Answer :

[tex]\dfrac{2}{3}t-3\leq1|\cdot3\\ 2t-9\leq3\\ 2t\leq12\\ t\leq6\\\\ \dfrac{3}{4}t-2>7|\cdot4\\ 3t-8>28\\ 3t>36\\ t>12\\\\ t\leq6 \vee t>12\\ \boxed{t\in(-\infty,6]\cup(12,\infty)} [/tex]

Other Questions