Question

Please solve this ratio question?

If `(a+b)/(xa+yb)=(b+c)/(xb+yc)=(c+a)/(xc+ya)` where `x+y!=0` and `a+b+c!=0`. Then each of the ratios is equal to?

Answer 1

`2/(x + y)`.

Explanation:

Use The Addendo Process.

If we have `a/b = c/d = e/f`... and so on,

Each ratio will be equal to `(a + c + e.......)/(b + d + f........)`

So, We have,

`(a + b)/(xa + yb) = (b + c)/(xb + yc) = (c + a)/(xc + ya)`

Using Addendo,

Each ratio = `(a + b + b + c + c + a)/(xa + yb + xb + yc + xc + ya)`

`= (2(a + b + c))/(xa + xb + xc + ya + yb + yc)`

`= (2(a + b + c))/(x(a + b + c) + y (a + b + c))`

`= (2cancel((a + b + c)))/(cancel((a + b + c))(x + y)) = 2/(x + y)`

Hence Explained, Hope This helps.