Dealing with a system of equations with more unknowns than equations by making some of the unknowns parameters and using Gaussian elimination to solve for the remaining ones.
2a + 3b = 4c
5a + 6b + 42 = 7d
Arbitrarily, make b and d parameters and solve for a and c:
[ 2 3 −4 0 | 0 ]
[ 5 6 0 −7 | −42 ]
[ 0 1 0 0 | b ]
[ 0 0 0 1 | d ]
[ 1 0 0 0 | (−6b+7d−42)/5 ]
[ 0 1 0 0 | b ]
[ 0 0 1 0 | (3b+14d−84)/20 ]
[ 0 0 0 1 | d ]
Thus,
a(b,d) = (−6b+7d−42)/5
c(b,d) = (3b+14d−84)/20