Solve the compound inequality
3y-6> 12 or 2y+6< 8

For 3y-6>12
+6>12

3y>12

3. 3

y>4

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

2y <2

2. 2

y <1

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

14 is the best answer can i get brainliest?