Question

Solve this plsss fast






plsss fast​

Answer 1

Given Question

Solve for x :

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

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

Given expression is

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

On taking LCM of (2 and 3 = 6) on LHS and Taking LCM of ( 2 and 6 = 6) on RHS, we get

$\rm :\longmapsto\:\dfrac{3(3x - 5) + 6x + 2(2x - 3)}{6} = \dfrac{5 - 9x}{6}$

$\rm :\longmapsto\:9x - 15 + 6x + 4x - 6 = 5 - 9x$

$\rm :\longmapsto\:(9x + 6x + 4x) - 15 - 6 = 5 - 9x$

$\rm :\longmapsto\:19x - 21 = 5 - 9x$

$\rm :\longmapsto\:19x + 9x = 5 + 21$

$\rm :\longmapsto\:28x = 26$

$\rm :\longmapsto\:x = \dfrac{26}{28}$

$\bf\implies \:x = \dfrac{13}{14}$

VERIFICATION

Consider LHS

$\rm :\longmapsto\:\dfrac{3x - 5}{2} + x + \dfrac{2x - 3}{3}$

On substituting the value of x, we get

$\rm \:  =  \: \dfrac{\dfrac{39}{14} - 5}{2} + \dfrac{13}{14} + \dfrac{\dfrac{13}{7} - 3}{3}$

$\rm \:  =  \: \dfrac{39- 70}{28} + \dfrac{13}{14} + \dfrac{13 - 21}{21}$

$\rm \:  =  \: \dfrac{ - 31}{28} + \dfrac{13}{14} - \dfrac{8}{21}$

$\rm \:  =  \: \dfrac{ - 31 + 26}{28} - \dfrac{8}{21}$

$\rm \:  =  \: \dfrac{ -5}{28} - \dfrac{8}{21}$

$\rm \:  =  \: \dfrac{ -15 - 32}{84}$

$\rm \:  =  \: - \: \dfrac{47}{84}$

Consider RHS

$\rm :\longmapsto\: \dfrac{5}{6} - \dfrac{3x}{2}$

On substituting the value of x, we get

$\rm \:  =  \: \dfrac{5}{6} - \dfrac{39}{28}$

$\rm \:  =  \: \dfrac{70 - 117}{84}$

$\rm \:  =  \: - \dfrac{47}{84}$

Hence, LHS = RHS

Hence, Verified

Answer 2

Answer:

Given Question

Solve for x :

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

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

Given expression is

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

On taking LCM of (2 and 3 = 6) on LHS and Taking LCM of ( 2 and 6 = 6) on RHS, we get

$\rm :\longmapsto\:\dfrac{3(3x - 5) + 6x + 2(2x - 3)}{6} = \dfrac{5 - 9x}{6}$

$\rm :\longmapsto\:9x - 15 + 6x + 4x - 6 = 5 - 9x$

$\rm :\longmapsto\:(9x + 6x + 4x) - 15 - 6 = 5 - 9x$

$\rm :\longmapsto\:19x - 21 = 5 - 9x$

$\rm :\longmapsto\:19x + 9x = 5 + 21$

$\rm :\longmapsto\:28x = 26$

$\rm :\longmapsto\:x = \dfrac{26}{28}$

$\bf\implies \:x = \dfrac{13}{14}$

VERIFICATION

Consider LHS

$\rm :\longmapsto\:\dfrac{3x - 5}{2} + x + \dfrac{2x - 3}{3}$

On substituting the value of x, we get

$\rm \:  =  \: \dfrac{\dfrac{39}{14} - 5}{2} + \dfrac{13}{14} + \dfrac{\dfrac{13}{7} - 3}{3}$

$\rm \:  =  \: \dfrac{39- 70}{28} + \dfrac{13}{14} + \dfrac{13 - 21}{21}$

$\rm \:  =  \: \dfrac{ - 31}{28} + \dfrac{13}{14} - \dfrac{8}{21}$

$\rm \:  =  \: \dfrac{ - 31 + 26}{28} - \dfrac{8}{21}$

$\rm \:  =  \: \dfrac{ -5}{28} - \dfrac{8}{21}$

$\rm \:  =  \: \dfrac{ -15 - 32}{84}$

$\rm \:  =  \: - \: \dfrac{47}{84}$

Consider RHS

$\rm :\longmapsto\: \dfrac{5}{6} - \dfrac{3x}{2}$

On substituting the value of x, we get

$\rm \:  =  \: \dfrac{5}{6} - \dfrac{39}{28}$

$\rm \:  =  \: \dfrac{70 - 117}{84}$

$\rm \:  =  \: - \dfrac{47}{84}$

Hence, LHS = RHS

Hence, Verified ✓