Question

Given the system of equations
$(b+c)(y+z)-ax=b-c$
$(c+a)(z+x)-by=c-a$
$(a+b)(x+y)-cz=a-b$
(where $a+b+c\ne 0$); then $x:y:z$ is given by

• A. $c-b:a-c:b-a$
• B. $b+c:c+a:a+b$
• C. $a:b:c$
• D. $\frac{a}{b}:\frac{b}{c}:\frac{c}{a}$

Answer 1

The correct option is A $c-b:a-c:b-a$

$(b+c)(y+z)-ax=b-c$ ..............(1)

$(c+a)(z+x)-by=c-a$ ..............(2)

$(a+b)(x+y)-cz=a-b$ .............(3)

Adding all three equation

$(x+y+z)(a+b+c)=0$

$\Rightarrow x+y+z=0$

$\Rightarrow y+z=-x$

Substituting this in the first (1) equation

$(b+c)(-x)-ax=b-c$

$x=\frac{c-b}{a+b+c}$

Similarly, we get

$y=\frac{a-c}{a+b+c},z=\frac{b-a}{a+b+c}$

Hence $x:y:z=(c-b):(a-c):(b-a)$