Basic algebra/fractions: derivation

Question

How do I do this derivation step? I don't understand why there is equality. The derivation is from my textbook.

$$mg-m\left(\frac{g}{1+\frac{M}{2m}}\right)=\frac{mg}{1+\frac{2m}{M}}$$

Answer 1

Have you tried writing it out?

$\displaystyle mg-m\left(\frac{g}{1+M/2m}\right) = \frac{mg(1+M/2m) - mg}{1 + M/2m} = \frac{M}{2m} \frac{mg}{1 + M/2m} = M\frac{mg}{2m + M} = \ ... $

Yes?

Answer 2

$$mg-m\left(\frac{g}{1+M/2m}\right)=mg\left(1-\frac{1}{1+M/2m}\right)$$ $$=mg\left(\frac{1+M/2m-1}{1+M/2m}\right)$$ $$=mg\left(\frac{M}{2m[1+M/2m]}\right)$$ $$=mg\left(\frac{M}{M[2m/M+1]}\right)$$ $$=\frac{mg}{2m/M+1}$$

Answer 3

Trying to get:

$$mg-m\left(\frac{g}{1+M/2m}\right)=\frac{mg}{1+2m/M}$$

first factor out the m and g:

$$mg\left(1-\frac{1}{1+\frac{M}{2m}}\right)$$

Need the same denominator to add fractions

$$mg\left(\frac{1+\frac{M}{2m}}{1+\frac{M}{2m}}-\frac{1}{1+\frac{M}{2m}}\right)$$

Subtract now and the ones cancel in the numerator.

$$mg\left(\frac{\frac{M}{2m}}{1+\frac{M}{2m}}\right)$$

which is the same as

$$mg\left(\frac{1}{\frac{2m}{M}\left(1+\frac{M}{2m}\right)}\right)$$

Multiply through on the denominator and the $\frac{M}{2m}$ cancels when multiplied with $\frac{2m}{M}$ but you also have to multiply the $1$ by the term so:

$$mg\left(\frac{1}{\frac{2m}{M}\left(1+\frac{M}{2m}\right)}\right)=\frac{mg}{1+2m/M}$$