Posted by alex on Tuesday, January 8, 2013 at 7:51pm.
What do want done ?
I can "solve" #2 :
sub the first into the 2nd ...
3(y-2) - y = 6
3y - 6 - y = 6
2y = 12
y = 6
then x = y-2 = 6-2 = 4
for the 1st and 3rd, I have no clue what the
xt1 and 2xt3 are supposed to be.
Is the t a new variable?
is it t^3 ?
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