How do you solve $\frac{1}{2} (m - 6) = 3 + \frac{m}{5}$ $1/2(m-6) = 3 + m/5$ ?

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Answer 1 · 2018-04-08T02:45:00

$m = 20$ $m = 20$

$\Rightarrow \frac{1}{2} (m - 6) = 3 + \frac{m}{5}$ $=> 1/2(m - 6) = 3 + m/5$

$\Rightarrow \frac{m - 6}{2} = 3 + \frac{m}{5}$ $=> (m - 6)/2 = 3 + m/5$

$\Rightarrow \frac{m}{2} - \frac{6}{2} = 3 + \frac{m}{5}$ $=> m/2 - 6/2 = 3 + m/5$

$\Rightarrow \frac{m}{2} - 3 = 3 + \frac{m}{5}$ $=> m/2 - 3 = 3 + m/5$

$\Rightarrow \frac{m}{2} - \frac{m}{5} = 3 + 3$ $=> m/2 - m/5 = 3 + 3$

$\Rightarrow m [\frac{1}{2} - \frac{1}{5}] = 6$ $=> m[1/2 - 1/5] = 6$

$\Rightarrow m [(\frac{1}{2} \times \frac{5}{5}) - (\frac{1}{5} \times \frac{2}{2})] = 6$ $=> m[(1/2 × 5/5) - (1/5 × 2/2)] = 6$

$\Rightarrow m [\frac{5}{10} - \frac{2}{10}] = 6$ $=> m[5/10 - 2/10] = 6$

$\Rightarrow m [\frac{5 - 2}{10}] = 6$ $=> m[(5 - 2)/10] = 6$

$\Rightarrow m (\frac{3}{10}) = 6$ $=> m(3/10) = 6$

$=> m = 6 × 10/(3) = cancel(3) × 2 × 10/cancel(3)$

$=> m = 20$

How do you solve 1/2(m-6) = 3 + m/512(m−6)=3+m51/2(m-6) = 3 + m/5?