ANSWERS: 6

Shouldn't you be doing your own homework? a = 5, b = 5, c = 9.

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

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....

This problem does not have a unique solution. In fact, it has infinitely many. Indeed, b=14c and a=10b=10(14c)=c4. Give c any numeric value you like (whole, integer, rational, real, complex, etc.) Then the difference 14c will be the corresponding value of b, while the difference c4  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.

I you want to know the exact value of a, b, and c, you have to have 3 equations.

17.
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