How do you solve #\frac { 3} { 5} | 10- 5x | + 7> 25#?

the first step is to simplify the expression as much as we can.
#3/5 |10-5x| +7 >25#

#3/5 |10-5x| >25-7#

# |10-5x| > (18*5)/3#

# |10-5x| > 30#

Now #|10-5x|# contains two different functions the f(x) when f(x) is positive or -f(x) when f(x) is negative.

the turning point can easily be found be equating to zero,

#10-5x=0#

#x=2#

so when x is less than 2 the function is #10-5x# and when x is greater than two the function is #-10+5x#

so to find the values that make all these conditions true we must equate.

#10-5x>30#

#-5x>20#

#x<-4#
so, check that this x value meets the requirement of being less than 2. (which it does)

and

#-10+5x>30#

#5x>40#

#x>8#
so, check that this x value meets the requirement of being greater than 2. (which it does)

so we have our two solutions.
This is true either when x is less then -4 or greater than 8.

it is easier if you visualise it as a graph,
graph{|10-5x| [-30.65, 34.3, 8.55, 41.05]}

Another way to solve this is using an algebraic technique.

#3/5|10-5x|+7>25#

#|10-5x|>30#

#|10-5x|>30# can be rewritten as #sqrt((10-5x)^2)>30#
on expansion,

#10-5x> +-sqrt(30^2)#

#10-5x>30# , #10-5x> -30#

#-5x>20# , #-5x> -40#

#x<-4# , #x>8#

(both sides are negative so no change of sign is needed for #x>8#)