6. (a) From the sum of 3x – y + 11 and – y – 11, subtract 3x – y – 11.
(b) From the sum of 4 + 3x and 5 – 4x + 2x², subtract the sum of 3x² – 5x and –x² + 2x + 5.

(a) Let's first find the sum of 3x – y + 11 and – y – 11

= 3x – y + 11 + (-y – 11)

= 3x – y + 11 – y – 11

= 3x – y – y + 11 – 11

= 3x – 2y

Now, subtracting 3x – y – 11 from the obtained expression 3x – 2y, we get

= 3x – 2y – (3x – y – 11)

= 3x – 2y – 3x + y + 11

= 3x – 3x – 2y + y + 11

= – y + 11

(b) Let's first find the sum of 4 + 3x and 5 – 4x + 2x²

= 4 + 3x + (5 – 4x + 2x²)

= 4 + 3x + 5 – 4x + 2x²

= 4 + 5 + 3x – 4x + 2x²

= 9 – x + 2x²

= 2x² – x + 9 … [1]

Then, let's find the sum of 3x² – 5x and – x² + 2x + 5

= 3x² – 5x + (-x² + 2x + 5)

= 3x² – 5x – x² + 2x + 5

= 3x² – x² – 5x + 2x + 5

= 2x² – 3x + 5 … [2]

Now, subtracting equation (2) from equation (1), we get

= 2x² – x + 9 – (2x² – 3x + 5)

= 2x² – x + 9 – 2x² + 3x – 5

= 2x² – 2x² – x + 3x + 9 – 5

= 2x + 4