I Help manipulating a Physics equation

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I'm reading my physics textbook and the example has an equation of...

mg-m(g/(1+M/2m)) = T

manipulated so that it equals...

T = mg/(1+2m/M)

I'm not sure how to get this I tried to distribute the m to get...

mg - mg/(1+M/2m)

and multiplying mg by ((1+M/2m)/(1+M/2m)) but could not get the result the book got.
 
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The approach is good so far. Make a single fraction out of it, then divide both numerator and denominator by M/(2m).
 
It seems i missed some parentheses; it should be...

mg - m(g/(1+(M/2m))) = T*

T = mg/(1+(2m/M))*

And...

mg - mg/(1+(M/2m))

respectively

Does that change the approach?
 
mfb said:
The approach is good so far. Make a single fraction out of it, then divide both numerator and denominator by M/(2m).

I made into one single fraction of...

(mg + (Mmg/2m) - mg)/(1+(M/2m))

canceled out the mg's and multiplied the top by the reciprocal of the bottom...

Mmg/2m x 1/(1+(M/2m)) = Mmg/(2m +2Mm/2m)

Took out M from the denominator...

Mmg/(M(2m/M+1)

Which is equal to...

mg/(1+(2m/M))

Did i do this correctly?
 
Michael Barilla said:
I'm reading my physics textbook and the example has an equation of...

mg-m(g/(1+M/2m)) = T

manipulated so that it equals...

T = mg/(1+2m/M)

I'm not sure how to get this I tried to distribute the m to get...

mg - mg/(1+M/2m)

and multiplying mg by ((1+M/2m)/(1+M/2m)) but could not get the result the book got.
Only looked briefly - did not read it;
Mere basic algebra. Nothing tough and nothing unusual.
 
Looks fine.
Brackets around 2m denominatiors would have been more important than the other brackets by the way.
 
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