(a+b)/2=(b+c)/3=(a+c)/7 a:b:c = ?

[Bob_kyt]

(a+b)/2=(b+c)/3=(a+c)/7 a:b:c = ?

The answer is 3 : -1 : 4
Is there a better solution or different way to solve this ?

Thanks / Bob

 

[Dr.Peterson]

Since your answer is correct, there is no better answer.

Since you didn't show how you got it, I have no idea whether my [method of] solution is better than or different from yours.

What was yours? (I myself set the three expressions equal to k, making a system of three linear equations, and solved for a, b, and c in terms of k.)

 

[Bob_kyt]

my friend's method-

set a+b=2k
b+c=3k
a+c=7k
2(a+b+c)=12k
a+b+c=6k
c=4k b=-1k a=3k

Is there a different way to solve this ? just can not used to it.

Thanks.

Bob

 

[MarkFL]

A different approach would be to multiply the original equation by 42 to get:

[MATH]21a+21b=14b+14c=6a+6c[/MATH]
From this, we can generate 3 unique equations:

[MATH]21a+21b=14b+14c\implies 3a+b-2c=0[/MATH]
[MATH]21a+21b=6a+6c\implies 5a+7b-2c=0[/MATH]
[MATH]14b+14c=6a+6c\implies -3a+7b+4c=0[/MATH]
The first two imply:

[MATH]3a+b=5a+7b\implies a+3b=0[/MATH]
The last two imply:

[MATH]2c-5a=3a-4c\implies 4a-3c=0[/MATH]
If we let \(a=1\) then:

[MATH]a:b:c=1:-\frac{1}{3}:\frac{4}{3}[/MATH]
or:

[MATH]a:b:c=3:-1:4[/MATH]

 

[Bob_kyt]

Thanks for the sharing. Bob