Write each specification as an absolute value inequality.
50 < k < 51

(a) define a variable for each unknown and write a system of equations to represent the situation.
call x the number of half-page advertisements and y the number of full-page adverstisements
=>
40x + 75 y is the money charged, and given that he has a budget of $ 800
40x + 75y ≤ 800
the number of advertisements is 13, so, the other equation is:
x + y = 13
you coul also state the x and y cannot be negative, i.e. x ≥ 0 and y ≥ 0
so the complete system is:
40x + 75y ≤ 800
x + y = 13
x ≥ 0, y ≥ 0
(b) how many half-page advertisements does michael purchase? show your work.
in a graph draw the lines
40x + 75y = 800 > shade the are between this line, in the first quadrant.
x + y = 13
the solution is the on the part of the line x + y = 13 that is inside in the shaded region.
an addtional fact that you need to know is that the best solution (optimus) is one vertex of the solution area (this is one of the edges of the segment that we determined as possible solutions). 
on of the edges is the point (0,13) and the other is the intercection point between the lines
40x + 75y = 800 and
x + y = 13.
find the solution of that system:
40 (13-y) + 75y = 800
520 - 40y + 75y = 800
35y = 800 - 520
35y = 280
y = 280/35
y = 8 = x = 13 - 8 = 5.
so the two possible solutions are
1) x = 5, y = 8
2) x = 13, y = 0 but with this solution he will not spend all the budget, because 40(13) +75(0) = 520.
so the solution is x = 5 and y = 8
that means that the number of half-page advertisements is 5
(c) how many full-page advertisements does michael purchase? show your work.
the work is above. there you got y = 8 which means that he purchases 8 full  advertisements.
look that 40(5) + 75 (8) =  200 + 600 = 800.
which means that he spends all the budget when purchases 5 half-page advertisements and 8 full-page advertisements.
we also assured that this is the optimal solution.