Question

Solve the following
$\frac{4m - 3}{4 } - \frac{3m + 4}{3} = \frac{1}{4} - m$
​

Answer 1

Given :-

$\quad \leadsto \quad \sf \dfrac{4m-3}{4} - \dfrac{3m+4}{3} = \dfrac{1}{4} - m$

To Find:-

Value of ' m '

Solution :-

Let's start with the begin :)

$\quad \leadsto \quad \sf \dfrac{4m-3}{4} - \dfrac{3m+4}{3} = \dfrac{1}{4} - m$

Take LCM on LHS ;

${ : \implies \quad \sf \dfrac{3 ( 4m - 3 ) - 4 ( 3m + 4 )}{4 \times 3} = \dfrac{1}{4} - m }$

${ : \implies \quad \sf \dfrac{12m - 9 - 12m - 16}{12} = \dfrac{1}{4} - m }$

${ : \implies \quad \sf \dfrac{\cancel{12m} - 9 - \cancel{12m} - 16}{12} = \dfrac{1}{4} - m }$

${ : \implies \quad \sf - \dfrac{25}{12} = \dfrac{1}{4} - m }$

Transpose ' m ' to LHS ;

${ : \implies \quad \sf m - \dfrac{25}{12} = \dfrac{1}{4}}$

Transpose " $\bf \cfrac{25}{12}$ " to RHS ;

${ : \implies \quad \sf m = \dfrac{1}{4} + \dfrac{25}{12}}$

${ : \implies \quad \sf m = \dfrac{3 + 25}{12}}$

${ : \implies \quad \sf m = \dfrac{28}{12}}$

${ : \implies \quad \sf m = \cancel{\dfrac{28}{12}}}$

${ : \implies \quad \sf m = \dfrac{14}{6}}$

${ : \implies \quad \sf m = \cancel{\dfrac{14}{6}}}$

${ : \implies \quad \bf \therefore \quad m = \dfrac{7}{3}}$

Henceforth , The Required Answer is ${ \pmb { \blue { \bf { \cfrac{7}{3}}}}}$ :)

Answer 2

Step-by-step explanation:

$\frac{4m \: - \: 3}{4} \: - \: \frac{3m \: + \: 4}{3} \: = \: \frac{1}{4} \: - \: m \\ \\ \frac{3(4m \: - \: 3) - 4(3m \: + \: 4)}{12} \: = \: \frac{1}{4} \: - \: m \\ \\ \frac{12m - 9 - 12m - 16}{12} \: = \: \frac{1}{4} - m \\ \\ \frac{ - 25}{12} \: = \: \frac{1}{4} - m \\ \\ m \: = \: \frac{1}{4} + \frac{25}{12} \: \\ \\ m = \frac{3 + 25}{12} \\ \\ m \: = \: \frac{28}{12} \\ \\ m \: = \: \frac{7}{3}$