.
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,
).
Solved.