Question

guys, please help me​

Answer 1

Given Question :-

Solve for x :

$\rm :\longmapsto\:\dfrac{2x}{3} - \dfrac{x}{4} = \dfrac{x}{5} - \dfrac{x}{6} + 23$

$\red{\large\underline{\sf{Solution-}}}$

Given equation is

$\rm :\longmapsto\:\dfrac{2x}{3} - \dfrac{x}{4} = \dfrac{x}{5} - \dfrac{x}{6} + 23$

On taking LCM of 3 and 4 on LHS and taking LCM of 5 and 6 on RHS, we get

$\rm :\longmapsto\:\dfrac{8x - 3x}{12} = \dfrac{6x - 5x + 690}{30}$

$\rm :\longmapsto\:\dfrac{5x}{12} = \dfrac{x + 690}{30}$

$\rm :\longmapsto\:30(5x) = 12(x + 690)$

On dividing by 6, on both sides, we get

$\rm :\longmapsto\:5(5x) = 2(x + 690)$

$\rm :\longmapsto\:25x = 2x + 690 \times 2$

$\rm :\longmapsto\:25x - 2x = 690 \times 2$

$\rm :\longmapsto\:23x = 690 \times 2$

$\rm :\longmapsto\:x = \dfrac{690 \times 2}{23}$

$\rm :\longmapsto\:x = 30 \times 2$

$\purple{\bf\implies \:\boxed{ \tt{ \: x = 60 \: }}}$

Verification :-

Given equation is

$\rm :\longmapsto\:\dfrac{2x}{3} - \dfrac{x}{4} = \dfrac{x}{5} - \dfrac{x}{6} + 23$

Consider LHS

$\rm :\longmapsto\:\dfrac{2x}{3} - \dfrac{x}{4}$

On substituting x = 60, we get

$\rm \: = \:\dfrac{2 \times 60}{3} - \dfrac{60}{4}$

$\rm \: = \:40 - 15$

$\rm \: = \:25$

Now, Consider RHS

$\rm :\longmapsto\:\dfrac{x}{5} - \dfrac{x}{6} + 23$

On substituting x = 60, we get

$\rm \: = \:\dfrac{60}{5} - \dfrac{60}{6} + 23$

$\rm \: = \:12 - 10 + 23$

$\rm \: = \:2 + 23$

$\rm \: = \:25$

So,

$\bf\implies \:LHS = RHS$

Hence, Verified