Question

If the equations $a(y+z)=x,b(z+x)=y$ and $c(x+y)=z$, where $a\ne -1$, $b\ne -1$, $c\ne -1$ admit non-trivial solution, then $(1+a{)}^{-1}+(1+b{)}^{-1}+(1+c{)}^{-1}$ is

• A. $2$
• B. $1$
• C. $\frac{1}{2}$
• D. none of these

Answer 1

The correct option is A $2$

$a(y+z)=x,b(z+x)=y,c(x+y)=z$

$\Rightarrow a=\frac{x}{y+z};b=\frac{y}{z+x};c=\frac{z}{x+y}$

Now, ${(1+a)}^{-1}+{(1+b)}^{-1}+{(1+c)}^{-1}=\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$

$=\frac{1}{1+\frac{x}{y+z}}+\frac{1}{1+\frac{y}{z+x}}+\frac{1}{1+\frac{z}{x+y}}$

$=\frac{y+z}{y+z+x}+\frac{z+x}{z+x+y}+\frac{x+y}{x+y+z}$

$=\frac{2(x+y+z)}{y+z+x}=2$