Question 1186113
Let the line parallel to 3x + 4y = 20 be given by  3x + 4y = C, for some constant C.

Then the distance between the two lines is given by {{{abs(C-20)/sqrt(3^2 + 4^2) = 5}}}, or equivalently, {{{abs(C-20) = 25}}}.


Solving this absolute-value equation leads to C - 20 = 25 or C - 20 = -25.

===> C  = 45 or C = -5.


Thus there are two lines which satisfy the given conditions, namely 3x + 4y = 45   and 3x + 4y = -5.