решите систему неравенств: 3х + 12<0 2x - 1<0

Question

решите систему неравенств:

3х + 12<0</p>

2x - 1<0</p>

Answer 1

\{ {{3x<-12} \atop {2x<1}}<=> \{ {{x<-4} \atop {x<0,5}}\\ Answer: (-\infty;-4)" alt="\{ {{3x+12<0} \atop {2x-1<0}}<=> \{ {{3x<-12} \atop {2x<1}}<=> \{ {{x<-4} \atop {x<0,5}}\\ Answer: (-\infty;-4)" align="absmiddle" class="latex-formula">

Answer 2

3x+12<0</p>

3x<-12</p>

x<-4</p>

2x-1<0</p>

2x<1</p>

x<0,5</p>

(-бесконечности;-4)