B + 1C = 100
(iii) 2A + 1B + 2C = 80
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Lets eliminate the B's. Take (ii) - 2*(i) to get
(iv) -3A -3C = -90
lets eliminate the B's. take (ii) - 2*(iii) to get
(v) -1A -3C = -60
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To eliminate C's, take (v) - (iv) to get
(vi) 2A = 30 so
A = 15 cables
C = 15 cables
B = 20 cables