2-x<3x+1<10x; 2-x<3x+1, 2-1<3x+x, 1<4x, ¼<x, that is, x>¼; 3x+1<10x, 1<7x, x>1/7.
Therefore, since ¼=7/28 and 1/7=4/28, ¼>1/7, because 7>4.
x is greater than ¼ so x is also greater than 1/7, making the solution x>¼.
Let's check using the original inequality. Let x=¼+k where k is a positive number, then x-2=k-1¾; 3x+1=¾+3k; 10x=2½+10k.
Now let's make k=¾ and see if the inequality is true:
x=¼+¾=1, 2-x=1; 3x+1=4; 10x=10 and 1<4<10 which is true.
If k=¼, x=½; 2-x=1½, 3x+1=2½; 10x=5 and 1½<2½<5 which is also true.
But if k=0, x=¼; 2-x=1¾, 3x+1=1¾, 10x=2½ and 1¾<1¾<2½, which is false, as we would expect because x has to be greater than ¼.