We have to solve 4*|8x - 8| < 32
4*|8x - 8| < 32
=> |8x - 8| < 8
=> |x - 1| < 1
-1 < (x - 1) < 1
-1 < (x - 1)
=> 0 < x
(x - 1) < 1
=> x < 2
The values of x lie in the interval (0, 2)
algebra1
We have to solve 4*|8x - 8| < 32
4*|8x - 8| < 32
=> |8x - 8| < 8
=> |x - 1| < 1
-1 < (x - 1) < 1
-1 < (x - 1)
=> 0 < x
(x - 1) < 1
=> x < 2
The values of x lie in the interval (0, 2)