given that a+b/5 = b+c/6 = c+a/7 = k,
find the ratio a:b:c.
I am sure you meant:
(a+b)/5 = (b+c)/6 = (c+a)7 = k
1st = 2nd: -----> 6a + 6b = 5b + 5c
b = 5c - 6a (#1)
1st and 3rd: ----> 7a+7b = 5a+5c
2a = 5c-7b (#2)
sub #1 into #2
2a = 5c - 7(5c-6a)
c = 4a/3
2nd and 3rd: 6a+6c = 7b+7c
c = 6a-7b (#3)
so 4a/3 = 6a - 7b
4a = 18a - 21b
21b = 14a
b = 2a/3
now..... finally
a : b : c = a : 2a/3 : 4a/3
= 1 : 2/3 : 4/3
= 3 : 2 : 4
thx...=]
To find the ratio a:b:c, we can start by solving the system of equations given by:
a + b/5 = b + c/6 = c + a/7 = k
Let's start by isolating one variable in each equation. We can isolate 'a' in the first equation:
a = k - b/5
Next, let's isolate 'b' in the second equation:
b = 5k - c/6
Finally, let's isolate 'c' in the third equation:
c = 6k - a/7
Now, we can substitute the values obtained for 'a', 'b', and 'c' in terms of 'k' into the other equations to get a system of equations involving only 'k'.
Substituting 'a' in terms of 'k' in the second equation:
5k - c/6 + c/6 = b
5k = b
Substituting 'b' in terms of 'k' in the third equation:
c = 6k - (k - b/5)/7
c = 6k - (k - 5k/5)/7
c = 6k - 4k/7
c = 30k/7
Now, we can substitute these expressions for 'b' and 'c' back into the first equation to get a single equation involving only 'k'.
a + b/5 = k
k - b/5 + b/5 = k
k = k
Since this equation holds true for any value of 'k', we have infinitely many solutions. This means we cannot find a unique ratio a:b:c.