Question 1118568
let x = number of movie passes.
let y = number of gift certificates.


you need at least 5 movie passes, so x >= 5.


you need at least 2 gift certificates, so y >= 2.


a movie pass costs 6 dollars and a gift certificate costs 10 dollars and the most ou can spend is 70 dollars, so 6x + 10y <= 70.


your system of requirements are:


x >= 5
y >= 2
6x + 10y <= 70


when x = 6 and y = 3, 6x + 10y <= 70 becomes 6*6 + 10*3 <= 70 which becomes 36 + 30 <= 70 which becomes 66 <= 70 which is true.


therefore, you can buy 6 movie passes and 3 gift certificates and still stay under the limit of 70 dollars.


the following graph shows that.


<img src = "http://theo.x10hosting.com/2018/061302.jpg" alt="$$$" >


in this graph, the opposite of the constraint inequalities are graphed.


the area of the graph that is not shaded is the region of feasibility.


6 movie passes and 3 gift certificates is the intersection of the line x = 6 and y = 3.


that intersection is in the portion of the graph that is not shaded.


this means that 6 movie passes and 3 gift certificates are within the constraints of the problem.


this was shown algebraically above the graph.