Find the value of $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}$ If $$x+y+z=7$$ and $$\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=\frac{7}{10}$$ Find the value of $$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}$$
I tried but I got nothing
 A: \begin{align*}
\frac{x}{y+z} + \frac{y}{z+x} + \frac{z}{x+y} &= \frac{7-(y+z)}{y+z} + \frac{7-(x+z)}{x+z} + \frac{7-(x+y)}{x+y} \\
&=\left( \frac{7}{y+z} - 1 \right)+ \left(\frac{7}{x+z} - 1 \right) + \left(\frac{7}{x+y} - 1 \right) \\
&=7 \left( \frac{1}{y+z} + \frac{1}{x+z}  +\frac{1}{x+y} \right) - 3\\
&=7 \left(\frac{7}{10} \right) - 3\\
&= \boxed{\frac{19}{10}}
\end{align*}
A: Ever heard of Ravi substitution? If we set $a=y+z,b=x+z,c=x+y$, we have:
$$a+b+c = 14,\qquad \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{7}{10} $$
and we are asked to find:
$$ \sum_{cyc}\frac{b+c-a}{2a} = \sum_{cyc}\left(\frac{a+b+c}{2a}-1\right) = -3+\frac{1}{2}(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).$$
A: Rewrite the final expression you are trying to solve for as: 
$\frac{(x+y+z) - (y+z)}{y+z} + \frac{(x+y+z) - (x+z)}{x+z} + \frac{(x+y+z) - (y+x)}{y+x}$
This is equal to: 
$\frac{(x+y+z) }{y+z} + \frac{(x+y+z) }{x+z} + \frac{(x+y+z) }{y+x} - 3$
Which simplifies to: 
$\frac{7}{y+z} + \frac{7 }{x+z} + \frac{7}{y+x} - 3 = 7 (\frac{1}{y+z} + \frac{1 }{x+z} + \frac{1}{y+x}) - 3$
This is equal to:
$7(\frac{7}{10}) - 3 = \frac{19}{10}$
A: Multiply both sides of $\frac{1}{x+y} + \frac{1}{y+z} + \frac{1}{z+x} = \frac{7}{10}$ by $x+y+z$. Then it becomes $\frac{x+y+z}{x+y} + \frac{x+y+z}{y+z} + \frac{x+y+z}{z+x} = \frac{49}{10}$. Then you can simplify $\frac{x+y+z}{x+y}$ into $1+\frac{z}{x+y}$, and so on, getting
$3+\frac{z}{x+y} + \frac{x}{y+z} + \frac{y}{z+x} = \frac{49}{10}$. Now subtract 3 from both sides.
