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anonymous
 5 years ago
In solving the system 4x + 6y = 1 and x + 1.5y = 0, Sue arrived at the equation 0 = 1 . Assuming Sue did not make a mistake, that means the system has:
a. (0, 1) as its only solution
b. infinitely many solutions
c. no solution
d. (1, 0) as its only solution
e. has (0, 1) and (1, 0) as its only solutions
anonymous
 5 years ago
In solving the system 4x + 6y = 1 and x + 1.5y = 0, Sue arrived at the equation 0 = 1 . Assuming Sue did not make a mistake, that means the system has: a. (0, 1) as its only solution b. infinitely many solutions c. no solution d. (1, 0) as its only solution e. has (0, 1) and (1, 0) as its only solutions

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when solving for the unknown variable you will get three possible answers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if solving for x and you get x= something, the you have an exact solution ( one solution)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if solving equations and your variables drop out and you get a true statement I.E 0=0, then you get infinite solutions.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you are solving the equation and the variables drop out and you get a false statement, I.E 1=2, then you have no solutions.
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