282
Question
Subtract 3a−2b+4c from the sum of −2a+b−5c and 3a−2b+c
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Solution
Sum of −2a+b−5c and 3a−2b+c = −2a+b−5c+3a−2b+c = a−b−4c
Subtracting 3a−2b+4c from a−b−4c
= a−b−4c−(3a−2b+4c)
=a−b−4c−3a+2b−4c
=−2a+b−8c
Sum of −2a+b−5c and 3a−2b+c = −2a+b−5c+3a−2b+c
= a−b−4c
Subtracting 3a−2b+4c from a−b−4c
= a−b−4c−(3a−2b+4c)
=a−b−4c−3a+2b−4c
=−2a+b−8c
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