Question 1014713
here's the graph that i made.
look below the graph for further comments.


<img src = "http://theo.x10hosting.com/2016/012002.jpg" alt="$$$" </>


your equations are:


x+4y<8
y>x-3


in the first equation, solve for y to get y<(8-x)/4.


your two equations become:


y < (8-x)/4 and y > x-3.


in the graph, the top dashed line is the equation of y = (8-x) and the bottom line is the equation of y = x-3.


the white area between those 2 dashed lines is where your solution lies.


it is the area beneath the line of y = (8-x)/4 and above the line of y = (x-3).


it has to satisfy both those criteria to be within the solution set.


your solution area lies in the region defined by x < 4.


when x is equal to 4, y = x-3 becomes y = 1 and y = (8-x)/4 becomes y = (8-4)/4 which becomes y = 4/4 which becomes y = 1.


the coordinate point of (4,1) is on both lines.


since the inequality says that the value of y has to be below the line of y = (8-x)/4 and above the line of y = x-3, then that point does not satisfy the requirements because it is on the line.


all of your solutions lie in the region where x < 4.


there are no values of y that satisfy the requirements of the problem when x > 4.


for example:


when x = 10, y < (8-x)/4 becomes y < -2/4 and y > x-3 becomes y > 7


y cannot be < -2/4 and > 7 at the same time, so the requirements of the problem are not satisfied when x = 10.


when x = 5, y < (8-x)/4 becomes y < 3/4 and y > x-3 becomes y > 2.


y cannot be < 3/4 and > 2 at the same time, so the requirements of the problem are not satisfied when x = 5.


when x = 4, y < (8-x)/4 becomes y < 1 and y > x-3 becomes y > 1.


y cannot be < 1 and > 1 at the same time, so the requirements of the problem are not satisfied with x = 4.


when x = 3, y < (8-x)/4 becomes y < 5/4 and y > x-3 becomes y > 0.


y can be < 5/4 and > 0 at the same time, so the requirements of the problem are satisfied with x = 3.


you will only get valid solutions when x < 4.