NIMCET 2009 — Mathematics PYQ

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

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

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

Adding all three equations:

$[-a+(c+a)+(a+b)]x+[(b+c)-b+(a+b)]y+[(b+c)+(c+a)-c]z=(b-c)+(c-a)+(a-b)$

Simplifying the coefficients:

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

Factoring out $(a+b+c)$:

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

Since we are given that $a+b+c\ne 0$, it must be that:

$x+y+z=0$

Now, let's substitute $y+z=-x$ into the first equation:

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

$-bx-cx-ax=b-c$

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

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

Similarly, substituting $z+x=-y$ into the second equation and $x+y=-z$ into the third:

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

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

Since $a,b,\text{ and }c$ are constants, we get exactly one specific value for $x,y,\text{ and }z$.

Conclusion:

The system has a single, specific set of values for the variables.

Correct Option:

(a) a unique solution