Question

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

Answer 1

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