Question

(6a+3b+3c-5)-(6a-4b+0c+8)​

Answer 1

Answer:

The sum of (6a+4b−c+3),(2b−3c+4),(11b−7a+2c−1) and (2c−5a−6) is (−6a+17b)

Step-by-step explanation:

(6a+4b−c+3)+(2b−3c+4)+(11b−7a+2c−1)+(2c−5a−6)

⇒  6a+4b−c+3+2b−3c+4+11b−7a+2c−1+2c−5a−6

Now arranging and combining the like terms,

⇒  (6a−7a−5a)+(4b+2b+11b)+(−c−3c+2c+2c)+(3+4−1−6)

⇒  −6a+17b

∴  The sum of (6a+4b−c+3),(2b−3c+4),(11b−7a+2c−1) and (2c−5a−6) is (−6a+17b)

Answer 2

Answer: 7B + 3C - 13

STEP BY STEP :-)

Let's simplify step-by-step.

6a+3b+3c−5−(6a−4b+0c+8)

Distribute the Negative Sign:

=6a+3b+3c−5+−1(6a−4b+0c+8)

=6a+3b+3c+−5+−1(6a)+−1(−4b)+−1(0c)+(−1)(8)

=6a+3b+3c+−5+−6a+4b+0+−8

Combine Like Terms:

=6a+3b+3c+−5+−6a+4b+0+−8

=(6a+−6a)+(3b+4b)+(3c)+(−5+0+−8)

=7b+3c+−13

Answer:

=7b+3c−13

Hope this helps:-)