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

 Dec 10, 2017

Best Answer 

 #2
avatar+8966 
+1

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

 Dec 10, 2017
edited by hectictar  Dec 10, 2017
edited by hectictar  Dec 10, 2017
edited by hectictar  Dec 10, 2017
edited by hectictar  Dec 10, 2017
edited by hectictar  Dec 10, 2017
edited by hectictar  Dec 10, 2017
edited by hectictar  Jun 4, 2019
 #1
avatar+11 
0

Interval or integral?

 Dec 10, 2017
 #2
avatar+8966 
+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
edited by hectictar  Dec 10, 2017
edited by hectictar  Dec 10, 2017
edited by hectictar  Dec 10, 2017
edited by hectictar  Dec 10, 2017
edited by hectictar  Dec 10, 2017
edited by hectictar  Dec 10, 2017
edited by hectictar  Jun 4, 2019
 #3
avatar+111438 
+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 ]

 

 

cool cool cool

 Dec 10, 2017

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