How do you solve $\frac{2}{3} n + 6 = \frac{1}{4} n - 3$ $2/3n+6=1/4n-3$ ?

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Answer 1 · 2016-07-12T14:08:04

The soln. is $n = - \frac{108}{5} = - 21.6$ $n=-108/5=-21.6$ .

Given that, $\frac{2}{3} n + 6 = \frac{1}{4} n - 3 .$ $2/3n+6=1/4n-3.$

Isolating the terms with variable $n$ $n$ on $L . H . S .$ $L.H.S.$ and the consts. on the $R . H . S .$ $R.H.S.$ , we get,

$\frac{2}{3} n - \frac{1}{4} n = - 6 - 3 = - 9 .$ $2/3n-1/4n=-6-3=-9.$

To get rid of fractions, we multiply the eqn. by $12$ $12$ , which is the $l . c . m .$ $l.c.m.$ of nos. in Drs. of $L . H . S .$ $L.H.S.$

$8 n - 3 n = - 108 \Rightarrow 5 n = - 108 \Rightarrow n = - \frac{108}{5}$ $8n-3n=-108 rArr 5n=-108 rArr n=-108/5$ .

Hence, the soln. is $n = - \frac{108}{5} = - 21.6$ $n=-108/5=-21.6$ .

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