Set Operation: Union - Problem 2 3,352 views
Solving a union of linear inequalities. So we have two statements here, the first thing we need to do is solve each of these inequalities as we would any to other inequalities. So for this one here we need to add one to both sides -3x is greater than 6 divide by -3, remember when we divide by a negative we have to flip our sign we end up with x is less than -2.
Going to this one over here, same thing we are solving for x add 3 to both sides 4x is less than 16 divide by 4, x is less than 4. Whenever I am solving a linear inequality union or intersection, I always make a number line. We have our -2 and we have our 4. X is less than -1 sorry x is less than -2 so this is going to be an open circle. And we are going down. Always feel when I’m doing two things on the same number line I always bring them up so I can just sort of see what’s each one is doing instead of having them on top of each other.
This one we have x is less than 4 again we have a open circle and this is going to be shaded down as well. For this we are dealing with the union this union carries down union is where at least one element is represented. So looking at this number line if we are looking at numbers less than -2 , we have both elements we have both inequalities satisfying that so that’s an union.
Looking in the middle between -2 and 4 just this one any inequalities represented there? That’s perfectly fine we only need one and lastly over here there’s nothing so this part isn’t in the union but everything else is. So our answer then is going to be everything less than 4 not including 4 because we don’t have an equal to so soft bracket soft bracket.