Find all real numbers $t$ such that $\frac{2}{3} t - 1 < t + 7 \le -2t + 15$. Give your answer as an interval.

Guest Dec 10, 2017

#2**+1 **

\(\frac{2}{3}t-1\,\text{<}\,t+7\le-2t+15 \\~\\ \text{Let's split it into two inequalities.} \\~\\ \frac{2}{3}t-1\,\text{<}\,t+7\\~\\ -1-7\,\text{<}\,t-\frac{2}{3}t\\~\\ -8\,\text{<}\,\frac{1}{3}t\\~\\ -24\,\text{<}\,t\\~\\ \text{and}\\~\\ t+7\le-2t+15\\~\\ t+2t\le15-7\\~\\ 3t\le8\\~\\ t\le\frac{8}{3}\\~\\ \text{So t is greater than -24 but less than or equal to 8/3.}\\~\\ \text{t is in the interval:}\quad (-24,\frac{8}{3}]\)

hectictar Dec 10, 2017

#2**+1 **

Best Answer

\(\frac{2}{3}t-1\,\text{<}\,t+7\le-2t+15 \\~\\ \text{Let's split it into two inequalities.} \\~\\ \frac{2}{3}t-1\,\text{<}\,t+7\\~\\ -1-7\,\text{<}\,t-\frac{2}{3}t\\~\\ -8\,\text{<}\,\frac{1}{3}t\\~\\ -24\,\text{<}\,t\\~\\ \text{and}\\~\\ t+7\le-2t+15\\~\\ t+2t\le15-7\\~\\ 3t\le8\\~\\ t\le\frac{8}{3}\\~\\ \text{So t is greater than -24 but less than or equal to 8/3.}\\~\\ \text{t is in the interval:}\quad (-24,\frac{8}{3}]\)

hectictar Dec 10, 2017

#3**+2 **

(2/3)t - 1 < t+ 7 ≤ -2t + 15

First, we have

(2/3) t - 1 < t + 7 multiply through by 3

2t - 3 < 3t + 21 subtract 21, 2t from both sides

-24 < t

Then, we have

t + 7 ≤ -2t+ 15 add 2t to both sides, subtract 7 from both sides

3t ≤ 8 divide both sides by 3

t ≤ 8/3

So......the answer is

(- 24, 8/3 ]

CPhill Dec 10, 2017