1
Question
Solve :
2a−3b−[4a−3b−{a−2c−(a−2b−c)}]]
2a−3b−[4a−3b−{a−2c−(a−2b−c)}]]
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Solution
2a−3b−[4a−3b−{a−2c−(a−2b−c)}]=2a−3b−[4a−3b−{a−2c−a+2b+c}]=2a−3b−[4a−3b−{2b−c}]=2a−3b−[4a−5b+c]=2a−3b−4a+5b−c=−2a+2b−c
=2a−3b−[4a−3b−{a−2c−a+2b+c}]
=2a−3b−[4a−3b−{2b−c}]
=2a−3b−[4a−5b+c]
=2a−3b−4a+5b−c
=−2a+2b−c
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