解不等式组0.3x+0.1(150-x)≤10.2.0.1x+0.1(150-x)≤10.2
问题描述:
解不等式组0.3x+0.1(150-x)≤10.2.0.1x+0.1(150-x)≤10.2
答
不等式两边同乘以10,0.3x+0.1(150-x)≤10.2变成3x+150-≤102,x≤-24
同理0.1x+0.1(150-x)≤10.2解得x为全体实数,
所以x≤-24.