To solve this inequality, we treat is a a regular equation. First, we add 8 to both sides.

\(\displaystyle 2x>18\).

We now divide both sides by 2 to get 

\(\displaystyle x>9\).

To evaluate, we will need to multiply both sides by one-fifth.

\(\displaystyle 5x\cdot \frac{1}{5}\leq-\frac{3}{5}\cdot\frac{1}{5}\)

The sign does not need to be switched unless we multiply or divide by a negative value.

The answer is:  \(\displaystyle x \leq -\frac{3}{25}\)

Upon solving for x, we find that x is larger than -2.  The left-hand term of the interval is -2 since it is the lower bound for our set, and it has a parenthesis around it because it is not included in our set (-2 is not greater than -2).  The right-hand term of the interval is positive infinity because any number larger than -2 is in the set.  There is always a parenthesis around infinity or negative infinity.

Upon expanding and solving for x, we find that x is greater than -1.  The left-hand term of the interval is -1 since it is the lower bound of our set, and has a parenthesis around it because it is not in our set (-1 is not greater than -1).  The right-hand term of the interval is infinity because any number larger than -1 is in our set, and there is always a parenthesis around infinity.

Upon solving for x, we find that 8 is less than or equal to x which is less than or equal to 10, or x is greater than or equal to 8 AND x is less than or equal to 10.  The left-hand term of our interval is 8 since it is the lower bound for our set, and there is a bracket around it since it is in our set (8 is greater than or equal to 8).  The right-hand term of our interval is 10 since it is the upper bound for our set, and there is also a bracket around it since it is in our set (10 is less than or equal to 10).

Upon solving for x, we find that 1 is less than x which is less than or equal to 5, or x is greater than 1 AND x is less than or equal to 5.  The left-hand term of our interval is 1 since it is the lower bound for our set, and there is a parenthesis around it since it is not in our set (1 is not greater than 1).  The right-hand term of our interval is 5 since it is the upper bound for our set, and there is a bracket around it since it is in our set (5 is less than or equal to 5).  Remember when dividing or multiplying an inequality by a negative number to switch the direction of the inequalities.

Upon solving for x, we find that -1 is less than or equal to x which is less than 2, or x is greater than or equal to -1 AND x is less than 2.  The left-hand term of our interval is -1 since it is the lower bound for our set, and there is a bracket around it since it is in our set (-1 is greater than or equal to -1).  The right-hand term of our interval is 2 since it is the upper bound for our set, and there is a parenthesis around it since it is not in our set (2 is not less than 2).  Remember when dividing or multiplying an inequality by a negative number to switch the direction of the inequalities.

Upon solving for x, we find that x is less than or equal to -2 OR x is greater than 1. Since x may be in either of those sets, we must take the union of the sets.  The first interval begins with negative infinity, since x can be anything less than -2.  The first interval ends at -2 and includes it in the set, so there is a bracket around it.  We join this interval with the second interval beginning with 1, and there is a parenthesis around 1 since it is not in the set.  The second interval ends with infinity since any number larger than 1 is in our set. There are always parenthesis around infinities.

Upon solving for x, we find that x is less than or equal to -7/2 OR x is greater than or equal to 1/2. Since x may be in either of those sets, we must take the union of the sets.  The first interval begins with negative infinity, since x can be anything less than -7/2.  The first interval ends at -7/2 and includes it in the set, so there is a bracket around it.  We join this interval with the second interval beginning with 1/2, and there is a bracket around 1/2 since it is in the set.  The second interval ends with infinity since any number larger than 1/2 is in our set. There are always parenthesis around infinities.