Find the values of $x,y$ and $z$ in these equations

Question

I'm stuck with these equations. Can somebody help me with solving it?

If $$(a+b) = \frac{x(y+z)}{x+y+z},\quad (b+c) = \frac{y(x+z)}{x+y+z} ,\quad (a+c) = \frac{z(x+y)}{x+y+z}$$ then find $x,y,z$.

$x,y,z$ in terms of $a,b,c$. Right? – Inceptio Apr 21 '13 at 16:09 — Inceptio, Apr 21 '13 at 16:09

Shaswata · Answer 1 · 2013-04-22T16:43:37.060

5

On adding all the 3 we get,

$$a+b+c=\frac{xy+yz+zx}{x+y+z}$$

Subtracting from the given equations we get $$c=\frac{yz}{x+y+z}$$ $$b=\frac{xz}{x+y+z}$$ $$a=\frac{xy}{x+y+z}$$

$$\implies \frac{y}{x}=\frac{c}{b}$$ $$\implies \frac{z}{x}=\frac{c}{a}$$ $$\implies \frac{z}{y}=\frac{b}{a}$$

$$a=\frac{y}{1+\frac{y}{x}+\frac{z}{x}}$$ $$\implies y = a(1+\frac{y}{x}+\frac{z}{x})=a(1+\frac{c}{b}+\frac{c}{a})=\frac{ab+bc+ca}{b}$$

I guess you can do the same for $x$ and $z$

edited Apr 22 '13 at 16:43

answered Apr 21 '13 at 16:28

Shaswata

Can you illustrate it further more,please. – Grazel Apr 21 '13 at 16:33
1

Nice effort. (+1) – Inceptio Apr 21 '13 at 16:39
@shaswata: Do you mean $c=\frac{yz}{x+y+z}$? – Dejan Govc Apr 21 '13 at 16:42
@shaswata-yes,got it.Thank you so much. @ Dejan Gove- It should be so. – Grazel Apr 21 '13 at 16:51

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