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Solve:
3x-5/2 +x+ 2x-3/3 = 5/6 - 3x/2​

Solve 3x52 X 2x33 56 3x2 class=

Sagot :

ItzChemistryy

Answer:

x = 26/45

Step-by-step explanation:

SalmonWorldTopper

Answer:

Given Question

Solve for x :

[tex]\rm :\longmapsto\:\dfrac{3x - 5}{2} + x + \dfrac{2x - 3}{3} = \dfrac{5}{6} - \dfrac{3x}{2} [/tex]

[tex] \green{\large\underline{\sf{Solution-}}}[/tex]

Given expression is

[tex]\rm :\longmapsto\:\dfrac{3x - 5}{2} + x + \dfrac{2x - 3}{3} = \dfrac{5}{6} - \dfrac{3x}{2} [/tex]

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

[tex]\rm :\longmapsto\:\dfrac{3(3x - 5) + 6x + 2(2x - 3)}{6} = \dfrac{5 - 9x}{6}[/tex]

[tex]\rm :\longmapsto\:9x - 15 + 6x + 4x - 6 = 5 - 9x[/tex]

[tex]\rm :\longmapsto\:(9x + 6x + 4x) - 15 - 6 = 5 - 9x[/tex]

[tex]\rm :\longmapsto\:19x - 21 = 5 - 9x[/tex]

[tex]\rm :\longmapsto\:19x + 9x = 5 + 21[/tex]

[tex]\rm :\longmapsto\:28x = 26[/tex]

[tex]\rm :\longmapsto\:x = \dfrac{26}{28} [/tex]

[tex]\bf\implies \:x = \dfrac{13}{14} [/tex]

VERIFICATION

Consider LHS

[tex]\rm :\longmapsto\:\dfrac{3x - 5}{2} + x + \dfrac{2x - 3}{3}[/tex]

On substituting the value of x, we get

[tex]\rm \:  =  \: \dfrac{\dfrac{39}{14} - 5}{2} + \dfrac{13}{14} + \dfrac{\dfrac{13}{7} - 3}{3}[/tex]

[tex]\rm \:  =  \: \dfrac{39- 70}{28} + \dfrac{13}{14} + \dfrac{13 - 21}{21}[/tex]

[tex]\rm \:  =  \: \dfrac{ - 31}{28} + \dfrac{13}{14} - \dfrac{8}{21}[/tex]

[tex]\rm \:  =  \: \dfrac{ - 31 + 26}{28} - \dfrac{8}{21}[/tex]

[tex]\rm \:  =  \: \dfrac{ -5}{28} - \dfrac{8}{21}[/tex]

[tex]\rm \:  =  \: \dfrac{ -15 - 32}{84}[/tex]

[tex]\rm \:  =  \: - \: \dfrac{47}{84}[/tex]

Consider RHS

[tex]\rm :\longmapsto\: \dfrac{5}{6} - \dfrac{3x}{2} [/tex]

On substituting the value of x, we get

[tex] \rm \:  =  \: \dfrac{5}{6} - \dfrac{39}{28} [/tex]

[tex] \rm \:  =  \: \dfrac{70 - 117}{84}[/tex]

[tex] \rm \:  =  \: - \dfrac{47}{84}[/tex]

Hence, LHS = RHS

Hence, Verified.

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