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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[@tomtke7351]

{[(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-gb9cd]

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.

[@russelllomando8460]

that was so cool. thanks.

[@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

[@heidicameron2015]

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

[@rayelee1301]

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

[@aryusure1943]

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?

[@panlomito]

K - (K - C) = K - K - (-C) = 0 + C

[@aryusure1943]

@@panlomito Got it! Thanks! A double minus = +.

[@kennethwright870]

1-1+ c/k = c/k

[@heidicameron2015]

-1?

[@samjones3106]

answer - KFC

[@heidicameron2015]

C over K.

[@saiedahmed5289]

C/K

[@aryusure1943]

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

[@panlomito]

k/k - (k-c) / k = (k - (k-c)) / k = (k - k - (-c)) / k = (k - k + c) / k and not (k - k - c) / k = -c/k

[@quigonkenny]

Guessing at some point you had a "-(...-c)" and didn't double the minus signs when you cleared the parentheses.

[@aryusure1943]

@@quigonkenny Exactly! You are right! I still struggle with the distribution operation when there is a minus sign before the parentheses.

[@harrymatabal8448]

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

[@raynewport9395]

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

[@terry_willis]

If you do that, won't you then get a K^2 component, moving you into the quadratic realm, complicating the process?

[@raynewport9395]

No, most of the k-c components disappear you immediately get ( (k) - (k-c) ) on top and just a k on the bottom@@terry_willis

[@margritkaminsky1470]

Zero