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Question

Subtract 4x2y+z from the sum of 3x+4y+z and 6x+7y8z

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Solution

Lets evaluate the sum of the given expressions 3x+4y+z and 6x+7y8z,

(3x+4y+z)+(6x+7y8z)=3x+4y+z6x+7y8z

=(3x6x)+(4y+7y)+(z8z)

=3x+11y7z (1)


Now, subtract 4x2y+z from the sum obtained in the above step,

Sum of [(3x+4y+z)+(6x+7y8z)](4x2y+z)

=(3x+11y7z)(4x2y+z) [ From (1) ]

=3x+11y7z4x+2yz

=(3x4x)+(11y+2y)+(7zz)

=7x+13y8z

Therefore, the required difference is 7x+13y8z .

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