Question

(x-3)/4 + (x-1)/5 - (x-2)/3 = 1

Answer 1

Ⓐⓝⓢⓦⓔⓡ

We have

$\blue{\tt{ \frac{x - 3}{4} + \frac{x - 1}{5} - \frac{x - 2}{3} = 1 }}$

$\blue{ \tt{ \frac{x - 3}{4} + \frac{x - 1}{5} = 1 + \frac{x - 2}{3} }}$

Taking LCM in both LHS and RHS

$\blue{ \tt{ \frac{5x - 15 + 4x - 4}{20} = \frac{3 + x - 2}{3} }}$

$\blue{ \tt{ \frac{9x - 19}{20} = \frac{x + 1}{3} }}$

Cross multiplying

$\blue{ \tt{3(9x - 19) = 20(x + 1)}}$

$\blue{ \tt{27x - 57 = 20x + 20}}$

$\blue{ \tt{27x - 20x = 20 + 57}}$

$\blue{ \tt{7x = 77}}$

$\boxed{ \purple{ \bf{x = 11}}}$

Answer 2

GIVEN :-

$\rm{ \dfrac{x - 3}{4} + \dfrac{x - 1}{5} - \dfrac{x - 2}{3} = 1 }$

TO FIND :-

value of x in

$\rm{ \dfrac{x - 3}{4} + \dfrac{x - 1}{5} - \dfrac{x - 2}{3} = 1 }$

SOLUTION :-

$\implies \rm{ \dfrac{x - 3}{4} + \dfrac{x - 1}{5} = 1 + \dfrac{x - 2}{3} }$

Taking LCM in both LHS and RHS

$\implies \rm{ \dfrac{5x - 15 + 4x - 4}{20} = \dfrac{3 + x - 2}{3} }$

$\implies\rm{ \dfrac{9x - 19}{20} = \dfrac{x + 1}{3} }$

now by Cross multiplying

$\implies \rm{3(9x - 19) = 20(x + 1)}$

$\implies \rm{27x - 57 = 20x + 20}$

$\implies \rm{27x - 20x = 20 + 57}$

$\implies \rm{7x = 77}$

$\implies \boxed{ \boxed{ \rm{x = 11}}}$