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