How do you solve and write the following in interval notation: #4+abs(3x+2) <=5#?

1 Answer
Jan 6, 2017

Answer:

#x in [-1, -1/3]#

Explanation:

#4+|3x+2| <= 5#

#=> |3x+2| <= 1#

The absolute value function #|a|# is defined as

#|a| = {(a if a>=0), (-a if a<0):}#

We will use this definition to consider all possible cases.


Case 1: #3x+2 >= 0#

Note that for this to occur, we must have #3x >= -2 => x >= -2/3#

Now, by the definition of the absolute value function, we have #|3x+2| = 3x+2# in this case. Substituting that into our initial inequality, we get

#3x+2 <= 1#

#=> 3x <= -1#

#=> (3x)/3<=(-1)/3#

Recall that when multiplying or dividing by a positive number, the direction of the inequality stays the same, and when multiplying or dividing by a negative number, we change the direction of the inequality

#=> x <= -1/3#

So, together with the initial restriction #x=>-2/3#, we have

#-2/3 <= x <= -1/3#

or, in interval notation, #x in [-2/3, -1/3]#


Case 2: #3x+2 < 0#

As above, we first solve for #x# to find that this case occurs when #3x < -2 => x < -2/3#

By the definition of the absolute value function, in this case we have #|3x+2| = -(3x+2)#. Again, we substitute this into the initial inequality.

#-(3x+2) <= 1#

#=> 3x+2 >= -1#

We multiplied by #-1# in the prior step, and so reversed the direction of the inequality.

#=> 3x >= -3#

#=> x >= -1#

Together with the initial restriction of #x < -2/3# in this case, we have

#-1 <= x < -2/3#

or, in interval notation, #x in [-1, -2/3)#


If we look at where the intervals of the solutions of each case lay, we find that they meet at and include #x=-2/3#, and so putting them together, we get

#-1 <= x <= -1/3#

or, in interval notation, #x in [-1, -1/3]#