Question 1208859
<font color=black size=3>
{{{1/3 < (x+1)/2 <= 2/3}}}


{{{2*(1/3) < 2*(x+1)/2 <= 2*(2/3)}}} Multiply all sides by 2


{{{2/3 < x+1 <= 4/3}}}


{{{2/3 - 1 < x+1-1 <= 4/3 - 1}}} Subtract 1 from all sides


{{{2/3 - 3/3 < x <= 4/3 - 3/3}}}


{{{-1/3 < x <= 1/3}}}


The answer in set builder notation would be *[tex \Large \left{\text{x} | \text{x}\in\mathbb{R}, \ -\frac{1}{3} < \text{x} \le \frac{1}{3}\right}]
That translates to "x is any real number between -1/3 and 1/3; excluding -1/3 but including 1/3".


The interval notation would be (-1/3, 1/3]
A curved parenthesis means we exclude the endpoint -1/3.
A square bracket is there to include the endpoint 1/3.


Here's the graph on a number line.
{{{
drawing(400,200,-6,6,-3,3,
line(-10,0,10,0),line(-5,0.04,-5,-0.04),line(-2.5,0.04,-2.5,-0.04),line(0,0.04,0,-0.04),line(2.5,0.04,2.5,-0.04),line(5,0.04,5,-0.04),locate(-5.1,-0.25,"-2"),locate(-2.6,-0.25,"-1"),locate(-0.05,-0.25,"0"),locate(2.45,-0.25,"1"),locate(4.95,-0.25,"2"),
red(
circle(-5/6,0,0.18),circle(-5/6,0,0.20),circle(-5/6,0,0.22),circle(-5/6,0,0.24),
circle(5/6,0,0.06),circle(5/6,0,0.08),circle(5/6,0,0.10),circle(5/6,0,0.12),circle(5/6,0,0.14),circle(5/6,0,0.16),circle(5/6,0,0.18),circle(5/6,0,0.20),circle(5/6,0,0.22),circle(5/6,0,0.24),
line(-5/6+0.2,0+0.08,5/6,0+0.08),
line(-5/6+0.2,0+0.06,5/6,0+0.06),
line(-5/6+0.2,0,5/6,0),
line(-5/6+0.2,0-0.08,5/6,0-0.08),
line(-5/6+0.2,0-0.06,5/6,0-0.06),
locate(-5/6-0.5,0.2+0.5,"-1/3"),
locate(5/6-0.2,0.2+0.5,"1/3")
)
)
}}}
Described in words: plot an open hole at -1/3 and a closed filled in circle at 1/3. Shade in between. 


Another way to express the diagram could be like this
<pre>
              -1/3     1/3
--|--------|-----O==|==●-----|--------|--
 -2       -1        0        1        2 
</pre>
</font>