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 \(2x^2 – x + 9\)
\(= 2x^2 – x + 9 – (2x^2 – 3x + 5)\)
\(= 2x^2 – x + 9 – 2x^2 + 3x – 5\)
\(= 2x^2 – 2x^2 – x + 3x + 9 – 5\)
\(= 2x + 4\)