This compound inequality

3y < 5y-2 < 7+y 

is EQUIVALENT to (or, if you want, is, BY THE DEFINITION)

the system of TWO INEQUALITIES

3y < 5y-2      (1)   and
5y-2 < 7+y     (2)

in the sense that the final solution set is THE INTERSECTION of the solution sets for two inequalities (1) and (2).


So, let us solve the inequality (1) first.


1)  3y < 5y - 2 <=== is equivalent to (subtract 3y from both sides) ===> 

    0 < 2y - 2  <=== is equivalent to (add 2 to both sides) ===> 

    2 < 2y      <=== is equivalent to (divide by 2 from both sides) ===>

    y > 1.

    So, the first inequality is solved, and its solution set is y > 1,  or,

    in the interval notation, the set (1,infinity).     (*)



Next, let us solve the second inequality 

2)    5 - 2y < 7 + y.  <=== It is equivalent to (add 2y to both sides) ===>

    5 < 7 + 3y      <===     is equivalent to (subtract 7 from both sides) ===>

    -2 < 3y         <===     is equivalent to (divide by 3 from both sides) ===>

     < y,          or, which is the same

    y > .

    Thus, the second inequality is solved, and its solution set is y > ,  or,

    in the interval notation, the set (,infinity).    (**)


3.  The intersection of both sets (*) and (**) is the set  y > 1,  or, in the interval notation, (1,).



Answer.  The solution of the given compound inequality is  y > 1,  or, in the interval notation, (1,).