From the sum of 2x2 + 3xy – 5 and 7 + 2xy – x2 subtract 3xy + x2 – 2.

Sum of the given term is (2x2 + 3xy – 5) + (7 + 2xy – x2)= 2x2 + 3xy – 5 + 7 + 2xy – x2= 2x2 – x2 + 3xy + 2xy – 5 + 7= x2 + 5xy + 2Now (x2 + 5xy + 2) – (3xy + x2 – 2)= x2 + 5xy + 2 – 3xy – x2 + 2= x2 – x2 + 5xy – 3xy + 2 + 2= 0 + 2xy + 4= 2xy + 4

From the sum of 2x2 + 3xy – 5 and 7 + 2xy – x2 subtract 3xy + x2 – 2.

Sum of the given term is (2x2 + 3xy – 5) + (7 + 2xy – x2)= 2x2 + 3xy – 5 + 7 + 2xy – x2= 2x2 – x2 + 3xy + 2xy – 5 + 7= x2 + 5xy + 2Now (x2 + 5xy + 2) – (3xy + x2 – 2)= x2 + 5xy + 2 – 3xy – x2 + 2= x2 – x2 + 5xy – 3xy + 2 + 2= 0 + 2xy + 4= 2xy + 4

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From the sum of 2x² + 3xy – 5 and 7 + 2xy – x² subtract 3xy + x² – 2.

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2xy+4

xy-4

12xy+14

xy-14

answer is A.

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Detailed Solution

Sum of the given term is (2x² + 3xy – 5) + (7 + 2xy – x²)
= 2x² + 3xy – 5 + 7 + 2xy – x²
= 2x² – x² + 3xy + 2xy – 5 + 7
= x² + 5xy + 2
Now (x² + 5xy + 2) – (3xy + x² – 2)
= x² + 5xy + 2 – 3xy – x² + 2
= x² – x² + 5xy – 3xy + 2 + 2
= 0 + 2xy + 4
= 2xy + 4