The answer: #{-14/5,2}#
Absolute value operation always return non-negative.
So,
#if (5y+2)/2>=0 ->abs((5y+2)/2) = (5y+2)/2#
#if (5y+2)/2<0 ->abs((5y+2)/2) = (-1) * ((5y+2)/2)#
Since we don't know the value of #y#, we need to check all these possibilities.
First
Assume that #(5y+2)/2 >= 0#
We need to find the range of #y#
#2 * ((5y+2)/2) >= 2 * 0#
#5y +2 >= 0#
#A: y>=-2/5#
We need to remember this.
Now, try to solve the given equation.
#(5y+2)/2 = 6#
#5y+2=12#
#5y=10#
#y = 2#
Remember expression #A#, #y=2# satisfies the inequality in expression #A#. So #2# is in the solution set.
Second
Assume that #(5y+2)/2<0#
#B: y < -2/5#
Now try to solve the given equation again, but this time the absolute value operation will return #(-1)# times of the input since the input is assumed negative.
#(-1) * (5y+2)/2 = 6#
#(5y+2)/2 = -6#
#5y+2 = -12#
#5y = -14#
#y = -14/5#
#y=-14/5# satisfies the inequality in the expression #B# so #-14/5# is also in our solution set.
Result: #{-14/5,2}#
