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Question
Solve by trial and error method:z−3=2z−5
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Solution
z−3=2z−5
At z=0, LHS=0−3=−3, RHS=0−5=−5
LHS≠RHS
At z=1, LHS=1−3=−2, RHS=2(1)−5=−3
LHS≠RHS
At z=2, LHS=2−3=−1, RHS=2(2)−5=−1
LHS=RHS
∴z=2 is the root
z−3=2z−5
At z=0, LHS=0−3=−3, RHS=0−5=−5
LHS≠RHS
At z=1, LHS=1−3=−2, RHS=2(1)−5=−3
LHS≠RHS
At z=2, LHS=2−3=−1, RHS=2(2)−5=−1
LHS=RHS
∴z=2 is the root
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