-1≤a+b≤1,1≤a-2b≤3,则a+3b的取值范围为_________________.

令a+3b=p(a+b)+q(a-2b)=(p+q)a+(p-2q)b所以p+q=1p-2q=3q=-2/3,p=5/3所以a+3b=(5/3)(a+b)-(2/3)(a-2b)-1≤a+b≤1-5/3≤(5/3)(a+b)≤5/31≤a-2b≤3-3≤-(a-2b)≤-1-2≤(-2/3)(a-2b)≤-2/3相加-11/3≤(5/3)(a+b)-(2/3)(...