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Question

Add: 3a4b+4c,2a+3b8c,a6b+c.

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Solution

(3a4b+4c)+(2a+3b8c)+(a6b+c)=3a4b+4c+2a+3b8c+a6b+c=(3a+2a+a)+(4b+3b6b)+(4c8c+c)(Combiningliketerms)=(3+2+1)a+(4+36)b+(48+1)c=6a7b3c
Hence, (3a4b+4c)+(2a+3b8c)+(a6b+c)=6a7b3c.

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