+166 \(3t-5
Adding \(4t\) gives \(7t-5<5t-2\leq15\).
Solving \(7t-5<5t-2\) and \(5t-2\leq15\) gives \(t<1.5\) and \(t\leq3.4\), respectively.
Combining these then shows that \(t<1.5\), so therefore in interval notation the answer is \((-\infty,1.5)\).
+166 Something went wrong with my Latex on line one, it's supposed to read \(3t-5 .
