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Subtract the sum of 3a2 – 2a + 5 and a2 – 5a – 7 from the sum of 5a2 -9a + 3 and 2a – a2 – 1

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Sum of 3a2 – 2a + 5 and a2 – 5a – 7

= 3a2 – 2a + 5 + a2 – 5a – 7

= 3a2 + a2 – 2a – 5a + 5 – 7

= 4a2 – 7a – 2

And sum of 5a2 – 9a + 3 and 2a – a2 - 1

= 5a2 – 9a + 3 + 2a – a2 – 1

= 5a2 – a2 – 9a + 2a + 3 – 1

= 4a2 – 7a + 2

Now, (4a2 – 7a + 2) – (4a2 – 7a – 2)

= 4a2 – 7a + 2 – 4a2 + 7a + 2

= 4a2 – 4a2 – 7a + 7a + 2 + 2

= 0 + 0 + 4 = 4

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