First, subtract #color(red)(7x)# and add #color(blue)(9)# to each side of the inequality to isolate the #x# term while keeping the inequality balanced:

#7x - color(red)(7x) + 8 + color(blue)(9) < 14x - color(red)(7x) - 9 + color(blue)(9)#

#0 + 17 < (14 - color(red)(7))x - 0#

#17 < 7x#

Now, divide each side of the equation by #color(red)(7)# to solve for #x# while keeping the equation balanced:

#17/color(red)(7) < (7x)/color(red)(7)#

#17/7 < (color(red)(cancel(color(black)(7)))x)/cancel(color(red)(7))#

#17/7 < x#

We can reverse or "flip" the entire inequality to state the result in terms of #x#:

#x > 17/7#

Because #x# can be any value from #17/7# but not including #17/7# to infinity we can write this in interval notation as:

#x = (17/7, +oo)#