How do you solve #-x<2x +3 <-8x+ 3#?

1 Answer
Mar 7, 2017

Answer:

#x in (-1,0)#

Explanation:

Remember that you can always add or subtract the same amount to both sides of an inequality and still maintain the inequality; and
you can multiply or divide both sides of an inequality by any value greater than zero and still maintain the inequality.

If #-x<2x+3<-8x+3#
then
#{:(-x<2x+3," and ", 2x+3 < -8x+3), (rarr0 < 3x+3,,rarr2x <-8x), (rarr -3 < 3x,,rarr10x < 0), (rarr -1 < x,, rarr x < 0) :}#