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# Solve \frac{1}{3} t - 5 < t - 2 \le -3t + 7. Give your answer as an interval.

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Solve $$\frac{1}{3} t - 5 < t - 2 \le -3t + 7.$$ Give your answer as an interval.

Apr 26, 2021

#1
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the solution is: -9/2 < t ≤ 9/4

so the interval is: $$(-\frac{9}{2},\:\frac{9}{4}]$$

Apr 26, 2021
edited by mworkhard222  Apr 26, 2021
edited by mworkhard222  Apr 26, 2021
edited by mworkhard222  Apr 26, 2021
edited by mworkhard222  Apr 26, 2021
#2
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$$\frac{1}{3} t - 5 < t - 2 \le -3t + 7\\ \frac{1}{3} t - 5 < t - 2\qquad and \qquad t - 2 \le -3t + 7\\ \frac{-2}{3} t < 3\qquad \qquad and \qquad \qquad 4 t \le 9\\ -2 t < 9\qquad \qquad and \qquad \qquad t \le 2.25\\ t > -4.5\qquad \qquad and \qquad \qquad t \le 2.25\\ -4.5$$

The Latex is not working so here is a pic of the preview:

Apr 30, 2021
edited by Melody  Apr 30, 2021