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 + 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