NIMCET 2009 Question Paper with solutions

Question : 5 of 120

Marks: +1, -0

a+b+c \neq 0

, then the system of equations

(b+c)(y+z)-a x=b-c

(c+a)(z+x)-b y=c-a

(a+b)(x+y)-c z=a-b

has

Solution:

By adding the three. given equations, we get

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

⇒ \;\; x+y+z=0

Here,

x=\;{c-b}/{a+b+c} , y=\;{a-c}/{a+b+c}

and

z=\;{b-a}/{a+b+c} , a+b+c ≠ 0

So, there are infinitely many solutions.

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