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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.

 May 27, 2019

Best Answer 

 #1
avatar+22273 
+2

Find all real numbers\(t\) such that \(\frac{2}{3} t - 1 < t + 7 \le -2t + 15\).

 

\(\begin{array}{|rcll|} \hline \dfrac{2}{3} t - 1 &<& t + 7 \\ \dfrac{2}{3} t &<& t +8 \\ \dfrac{2}{3} t-t &<& 8 \\ -\dfrac{1}{3} t &<& 8 \\ \dfrac{1}{3} t &>& -8 \\ -8 &<&\dfrac{1}{3} t \\ -8\cdot 3 &<& t \\ \mathbf{ -24 } &\mathbf{<}& \mathbf{t} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline t + 7 &\le& -2t + 15 \\ t &\le& -2t + 8 \\ 3t &\le& 8 \\ \mathbf{t } & \mathbf{\le} & \mathbf{\dfrac{ 8 } {3} } \\ \hline \end{array}\)

 

\(\mathbf{ -24 < t \le \dfrac{ 8 } {3} } \)

 

laugh

 May 27, 2019
 #1
avatar+22273 
+2
Best Answer

Find all real numbers\(t\) such that \(\frac{2}{3} t - 1 < t + 7 \le -2t + 15\).

 

\(\begin{array}{|rcll|} \hline \dfrac{2}{3} t - 1 &<& t + 7 \\ \dfrac{2}{3} t &<& t +8 \\ \dfrac{2}{3} t-t &<& 8 \\ -\dfrac{1}{3} t &<& 8 \\ \dfrac{1}{3} t &>& -8 \\ -8 &<&\dfrac{1}{3} t \\ -8\cdot 3 &<& t \\ \mathbf{ -24 } &\mathbf{<}& \mathbf{t} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline t + 7 &\le& -2t + 15 \\ t &\le& -2t + 8 \\ 3t &\le& 8 \\ \mathbf{t } & \mathbf{\le} & \mathbf{\dfrac{ 8 } {3} } \\ \hline \end{array}\)

 

\(\mathbf{ -24 < t \le \dfrac{ 8 } {3} } \)

 

laugh

heureka May 27, 2019
 #2
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Ok! Thank you so much! I get it now!

 May 27, 2019

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