Question

Solve each of the following equations. Also, verify the result in each case.
$m-\frac{m-1}{2}=1-\frac{m-2}{3}$

Answer 1

$m-\frac{m-1}{2}=1-\frac{m-2}{3}$
=> $\frac{2m-m-(-1)}{2}=\frac{3-m-(-2)}{3}$
=> $\frac{m+1}{2}=\frac{3-m+2}{3}$
=> $\frac{m+1}{2}=\frac{5-m}{3}$
=> $\frac{m}{2}+\frac{1}{2}=\frac{5}{3}-\frac{m}{3}$
Transposing m/3 to LHS and 1/2 to RHS, we get
=> $\frac{m}{2}+\frac{m}{3}=\frac{5}{3}-\frac{1}{2}$
=> $\frac{3m+2m}{6}=\frac{10-3}{6}$
Multiplying both sides by 6, we get
=> $\frac{5m}{6}\times 6=\frac{7}{6}\times 6$
=> 5m = 7
Dividing both sides by 5, we get
=> $\frac{5m}{5}=\frac{7}{5}$
=> m = $\frac{7}{5}$
Verification:
Substituting m =$\frac{7}{5}$ on both sides, we get
$\frac{7}{5}-\frac{{\displaystyle \frac{7}{5}}-1}{2}=1-\frac{{\displaystyle \frac{7}{5}}-2}{3}$
$\frac{7}{5}-\frac{{\displaystyle \frac{7-5}{5}}}{2}=1-\frac{{\displaystyle \frac{7-10}{5}}}{3}$
$\frac{7}{5}-\frac{2}{5\times 2}=1-\frac{-3}{5\times 3}$
$\frac{7}{5}-\frac{1}{5}=1+\frac{1}{5}$
$\frac{6}{5}=\frac{6}{5}$
LHS = RHS
Hence, verified.