![\dfrac{2}{3}t-3\leq1|\cdot3\\ 2t-9\leq3\\ 2t\leq12\\ t\leq6\\\\ \dfrac{3}{4}t-27|\cdot4\\ 3t-828\\ 3t36\\ t12\\\\ t\leq6 \vee t12\\ \boxed{t\in(-\infty,6]\cup(12,\infty)} ](/tpl/images/0272/1871/d3149.png)
Can someone with question 51 i need to break it down
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