Fraction Skills are a MUST in Algebra! Can you do this? (complex fraction)

How to simplify a complex fraction and rational expression.

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[tomtke]

{[(K)/(K-C)-(1)]/[(K)/(K-C)]}

top bracket (numerator)
[(K)/(K-C)-(1)]

bottom bracket (denominator)
[(K)/(K-C)]

top bracket (numerator)
[(K)/(K-C)-(1)]
[(K)/(K-C) - (K-C)/(K-C)]
[(K - K + C)/(K-C)]
[(C)/(K-C)]

bottom bracket (denominator)
[(K)/(K-C)]

The rule of dividing by a fraction is to 'invert and multiply.'

i.e. new equation becomes

[(C)/(K-C)] × [(K-C)/(K)]

rearranged
(K-C)/(K-C) × (C)/(K)
or
simply
C/K

alternately ----
substitute
(K)/(K-C) = B

yielding

(B - 1)÷B
=
1- (1/B)

recall
B = (K)/(K-C)
so 1/B
= (K-C)/(K)

1-(1/B)
= 1 - (K-C)/K
= 1 - (K/K) + (C/K)
= 1 - 1 + (C/K)
= C/K✔️

[KW-gbcd]

a = K/(K - C); (a - 1)/a = 1 - 1/a = 1 - (K - C)/K = 1 - (1 - C/K) = C/K.
I used the variable "a" to eliminate some confusion.

[Kleermaker]

I did it like this: K/K -C - 1/1/K/K - C = (K - C /K)(K/K - C - 1/1) = K^2 - CK/K^2 - CK - (K - C)/K = 1/1 - (K - C)/K = 1/1 - K + C/K = K - K + C/K = C/K.

[panlomito]

Nice ! a = k / (k-c) then a - 1 / a = a/a - 1/a = 1 - 1/a = k/k - (k-c) / k = (k - k + c)/k = c/k

[rayelee]

Just multiply the inverse of the denominator by the numerator and it gets simple very quick. C/K

[heidicameron]

I figured it out with your help where were you when I was in school??

[aryusure]

OK! Distribution with the numerator. But K - (K - C) - if we remove the parentheses- looks like K - K = 0 and we are left with minus C. Hence, my mistake.
So why I can't do that?
What's the rule I missed here?

[aryusure]

I got minus C over K. Close! Let's see what mistake I made.

[harrymatabal]

(I -k+c)/k, I maybe wrong didn't even take 1 minute

[raynewport]

Too confusing. Just multiply the top and bottom of the expression by (k-c) and the answer falls out