two CD's and three tapes cost $31. three CD's and two tapes cost $29. find the cost of each CD and each tape.
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Let C be the cost of CDs and T be the cost of tapes
2C+3T=31
3C+2T=29
From these we can calculate
6C+9T=93
6C+4T=58
Subtract one from the other
5T=35
T=7
Sub into the original equations
2C+3*7=31
2C=10
C=5
2C+3T=31
3C+2T=29
From these we can calculate
6C+9T=93
6C+4T=58
Subtract one from the other
5T=35
T=7
Sub into the original equations
2C+3*7=31
2C=10
C=5
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CD = c
tape = p
2c + 3p = 31
2c = 31 - 3p
c = (31-3p) / 2
3c + 2p = 29
3 ((31 - 3p) / 2 ) + 2p = 29
46.5 - 4.5p + 2p = 29
46.5 - 2.5p = 29
-2.5p = -17.5
p = 7
2c + 3p = 31
2c + 3(7) = 31
2c + 21 = 31
2c = 10
c = 5
tape = p
2c + 3p = 31
2c = 31 - 3p
c = (31-3p) / 2
3c + 2p = 29
3 ((31 - 3p) / 2 ) + 2p = 29
46.5 - 4.5p + 2p = 29
46.5 - 2.5p = 29
-2.5p = -17.5
p = 7
2c + 3p = 31
2c + 3(7) = 31
2c + 21 = 31
2c = 10
c = 5
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well, distribute 2 in 1st equat, -3 in 2nd
4c+6t=62 2(-25)+3t=31
-3c-6t=-87 -50+3t=31
c=-25 3t=81
t=27
4c+6t=62 2(-25)+3t=31
-3c-6t=-87 -50+3t=31
c=-25 3t=81
t=27