Так как a = 10, то a>1. Решаем системой, знаки не меняются:
3x-1} \atop {8x-16>0}} \right.\\ \left \{ {{8x-3x>16-1} \atop {8x>16}} \right. \\ \left \{ {{5x>15} \atop {x>2}} \right.\\ \left \{ {{x>3} \atop {x>2}} \right. " alt=" \left \{ {{8x-16>3x-1} \atop {8x-16>0}} \right.\\ \left \{ {{8x-3x>16-1} \atop {8x>16}} \right. \\ \left \{ {{5x>15} \atop {x>2}} \right.\\ \left \{ {{x>3} \atop {x>2}} \right. " align="absmiddle" class="latex-formula">
Ответ: x∈(3;∞).