So I have the task to prove that (c+1)/(c^2-1)-(c+1)/(1-c) can be simplified to (c+2)/(c-1)and I have simplified the first term to 1/(c-1) and I am then left with
-(c+1)/(-c+1)+1/(c-1) and to my understanding will i factor the term "-c+1" to -(c-1), but what I don't get is how I can then remove the - infront of (c+1)/(-c+1) and end up with equal denominators to combine the fractions
+223 I´m not sure how to explain it but I´ll give you my solution:
(c+1)/(c^2-1)-(c+1)/(1-c)=
= 1/(c-1) - (c+1)/(1-c) (conjugates)
= (1-c)/(c-1)(1-c) - (c+1)(c-1)/(c-1)(1-c) (common denominators)
= (1-c)+(-(c-1)(c+1))/ -(c-1)(c-1) the red text is replaced with 1 due to divison
= (1+ c+1)/(c-1)
=(c+2)/(c-1)
VSB
Good luck, if i have been unclear about something, ask me or someone else here:)
+223 I´m not sure how to explain it but I´ll give you my solution:
(c+1)/(c^2-1)-(c+1)/(1-c)=
= 1/(c-1) - (c+1)/(1-c) (conjugates)
= (1-c)/(c-1)(1-c) - (c+1)(c-1)/(c-1)(1-c) (common denominators)
= (1-c)+(-(c-1)(c+1))/ -(c-1)(c-1) the red text is replaced with 1 due to divison
= (1+ c+1)/(c-1)
=(c+2)/(c-1)
VSB
Good luck, if i have been unclear about something, ask me or someone else here:)
+129899 (c+1)/(c^2-1)-(c+1)/(1-c) =
1/ (c - 1) - (c + 1)/ (1 - c) factor the negative out of the second denominator
1/(c -1) (+) - (c + 1)/ [- (c -1)] notice that the negatives in the numerator and denominator "cancel' with each other
And we're left with
1/(c -1) + (c + 1)/(c -1) =
(1 + c + 1) / (c -1) =
(c + 2)/(c -1)
Factoring out a negative is a good "trick" in some cases !!!
