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Answer :
6.\(m-\frac{m-1}2=1-\frac{m-2}3\)
Given that,\(m-\frac{m-1}2=1-\frac{m-2}3\)
LCM of denominators 3 and 4=6
So, \(m×6-\frac{m-1}2×6=1×6-\frac{m-2}3×6 \quad \)[Multiplying both the sides by 6]
\(\Rightarrow 6m – (m – 1) × 3 = 6 – (m – 2) × 2\)
\(\Rightarrow 6m – 3m + 3 = 6 – 2m + 4 \quad \)[Solving the brackets]
\(\Rightarrow 3m + 3 = 10 – 2m \)
\(\Rightarrow 3m + 2m = 10 – 3 \)[Transposing 2m to LHS and 3 to RHS]
\(\Rightarrow 5m = 7\)
\(\Rightarrow m = \frac { 7 }{ 5 } \)[Transposing 5 to RHS]
Hence \(m = \frac { 7 }{ 5 } \) is the required solution.