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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.

Aug 28, 2019

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$$\dfrac 2 3 t - 1 < t + 7 \leq -2t+15\\ \dfrac 2 3 t -8< t \leq -2t+8\\~\\ \text{left side}\\ \dfrac 2 3 t -8 < t\\ -\dfrac 1 3 t < 8\\ t > -24$$

$$\text{Right side}\\ t \leq -2t+8\\ 3t\leq 8\\ t\leq \dfrac 8 3$$

$$\text{final result}\\ -24 < t \leq \dfrac 8 3 \\ t \in \left(-24, \dfrac 8 3 \right]$$

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Aug 28, 2019
edited by Rom  Aug 29, 2019