Question 977950
5x + 3y = 7
y - 4 > 0

Graph as 3y=-5x+7;;  y= -(5/3)x + (7/3)
That is a line.

The other line is y>4.  That is a horizontal line where y=4.
The solution to this problem, assuming that both equations have to be satisfied, lies ON the line ABOVE y=4, not including the point where y=4.  

{{{graph(300,300,-10,10,-10,10,4,(-5x/3)+(7/3))}}}