[tex]\frac{3}{4}(5x-y) < 3\ \ \ /\cdot\frac{4}{3}\\\\\frac{4}{3}\cdot\frac{3}{4}(5x-y) < \frac{4}{3}\cdot3\\\\5x-y < 4\ \ \ /-5x\\\\-y < 4-5x\ \ \ /\cdot(-1) < 0\ then\ "<"\ change\ to\ ">"\\\\y > 5x-4[/tex]
[tex]------------------------\\y=5x-4\\\\for\ x=0\to y=5\cdot0-4=0-4=-4\to A(0;-4)\\\\for\ x=2\to y=5\cdot2-4=10-4=6\to B(2;\ 6)\\------------------------\\\\look\ at\ the\ picture[/tex]
