How do you solve and write the following in interval notation: #x + 2 > -3# and #x + 2 <5#?

1 Answer
Dec 21, 2016

Answer:

#x in ("-"5, 3)#.

Explanation:

Each one of these two inequalities has a set of #x#-values that make it true. Solving the two inequalities together means finding the #x#-values that are in both sets.

From the first inequality, we have

#x+2> "-"3 => x+2- color(blue)(2)>"-3" - color(blue)(2)#

#color(white)(x+2> "-"3) => x"               ">"-5"#

Alright; so the first inequality is true for every #x# greater than -5:
#x in ("-5", oo)#

Similarly, we can solve the second inequality as follows:

#x+2<5 => x+2-color(blue)2 < 5 - color(blue)2#

#color(white)(x+2<5) => x"              "<3#

Okay, so the second inequality is true for every #x# less than 3:
#x in ("-"oo, 3)#

We then ask ourselves: where do these two sets overlap? In other words, what values of #x# make both inequalities true? We can graph both sets on a number line to help us see this:

#"                   o————————————>       "x>"-5"#
#"<————————————o                          "x<3#
<————— -5 ————— 0 ——— 3 —————>

From the number line above, we see that the two solutions overlap between #x="-5"# and #x=3# (not including the endpoints). In interval notation, this is written by combining the two individual solution intervals:

#x in ("-5", oo) nn ("-"oo, 3)#

which, as we can see from the number line, simplifies to

#x in ("-"5, 3)#.