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?
We can write
3a + 4b = 7
9a + 12b = k
We have infinite solutions when one equation is a multiple of the other.....so.....
Note that multiplying the first equation through by 3 we get.....
3 (3a + 4b) = 3 * 7
9a + 12b = 21 = k