What is the compleate solution? I can't solve.Please follow the rules of posting in this forum, as enunciated at:
READ BEFORE POSTING
Please share your work/thoughts about this assignment.
Hint:
Using the first (set) of equation/s, calculate: a*b*c = ?
You are given:What is the compleate solution? I can't solve.
We don't give complete solutions. Rather we help you get the solution.What is the compleate solution? I can't solve.
This problem is very strange. I am embarrassed that not only can I not do this problem but I do not see any errors in my work.
Here goes: \(\displaystyle \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = \dfrac{4}{abc} + \dfrac{4}{abc} + \dfrac{4}{abc} = \dfrac{12}{abc} = \dfrac{1}{2} \\ so \\abc =24\)
Since \(\displaystyle ab = 4\), we get \(\displaystyle c=6\) But
It is ok with that.
Try to have a look at my solution
Hi! Common Denominator: (bc+ac+ab)/abc=1/2
Next abc=16 and a=4/b so c=4 then if a=4/c then a=1 and b=4. So the answer is 1+4+4=9
Your Youtuber,
Math Science by Daniel Dallas
MATHScience Daniel Dallas, you're incorrect. greg1313 explained how you are wrong in the Math Help Forum about this.
This problem is faulty/impossible.
Strange is correctThis problem is very strange. I am embarrassed that not only can I not do this problem but I do not see any errors in my work.
Here goes: \(\displaystyle \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = \dfrac{4}{abc} + \dfrac{4}{abc} + \dfrac{4}{abc} = \dfrac{12}{abc} = \dfrac{1}{2} \\ so \\abc =24\)
Since \(\displaystyle ab = 4\), we get \(\displaystyle c=6\) But \(\displaystyle ac = 4\) so \(\displaystyle b=6\) (similarly we get \(\displaystyle a=6\)). But this contradicts \(\displaystyle ab =ac=bc = 4\)
2nd method. \(\displaystyle abc=24\) Now \(\displaystyle \dfrac{ab}{bc}=\dfrac{a}{c} = 1\) So \(\displaystyle a=c\) Same argument that \(\displaystyle b=c\) so \(\displaystyle a=b=c\). This yields that \(\displaystyle a=b=c=2\) (assuming positive numbers). Then we get \(\displaystyle \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} =\dfrac{1}{2}\) which is also not true.
Now if a, b and c were all -2, then the sum would not be 1/2. If only one was -2 and the others both 2, than that would contradict that the product of any 2 would be 4.
What is going on here?
How are you getting abc=8?Strange is correct
abc = \(\displaystyle \sqrt{64}\) = 8
Here goes: \(\displaystyle \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \)
\(\displaystyle = \dfrac{bc + ab + ca}{abc}
= \dfrac{3*4}{abc} = \dfrac{12}{8} \ \ne {\frac{1}{2}}\)......................... The given value
So the problem cannot be solved as posted.
a*b = b*c = a*c = 4How are you getting abc=8?