Question

Solve the linear equation:
\[m - \dfrac{{m - 1}}{2} = 1 - \dfrac{{m - 2}}{3}\]
$(a){\text{ }}m = \dfrac{7}{2}$
$(b){\text{ }}m = \dfrac{3}{4}$
$(c){\text{ }}m = \dfrac{7}{5}$
$(d){\text{ }}m = \dfrac{4}{7}$

Answer 1

Hint: In the above given question, the linear equation is to be solved. It can be solved through many ways, one of them is shown below. Another way to solve the same equation can also be possible by evaluating both the LHS as well as the RHS separately and then equating the final results so obtained.

We have the given equation in the question as,
 \[m - \dfrac{{m - 1}}{2} = 1 - \dfrac{{m - 2}}{3}\]
Taking LCM on LHS and RHS, we get,
$ \Rightarrow \dfrac{{2m - m + 1}}{2} = \dfrac{{3 - m + 2}}{3}$
\[ \Rightarrow \dfrac{{m + 1}}{2} = \dfrac{{5 - m}}{3}\]
By cross-multiplication, we get,
\[ \Rightarrow 3(m + 1) = 2(5 - m)\]
\[ \Rightarrow 3m + 3 = 10 - 2m\]
By transposing 3 to RHS and $ - 2m$ to LHS, we get the following equation,
\[ \Rightarrow 3m + 2m = 10 - 3\;\]
\[ \Rightarrow 5m = 7\]
\[ \Rightarrow m = \dfrac{7}{5}\]
Hence, the solution is$(c){\text{ }}m = \dfrac{7}{5}$.

Note: Whenever we face such types of problems the key point is to have a good grasp of the handling of the linear equations. We have to equate both LHS and RHS terms. Various steps may be involved in this calculation including taking LCM, transposition and cross multiplication. In the end, the value of the variable, for which all the calculations were made, will be obtained.