To solve this problem, we need to determine the intersection and union of the sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex] as defined:
[tex]\[
E = \{ y \mid y \leq 3 \}
\][/tex]
[tex]\[
F = \{ y \mid y < 9 \}
\][/tex]
### Intersection [tex]\( E \cap F \)[/tex]
The intersection of two sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex] consists of all elements that are in both [tex]\( E \)[/tex] and [tex]\( F \)[/tex]. In this context, it means we need to find all [tex]\( y \)[/tex] values that satisfy both conditions: [tex]\( y \leq 3 \)[/tex] and [tex]\( y < 9 \)[/tex].
- For [tex]\( E \)[/tex]: [tex]\( y \leq 3 \)[/tex]
- For [tex]\( F \)[/tex]: [tex]\( y < 9 \)[/tex]
Since [tex]\( y \leq 3 \)[/tex] implies [tex]\( y < 9 \)[/tex] (because 3 is less than 9), the intersection [tex]\( E \cap F \)[/tex] must be all [tex]\( y \)[/tex] such that [tex]\( y \leq 3 \)[/tex].
Thus, in interval notation:
[tex]\[ E \cap F = (-\infty, 3] \][/tex]
### Union [tex]\( E \cup F \)[/tex]
The union of two sets [tex]\( E \)[/tex] and [tex]\( F \)[/tex] consists of all elements that are in either [tex]\( E \)[/tex] or [tex]\( F \)[/tex], or both. We need to find all [tex]\( y \)[/tex] values that satisfy either of the conditions [tex]\( y \leq 3 \)[/tex] or [tex]\( y < 9 \)[/tex].
- For [tex]\( E \)[/tex]: [tex]\( y \leq 3 \)[/tex]
- For [tex]\( F \)[/tex]: [tex]\( y < 9 \)[/tex]
The union will thus include all [tex]\( y \)[/tex] values that are in [tex]\( E \)[/tex] (up to and including 3), and all [tex]\( y \)[/tex] values that are in [tex]\( F \)[/tex]. Practically, this means we take the larger of the two intervals:
- [tex]\( E \cup F \)[/tex] will be [tex]\( y < 9 \)[/tex], as this includes all required elements and extends beyond [tex]\( y \leq 3 \)[/tex].
Thus, in interval notation:
[tex]\[ E \cup F = (-\infty, 9) \][/tex]
Therefore, the solutions are:
[tex]\[
E \cap F = (-\infty, 3]
\][/tex]
[tex]\[
E \cup F = (-\infty, 9)
\][/tex]
