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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 >
 < y,          or, which is the same
    y >  .
    Thus, the second inequality is solved, and its solution set is y >
.
    Thus, the second inequality is solved, and its solution set is y >  ,  or,
    in the interval notation, the set (
,  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,
,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,
).
Answer.  The solution of the given compound inequality is  y > 1,  or, in the interval notation, (1, ).
).
Solved.