How do you solve and graph #-5 < (t + 15)/3 ≤ 9#?

1 Answer
Jul 13, 2015

Answer:

#-30 < t <= 12# or
#t in (-30,12]#

Explanation:

There are two inequalities here combined into one line. Separately, they look like this:
(a) #-5 < (t+15)/3# and
(b) #(t+15)/3 <= 9#.

Generally speaking, to solve an inequality, like #f(x) < a#, you have to transform it to a form, where an unknown is directly bounded by some value, like #x < b#.
It's important that the latter inequality is completely equivalent to the former, that is from #f(x) < a# follows #x < b# and from #x < b# follows #f(x) < a#.

In this case we can apply a few invariant transformations to the original inequality to obtain this final result as follows.

Step 1. Both parts of an inequality can be multiplied by the same positive number. So, let's multiply each inequality by #3# getting
(a) #-15 < t+15# and
(b) #t+15 <= 27#.

Step 2. The same number can be added to both parts of an inequality. Let's add #-15# to both parts:
(a) #-30 < t#
(b) #t <= 12#

Basically, we have completed our task. We know now that the condition #-30 < t <= 12# is equivalent to an original inequality.
The latter form constitutes the solution.

Graphically, it's represented by an interval of real values of variable #t# bounded on the left by a number #-30# (not included into the interval since the sign of inequality is #<#) and bounded on the right by a number #12# (included into the interval since the sign of inequality is #<=#).

In short, it can be written as
#t in (-30,12]#
(notice parenthesis used for the left boundary and square bracket for the right, signifying not including and including the border into the interval).

We can recommend to study this material at Unizor by following the menu items Algebra - Introduction to Inequalities.