Inactive Tutor answered • 10/22/22
(+8)(x+1) >0
8x+8 >0
x+1 >0
x >-1
That's assuming that ( +8)(x+1)>0 didn't leave out another variable in the space between the 1st parenthesis and +8. It's an odd way to write 8, as + is implied if it's alone. If there was an x before the +8 then
(x+1)(x+1)>0
x+1>0
x>-1 is one solution, taking the square root of both sides
also
-(x+1)>0 by taking the negative square root of both sides
x+1<0 as the inequality reverses direction when multiplied by a negative number
x <-1 is a 2nd solution
putting them together the full solution is
x = any real number except -1
in interval notation (-infinity,-1)U(-1, infinity)
or maybe you left out a 2nd y before ( +8)??
inequality( +8)(x+1)>0 is odd in running together the "y" of inequality right next to the 1st parenthesis
If you meant the
inequality y( +8)(x+1)>0
or maybe
(y+8)(x+1)>0
it has different solutions
