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

(a) 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, subtract 3x – y – 11 from 3x – 2y = 3x – 2y – (3x – y – 11)
= 3x – 2y – 3x + y + 11
= 3x – 3x – 2y + y + 11
= -y + 11
(b) sum of 4 + 3x and$$5 – 4x + 2x^2,$$
$$= 4 + 3x + (5 – 4x + 2x^2)$$
$$= 4 + 3x + 5 – 4x + 2x^2$$
$$= 4 + 5 + 3x – 4x + 2x^2$$
$$= 9 – x + 2x^2$$
$$= 2x^2 – x + 9$$
Now sum of $$3x^2 – 5x$$ and$$-x^2 + 2x + 5.$$
$$= 3x^2 – 5x + (-x^2 + 2x + 5)$$
$$= 3x^2 – 5x – x^2 + 2x + 5$$
$$= 3x^2 – x^2 – 5x + 2x + 5$$
$$= 2x^2 – 3x + 5$$
Now, we have to subtract $$2x^2 – 3x + 5$$ from $$2x^2 – x + 9$$
$$= 2x^2 – x + 9 – (2x^2 – 3x + 5)$$
$$= 2x^2 – x + 9 – 2x^2 + 3x – 5$$
$$= 2x^2 – 2x^2 – x + 3x + 9 – 5$$
$$= 2x + 4$$