1

1<3x+2

To solve the inequation above, we have to add the value

 (-1) in order to cancel the free term, 1, from the left side.

1-1<3x+2-1

0<3x+1

We'll move the unknown term in the left side of the inequality:

-3x<1

We'll multiply the inequality with the value (-1), therefore the inequality will become opposite :

3x>-1

x>-(1/3)

That means that  the solution of the first inequation will be the interval (-1/3, infinity).

3x+2<12

To solve the inequation above, we have to add the value

 (-12) in order to cancel the free term, 12, from the right side.

3x+2-12<12-12

3x-10<0

We'll move the value (-10), with the opposite sign, to the right side:

3x<10

x<10/3

The solution of the second inequality is the interval of values (- infinity, 10/3).

It's important to not miss the aspect of simultaneity of both inequations, so that the common solution of the double inequation is found by intersecting ranges of values:

(-1/3, infinity)intersected(- infinity,10/3)=(-1/3, 10/3)