If $a,b,c$ be distinct (non-zero) such that
$ a+ \frac{1}{b}= b+ \frac{1}{c}= c + \frac{1}{a} = k$
how do I find the value of k ?
Asked
Active
Viewed 150 times
-3
-
Please rewrite the equations with MathJax. I, at least, don't understand what you mean by the numbers. What is "1b" and "1c", etc ? – Matti P. Jun 06 '19 at 09:45
-
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Jun 06 '19 at 09:48
-
1Possibly related : https://math.stackexchange.com/questions/1917588/algebra-problem-a-1-b-b-1-c-c-1-a-t – Arnaud D. Jun 06 '19 at 09:50
-
thanks everyone, I'll try .I need time to get used to this – JinSeok Jun 06 '19 at 09:57
-
If you negate all of $a,b,c$, you get the opposite value of $k$ too. And $k=0$ is clearly impossible since $a,b,c$ would then need to have pairwise opposite signs... – hmakholm left over Monica Jun 06 '19 at 09:59
-
from value of $abc$ , how to find $k$? – JinSeok Jun 06 '19 at 10:18
-
@user, my answer to that question starts by finding your $k$. – hmakholm left over Monica Jun 06 '19 at 10:19
1 Answers
1
Hint: From the second equation we get $$b=k-\frac{1}{c}$$ so we get $$a(kc-1)+c=k(kc-1)$$ and wwith $$c=k-\frac{1}{a}$$ and we get $$a(k(k-\frac{1}{a})-1)+k-\frac{1}{a}=k(k(k-\frac{1}{a})-1)$$ simplifying and factorizing we get $$(k-1)(k+1)(a^2-ka+1)=0$$ Can you proceed?
Dr. Sonnhard Graubner
- 95,283
-
-
-
Yes. solved . after problem solved, is there something I have to do? this is my first time MSE. – JinSeok Jun 06 '19 at 10:33
-
-
I tried. but MSE says "Votes cast by those with less than 15 reputation are recorded " .... I just came here and this is the first question. so I have only one reputaion. so I can't vote.. – JinSeok Jun 06 '19 at 10:37
-