How do you solve #5> \frac { 2j - 5} { 3} > - 5#?

2 Answers
Nov 12, 2017

See a solution process below:

Explanation:

First, multiply each segment of the system of inequalities by #color(red)(3)# to eliminate the fraction while keeping the system balanced:

#color(red)(3) xx 5 > color(red)(3) xx (2j - 5)/3 > color(red)(3) xx -5#

#15 > cancel(color(red)(3)) xx (2j - 5)/color(red)(cancel(color(black)(3))) > -15#

#15 > 2j - 5 > -15#

Next, add #color(red)(5)# to each segment to isolate the #j# term while keeping the system balanced:

#15 + color(red)(5) > 2j - 5 + color(red)(5) > -15 + color(red)(5)#

#20 > 2j - 0 > -10#

#20 > 2j > -10#

Now, divide each segment by #color(red)(2)# to solve for #j# while keeping the system balanced:

#20/color(red)(2) > (2j)/color(red)(2) > -10/color(red)(2)#

#10 > (color(red)(cancel(color(black)(2)))j)/cancel(color(red)(2)) > -5#

#10 > j > -5#

Or

#j > -5# and #j < 10#

Or. in interval notation:

#(-5, 10)#

Nov 12, 2017

We first simplify this.

Explanation:

Multiplying every term by 3 (to get rid of the fraction) won't change anything:
#15>2j-5> -15#

Then we may add #5# to all terms:
#15+5>2j-cancel5+cancel5> -15+5->20>2j> -10#

Now we divide by #2#:
#20/2>(cancel2y)/cancel2> (-10)/2->10>y> -5#

Or, as we normally write it: #-5 < y<10#