How to solve?

[Blala]

 

[Subhotosh Khan]

View attachment 15157

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 = ?

 

[Blala]

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 = ?

What is the compleate solution? I can't solve.

 

[Subhotosh Khan]

What is the compleate solution? I can't solve.

You are given:

a*b = b*c = c*a = 4

Now calculate:

(a*b) * (b*c) * (c*a) = ?

 

[Jomo]

What is the compleate solution? I can't solve.

We don't give complete solutions. Rather we help you get the solution.
Try to get a common denominator and show us your work
Do you know what the common denominator would be?

 

[Jomo]

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 \(\displaystyle ac = 4\) 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?

 

[lookagain]

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.

 

[MATHScience Daniel Dallas]

Common Denominator: (bc+ac+ab)/abc=1/2
Next abc=24 . However if You express abc through initial conditions like a=4/b, b=4/c and c=4/a and multiply abc you will have that abc=64/(abc) from where abc=+-8, wich is a contradiction to original statement about abc=24.

Your Youtuber,
Math Science by Daniel Dallas

 

[Subhotosh Khan]

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 \(\displaystyle ac = 4\) 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?

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.

 

[Jomo]

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.

How are you getting abc=8?

 

[Subhotosh Khan]

How are you getting abc=8?

a*b = b*c = a*c = 4

(a*b) * (b*c) * (a*c) = 4 * 4 * 4 = 64

(a * b * c)² = 8²

continue.....