Expert Answers
pohnpei397 eNotes educator| Certified Educator

What you need to do here is to treat this just as if it were an equation rather than an inequality.  What that means is that you can solve for X just like if it were an equation.

So, what you need to do first is get rid of the 2 in the middle.  You do that by subtracting it from all three sides.  That gets you


So then you divide everything by 3 to solve for X.

That gives you


So we know that x is some number that is between -.33 and 3.33.

jess1999 | Student

1 < 3x + 2 < 12

Basically we have to get the " x " alone first. So subtract two on all 3 sides

By subtracting, your equation should look like

-1 < 3x < 10 now divide all 3 sides by 3

By dividing all 3 sides by 3 your equation should be

-1/3 < x < 10/3    also     -.33 < x < 3.33 which is your answer

sorry i had a typo the second equation onward should look like this: -1<3X<10

and -1/3<X<10/3


subtract 2 from both sides



Wiggin42 | Student

1 < 3x + 2 < 12

Treat it just like an equation. 

Isolate the variable by undoing the addition first: 

1 - 2 < 3x + 2 - 2 < 12 - 2

-1 < 3x < 10

Undo the multiplication next

-1/3 < x < 10/3

giorgiana1976 | Student


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

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



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


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



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


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

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



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



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)