solve 4<2y-2<10?

Answer 1

You didn't say whether you wanted integer solutions or real solutions. Either way, there will be more than one solution.

The integer solutions are: y = 4 or y = 5

The real solutions are: 3 < y < 6
(which means that ANY real number greater than 3 and less than 6 is a solution.)

Answer 2

The best way to answer this form of question is to split the one equation into two equation and make the <>sign into = sign and solve for y ie
from
4<2y-2<10 into
4=2y-2 and 2y-2=10 therefore
4+2=2y (bring 2 over to left side subtraction become addition)
6=2y (bring 2 to left side multiplication become division)
6/2=y
3=y and

2y-2=10
2y=10+2
2y=12
y=12/2
y=6

make the = sign back to <> again

36

The answer is 33 and y<6, the equation of 4<2y-2<10 will hold to be true. As a digit, y should be 4 or 5. As a decimal , y should be 3.00001 to 5.99999.

If you substitue the answer of 4 or 5 into the equation,
2y-2
=2x4-2
=6 which is make 4<6<10 True

2y-2
=2x5-2
=8 which make 4<8<10 True

Try substitue other value in decimal, the state will hold true.