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Verified Answer :

LMM735

for 3y-6>12
+6>12
--------------------
3y>12
---- ----
3. 3
----------
y>4




for 2y+6 <8
-6. -6
---------------------
2y <2
---- ----
2. 2
-------------
y <1

apologiabiology

ok so solve for y in each

3y-6>12
add 6
3y>18
divide 3
y>6

other
2y+6<8
minus 6
2y<2
divide 2
y<1

we have
y>6 or y<1
'or' means it can be either of them

so y is any number less than 1 or greater than 6
basically, it is all real numbers except for 1≤y≤6, or the numbers from 1 to 6 including 1 and 6

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