How do you solve #0\leq \frac { x + 3} { 2} < 6#?

Question

760 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-09T01:31:58

See the entire solution process below:

First, multiply each segment of the system if inequalities by #color(red)(2)# in order to eliminate the fraction while keeping the system balanced:

#color(red)(2) xx 0 <= color(red)(2) xx (x + 3)/2 < color(red)(2) xx 6#

#0 <= cancel(color(red)(2)) xx (x + 3)/color(red)(cancel(color(black)(2))) < 12#

#0 <= x + 3 < 12#

Now, subtract #color(red)(3)# from each segment of the system of inequalities to solve for #x# while keeping the system balanced:

#0 - color(red)(3) <= x + 3 - color(red)(3) < 12 - color(red)(3)#

#-3 <= x + 0 < 9#

#-3 <= x < 9#