For a certain value of k, the system
3a + 4b = 7,
6a + 4b = k - 3a - 8b
has infinitely many solutions (a,b). What is k?
+486 To have infinitely many solutions, we must have that the two equations are multiples of each other. Moving all variables to the left of the second equation gives $9a+12b=k$. Aha! The second equation is $3$ times the first! So we need $k$ to be $3(7)=\boxed{21}$
