by 
munna
ANSWERS: 11
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Shouldn't you be doing your own homework? a = 5, b = 5, c = 9.
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a = 6 b = 4 c = 10 Well, there are several answers really... *shrugs* a = 1 b = 9 c = 5 a = 7 b = 3 c = 11 etc : P
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Not really sure. What I think is: b can be any number from, 0 to 10 a=10, b=0, c=14 a=9, b=1, c=13 a=8, b=2, c=12 a=7, b=3, c=11 a=6, b=4, c=10 a=5, b=5, c=9 a=4, b=6, c=8 a=3, b=7, c=7 a=2, b=8, c=6 a=1, b=9, c=5 a=0, b=10, c=4 Edit: No, there are multiple ways to calculate that, a=-1, b=11, c=3 a=-2, b=12, c=2 a=-3, b=13, c=1 a=-4, b=14, c=0 and so on....
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This problem does not have a unique solution. In fact, it has infinitely many. Indeed, b=14-c and a=10-b=10-(14-c)=c-4. Give c any numeric value you like (whole, integer, rational, real, complex, etc.) Then the difference 14-c will be the corresponding value of b, while the difference c-4 - the value of a. Just check for yourself that this will work for any value of c. In general, a system of simultaneous linear equations that has fewer equations than variables has infinitely many solutions.
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I you want to know the exact value of a, b, and c, you have to have 3 equations.
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17.
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You need 3 equations. Two is not enough. a = 1 b = 9 c = 6 a = 2 b = 8 c = 5 etc.
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It's easy as 1, 2, 3.
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To enable a person to read and write.
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There isn't a single correct answer. There are several.
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You have 2 equations but 3 variables. Hence, infinite solutions. For unique solutions it should be number of variables equal to number of equations. So there shouldn't be more than 2 variables in order to guarantee unique solutions.
Anonymous 
Denmarks 
mushroom 
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