You have a compound inequality,
[tex]\dfrac{18-x}5 > 3 \lor \dfrac{x+7}2 < 4[/tex]
which is the same as saying
[tex]\dfrac{18-x}5 > 3 \text{ or } \dfrac{x+7}2 < 4[/tex]
(the [tex]\lor[/tex] symbol means "or") which in plain English translates to
"x belongs to some set of numbers such that the quantity (18 - x)/5 is larger than 3, or the quantity (x + 7)/2 is smaller than 4"
Solve the two inequalities separately. On their own, there are no hidden tricks to solving them.
(18 - x)/5 > 3
Multiply both sides by 5 :
18 - x > 15
Isolate x :
18 - 15 > x
x < 3
(x + 7)/2 < 4
Multiply both sides by 2 :
x + 7 < 8
Isolate x :
x < 8 - 7
x < 1
So the solution to the compound inequality is x < 3 or x < 1, meaning "the set of real numbers x that are either smaller than 3 or smaller than 1". But if x < 1, then x is automatically smaller than 3, so the "or" solution can be condensed into a single inequality, x < 3.
The thing about "or" statements in math is that the compound condition, "(condition 1) or (condition 2)", is met if at least one of condition 1 or condition 2 is met.
