Question

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

Answer 1

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)`

Answer 2

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`