For a certain value of $k,$ the system \begin{align*} 3a + 4b &= 7,\\ 6a + 4b &= k- 4b \end{align*}has infinitely many solutions $(a,b).$ What is $k$?
3a + 4b = 7 multiply through by 2
6a + 8b = 7
Tthe second rearranges as
6a + 8b = k
The system will have infinite solutions when the two equations are exactly the same
So...k = 7
