A) \[\frac{c+b}{a+b+c}\]
B) \[\frac{c-b}{a+b+c}\]
C) \[\frac{a-b}{a+b+c}\]
D) \[\frac{a+b}{a+b+c}\]
E) \[\frac{b-c}{a+b+c}\]
Correct Answer: B
Solution :
The given system of equation is \[(b+c)(y+z)-ax=b-c\] \[(c+a)(z+x)-by=c-a\] \[(a+b)(x+y)-cz=a-b,\] where \[a+b+c\ne 0\] \[\Rightarrow \] \[-ax+(b+c)y+(b+c)z=b-c\] ...(i) \[(a+c)x-by+(a+c)z=c-a\] ...(ii) \[(a+b)x+(a+b)y-cz=a-b\] ...(iii) On adding Eqs. (i),(ii) and (iii), \[(a+b+c)x+(a+b+c)y+(a+b+c)\text{ }z=0\] \[\Rightarrow \] \[(x+y+z)(a+b+c)=0\] \[\because \] \[a+b+c\ne 0\] \[\Rightarrow \] \[x+y+z=0\] ...(iv) Multiply by (b + c) in Eq. (iv), then subtract them from Eq. (i), \[-ax-(b+c)x=b-c\] \[-(a+b+c)x=-(c-b)\] \[\Rightarrow \] \[x=\frac{c-b}{a+b+c}\]
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