Graph the line of ##(3x+4y=12)## using two arbitrary solutions; then shade in the side of the line containing the point ##(0,0)##

##3x+4y<=12## includes all points on the line ##3x+4y=12##

The x and y intercepts provide a convenient pair of points on this line (although any other pairs would work). ##color(white)(“XXX”)x=0 rarr y=4## ##color(white)(“XXX”)y=0 rarr x=3## so the pairs ##(0,4)## and ##(3,0)## are on this line. Plot these point and draw a line through them.

Since the inequality is true for ##(x,y)=(0,0)## the point ##(0,0)## must be on the selected side of the line just drawn. graph{3x+4y <= 12 [-10, 10, -5, 5]}