Question

Refer to attached image. Urgently needed.

Answer 1

this is the way by which you can solve

Answer 2

Solution :

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Given :

To find the value of y,

if,

⇒ $\frac{3y- \frac{5}{6} }{4} + 1 =  \frac{2y - \frac{1}{3} }{3} + 5$

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By equating both the equations,

We get,.

⇒ $\frac{ \frac{18y - 5}{6} }{4} + 1 = \frac{ \frac{6y - 1}{3} }{3} + 5$

⇒ $[(\frac{18y - 5}{6}) \div 4] + 1 = [(\frac{6y-1}{3}) \div 3] + 5$

⇒ $[(\frac{18y - 5}{6})( \frac{1}{4} )] + 1 = [(\frac{6y-1}{3})( \frac{1}{3}) ] + 5$

⇒ $\frac{18y - 5}{24} + 1 = \frac{6y-1}{9} + 5$

⇒ $\frac{18y - 5}{24} - \frac{6y - 1}{9} = 5 - 1$

By taking LCM{24 & 9} = 72,

⇒ 24 × 3 = 72

⇒ 9 × 8 = 72,.

So,.

⇒ $\frac{(18y - 5)(3)}{(24)(3)} - \frac{(6y-1)(8)}{(9)(8)} = 4$

⇒ $\frac{54y - 15}{72} - \frac{48y - 8}{72} = 4$

⇒ $\frac{54y - 15 - (48y - 8)}{72} = 4$

⇒ $\frac{54y - 15 - 48y + 8}{72} = 4$

⇒ $\frac{6y - 7}{72} = 4$

⇒ $6y - 7 = 4(72)$

⇒ 6y - 7 = 298

⇒ 6y = 298 + 7

⇒ 6y = 305

⇒ $y = \frac{305}{6}$

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 I'm not sue about my answer,.

⇒ I Hope it's correct,.
.
⇒ If corect,. ⇒⇒ Hope it Helps !!