There may be a better way than this but I think you could use graph to see it:
(1) x + y ≥ a
(2) x − y ≤ −1
Hence we have
y ≥ a - x
And y ≥ x + 1
Hence y would be in the region about either line (above a-x to the left of intersection and above x+1 to the right)
a) When a<0, there is no minimum to x + ay since you can keep the same x but make y larger and larger
b) When a>0, consider the minimum of x + ay : for the same x the value of y has to be minimised and hence it would be on the line a-x or x+1
b1) if 0<a<1, on the branch y=a-x, x+ay becomes x+a(a-x) = (1-a)x + a^2, again this has no minimum as 1-a>0 and as we keep decreasing x the sum x+ay keeps decreasing
b2) if a >1, then on the y=a-x branch (1-a)x + a^2 would be minimised when x largest as 1-a<0
Also on the y=x+1 branch x+ay becomes x(1+a)+a which is minimised when x is minimised since 1+a>0
Therefore the minimum can only be obtained in scenario b2 and that is exactly at the intersection of the 2 branches (you can then solve for that in terms of a and then equate to 7 you would get the answer noting a>1)