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row reduce x + 2y = 1 and 3x + 6y = a
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each term has been multipied by 3 so a = 1 x 3
or if you what to go the other way... in the 2nd equation each term has been divided by 3 to get the 1st equation so 1 = a/3
Thanks campbel_st ... I have (1 2|1) and below that (3 6|a) to give me the augmented matrix and yes I reduced it to (1 2|1) and below that (0 0|a-3) but that is where I was flummoxed since it looks like no solution .... 0 = a-3?
I think I have it ... since the two equations look like multiples of each other two lines that are parallel, then for what value of a is this true and as you pointed out campbell_st it is the value where a = 3 that we have no solution ... sorry for being so slow ... but I think I got there eventually.
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