Want to have some fun /sci/? I am having a problem with th

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here is the section in particular for those too lazy

>>8562985
whats the problem really? just solve that equation with d in it for d. then use that equation to replace d in (b+c)/d and you get what you are looking for.

>>8562999
From the 2 above: lhs=[b+c]/d = [a+b]/c
lhs*d = b+c
lhs*c = a+b
lhs*d +2lhs*c = b+c + 2(a+b)
lhs*(2c+d) = 2a+3b+c
lhs = [2a+3b+c]/(2c+d)

>>8563002
That's a little vague. What do you mean by 'that equation'?

>>8563002
addendum:

i mean you have:
[math] \frac{2*(a+b)}{2c} = \frac{a-c}{c-d} [/math]
solve for d you get:
[math] \frac{c-d}{a-c} = \frac{2c}{2(a+b)} [/math]
=>
[math] c-d = \frac{2c}{2(a+b)}(a-c) [/math]
=>
[math] c - \frac{2c}{2(a+b)}(a-c) = d [/math]
then use that expression for d to replace it in [math] \frac{b+c}{d} [/math]

>>8563009
But why is it literally just the numerator added on the denominator?

I understand how you got that, it makes sense. It's just, is that something you can do in a situation like this, algebraically? Just add the numerators and denominators??

>>8563024
If r = N/D = n/d
then r = [xN+yn]/[xD+yd]

Usually this is mentioned with proportions, N:D = n:d then N:D = xN+yn : xD+yn

>>8563042
Thanks man, this makes sense now that you put it that way. I love you guys.