Find all real numbers $t$ such that $\frac{2}{3} t - 1 < t + 7 \le -2t + 15$. Give your answer as an interval.
+9479 \(\frac{2}{3}t-1\,\text{<}\,t+7\le-2t+15 \\~\\ \text{Let's split it into two inequalities.} \\~\\ \frac{2}{3}t-1\,\text{<}\,t+7\\~\\ -1-7\,\text{<}\,t-\frac{2}{3}t\\~\\ -8\,\text{<}\,\frac{1}{3}t\\~\\ -24\,\text{<}\,t\\~\\ \text{and}\\~\\ t+7\le-2t+15\\~\\ t+2t\le15-7\\~\\ 3t\le8\\~\\ t\le\frac{8}{3}\\~\\ \text{So t is greater than -24 but less than or equal to 8/3.}\\~\\ \text{t is in the interval:}\quad (-24,\frac{8}{3}]\)
+9479 \(\frac{2}{3}t-1\,\text{<}\,t+7\le-2t+15 \\~\\ \text{Let's split it into two inequalities.} \\~\\ \frac{2}{3}t-1\,\text{<}\,t+7\\~\\ -1-7\,\text{<}\,t-\frac{2}{3}t\\~\\ -8\,\text{<}\,\frac{1}{3}t\\~\\ -24\,\text{<}\,t\\~\\ \text{and}\\~\\ t+7\le-2t+15\\~\\ t+2t\le15-7\\~\\ 3t\le8\\~\\ t\le\frac{8}{3}\\~\\ \text{So t is greater than -24 but less than or equal to 8/3.}\\~\\ \text{t is in the interval:}\quad (-24,\frac{8}{3}]\)
+129849 (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 ]
