Question

If $x=by+cz+du,y=cz+ax+du,z=ax+by+du,u=ax+by+cz.$ Then the value of the expression.
$\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d}$ is

• A. $2$
• B. $\frac{1}{2}$
• C. $1$
• D. $3$

Answer 1

The correct option is C $1$

Given

$x=by+cz+du....\left(1\right)$
$y=ax+cz+du....\left(2\right)z=ax+by+cz....\left(3\right)u=ax+by+cz....\left(4\right)$

adding all the equation we get

$x+y+z+u=3ax+3by+3cy+3du$

comparing the coefficient of both the sides we get

$a=b=c=d=\frac{1}{3}$

$\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}+\frac{d}{d+1}=\frac{\frac{1}{3}}{\frac{4}{3}}+\frac{\frac{1}{3}}{\frac{4}{3}}+\frac{\frac{1}{3}}{\frac{4}{3}}+\frac{\frac{1}{3}}{\frac{4}{3}}=4\times \frac{1}{4}=1$

so C is the correct option