0\medskip\\4(x+3)<5(2x-3)\medskip\\4x+12<10x-15\medskip\\6x>27\medskip\\x>\dfrac{27}{6}\medskip\\x>4{,}5" alt="\dfrac{x+3}{5}<\dfrac{2x-3}{4}\mid\cdot~20>0\medskip\\4(x+3)<5(2x-3)\medskip\\4x+12<10x-15\medskip\\6x>27\medskip\\x>\dfrac{27}{6}\medskip\\x>4{,}5" align="absmiddle" class="latex-formula">
Ответ. 
4{,}5" alt="x>4{,}5" align="absmiddle" class="latex-formula">