Multiply each side of the inequality by #color(blue)((5/-4)# to solve for #x# while keeping the inequality balanced. However, because we are multiplying or dividing an inequality by a negative number we must reverse the inequality operator:

#color(blue)(5/-4) xx (-2)/5 color(red)(>) color(blue)(5/-4) xx (-4)/5x#

#(-10)/-20 color(red)(>) color(blue)(color(black)(cancel(color(blue)(5)))/color(black)(cancel(color(blue)(-4)))) xx color(blue)(cancel(color(black)(-4)))/color(blue)(cancel(color(black)(5)))x#

#1/2 > x#

Or, to state the inequality in terms of #x# we can reverse or "flip" the entire inequality:

#x < 1/2#

And, in interval notation:

#(-oo, 1/2)#