Question

Add: 6p + 4q − r + 3, 2r − 5p − 6, 11q − 7p + 2r − 1 and 2q − 3r + 4​

Answer 1

Answer:

Given: 6p+4q−r+3,2r−5p−6,11q−7p+2r−1 and 2q−3r+4

To add the given terms we have to add coefficient of like terms:

∴(6p+4q−r+)3+(2r−5p−6)+(11q−7p+2r−1)+(2q−3r+4)

=(6−5−7+0)p+(4+0+11+2)q+(−1+2+2−3)r+(3−6−1+4)

=−6p+17q+0r+0

=17q−6p

Answer 2

Answer:

-6p + 17q

Step-by-step explanation:

(6p + 4q - r + 3) + (2r - 5p - 6) + (11q - 7p +2r - 1) + (2q - 3r + 4)

6p + 4q - r + 3 + 2r - 5p - 6 + 11q - 7p + 2r - 1 + 2q - 3r + 4

6p - 5p -7p + 4q + 11q + 2q - r + 2r + 2r - 3r + 3 - 6 - 1 + 4

-6p + 17q

All r and constants terms gets cancelled out