To solve for this problem, you can do it similar to any regular function, but this time with the inequality (#>#, which is greater than, or #<#, which is less than). First, begin by multiplying the factor out with #2#:
#2(x+5) > 8x-8 => 2x+10>8x-8#
Then solve for #x# by method of subtracting the terms:
#2x+10 > 8x-8 =>#subtract #2x# and #-8# to both sides,
#18 > 6x =>#divide #6# to both sides,
#3 > x#, #=> x < 3#
Thus, for the inequality of the equation to work, the variable #x# must be less than (and not equal to) #3#, unless it is #x<=3# (note the symbol #<=# vs. #<#).
Now there will be cases where you may have to switch the inequity. If you end up with the step
#18 > -6x# or any other negative number multiplied by #x#,
then you have to switch the #># to #<#, since the result is a negative number:
#18 > -6x=>#divide the -6 on both sides and switch the inequity,
#-3 < x# #=> x > -3#.
Hope some of the key points help out when doing inequalities!
