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For P = {6, 12, 13, 15}, Q = {2, 7, 11), and R = {4, 7, 9, 11}, find PU (QNR).

1 Answer

1 vote

The answer is:

P U (Q n R) = {6, 7, 11, 12, 13, 15}

The question gives us 3 sets: P, Q and R and asks us to find:


P\cup(Q\cap R)

We should solve the sets in the bracket first because a bracket signifies relationship priority between sets Q and R comes before any relationship with P

Q n R means we are to find the intersection of Sets Q and R. This means that we should find which elements are present in both sets only.

If set Q contains {2, 7, 11} and set R contains {4, 7, 9, 11}

Therefore, the elements (or numbers) common to both sets are 7 and 11. Thus, this is given by:


Q\cap R=\mleft\lbrace7,11\mright\rbrace

Now that we have established what is in the bracket, we can now proceed to explore the relationship with set P.

P U K is a union of sets P and K. This means that you can include all the elements (or numbers) of both sets P and K.

Therefore, we can solve P U (Q n R) as:


\begin{gathered} Q\cap R=\mleft\lbrace7,11\mright\rbrace\text{ (from previous calculations.} \\ \\ \therefore P\cup(Q\cap R)=\mleft\lbrace6,12,13,15\mright\rbrace\cup\mleft\lbrace7,11\mright\rbrace \\ \\ P\cup(Q\cap R)=\mleft\lbrace6,7,11,12,13,15\mright\rbrace \end{gathered}

Therefore, the final answer is:

P U (Q n R) = {6, 7, 11, 12, 13, 15}

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User Necktwi
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