First, add #color(red)(9x)# and subtract #color(blue)(10)# from each side of the inequality to isolate the #x# term while keeping the inequality balanced:
#-color(blue)(10) + 15 - 9x + color(red)(9x) < -color(blue)(10) + 10 - 4x + color(red)(9x)#
#5 - 0 < 0 + (-4 + color(red)(9))x#
#5 < 5x#
Now, divide each side of the inequality by #color(red)(5)# to solve for #x# while keeping the inequality balanced:
#5/color(red)(5) < (5x)/color(red)(5)#
#1 < (color(red)(cancel(color(black)(5)))x)/cancel(color(red)(5))#
#1 < x#
If we want to state the solution in terms of #x# first we can reverse or "flip" the entire inequality:
#x > 1#
