Please solve this ratio question?

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?

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Answer 1 · 2018-03-11T05:27:40

$2/(x + y)$ .

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.