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**Pupil** While doing problem sets, I came across this question:

Given the linear system:

$\displaystyle x + 2y = 10$

$\displaystyle 3x + (6+t)y = 30$

a) Determine a particular value of t so that the system has infinitely many solutions.

b) Determine a particular value of t so that the system has a unique solution.

According to the book, the value for t so that the system has infinitely many solutions is t = 0, and for a unique solution any real number that isn't 0.

That would imply that:

$\displaystyle x + 2y = 10$

$\displaystyle 3x + 6y= 30$

has infinitely many solutions, but if you multiply the first equation by 3 and then proceed to eliminate, nothing would remain. So, wouldn't this be a linear system with no solution (inconsistent system) rather than infinitely many? And for any nonzero number, wouldn't it likewise have no solution?