Answer:
[tex]\sf 1. \quad \left(Y \cap X'\right) \cup Z = \{b,c,d,e,f\} \\ [/tex]
[tex]\sf 2. \quad Y\;\!'-X = \{d, f\} \\ [/tex]
Step-by-step explanation:
[tex] \\ \large \underline{ \sf{ \pmb{Set \: notation : }}} \\ [/tex]
[tex]\begin{gathered}\begin{array}{|c|c|l|} \cline{1-3} \sf Symbol & \sf N\:\!ame & \sf Meaning \\\cline{1-3} \{ \: \} & \sf Set & \sf A\:collection\:of\:elements\\\cline{1-3} \cup & \sf Union & \sf A \cup B=elements\:in\:A\:or\:B\:(or\:both)}\\\cline{1-3} \cap & \sf Intersection & \sf A \cap B=elements\: in \:both\: A \:and \:B} \\\cline{1-3} \sf ' \:or\: ^c & \sf Complement & \sf A'=elements\: not\: in\: A \\\cline{1-3} \sf - & \sf Difference & \sf A-B=elements \:in \:A \:but\: not\: in \:B}\\\cline{1-3} \end{array}\end{gathered}[/tex]
[tex] \sf{ \pmb{Given \: sets:}} \\ [/tex]
- [tex] \sf{X = {a, c, e, g}} \\ [/tex]
- [tex] \sf{Y = {a, b, c}} \\ [/tex]
- [tex] \sf{Z = {b, c, d, e, f}} \\ [/tex]
[tex] \large \sf{Therefore,} \\ [/tex]
- [tex] \sf{Universal \: set = {a, b, c, d, e, f, g}} \\ [/tex]
- [tex] \sf{X' = not \: in \: X = {b, d, f}} \\ [/tex]
- [tex] \sf{Y' = not \: in \: Y = {d, e, f, g}} \\ [/tex]
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[tex] \large \sf{ \pmb{Problem \: 1:}} \\ [/tex]
[tex]\begin{gathered}\begin{aligned}\sf \left(Y \cap X'\right) \cup Z & = \sf \left(\{a,b,c\} \cap \{b,d,f\}\right) \cup \{ b,c,d,e,f \}\\& = \sf \{b\} \cup \{b,c,d,e,f\}\\& = \sf \{b,c,d,e,f\}\end{aligned}\end{gathered} \\ [/tex]
[tex] \large \sf{ \pmb{Problem \: 2:}} \\ [/tex]
[tex]\begin{gathered}\begin{aligned}\sf Y'-X & = \sf \{d,e,f,g\} - \{a,c,e,g\}\\& = \sf \{d,f\}\end{aligned}\end{gathered} [/tex]
