Question

−5 < 4x + 3 ≤ 14 how to solve this

Answer 1

`\bf -5<4x + 3 \leqslant 14\implies \begin{cases} -5<4x+3\\ 4x+3 \leqslant 14 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ -5<4x+3\implies -8 < 4x\implies \cfrac{-8}{4}<x\implies \boxed{-2<x} \\\\[-0.35em] ~\dotfill\\\\ 4x+3\leqslant 14\implies 4x\leqslant 11\implies \boxed{x\leqslant \cfrac{11}{4}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill -2<x\leqslant \cfrac{11}{4}~\hfill`

you could also do it as a triplet at once

`\bf -5<4x+3\leqslant 14\implies -8<4x\leqslant 11\\\\\\ \cfrac{-8}{4}<x\leqslant \cfrac{11}{4}\implies -2<x\leqslant \cfrac{11}{4}`