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# 6.$$m-\frac{m-1}2=1-\frac{m-2}3$$

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.