Question

Solve the following equations :
$(b+c)(y+z)-ax=b-c,(c+a)(z+x)-by=c-a,(a+b)(x+y)-cz=a-b,$
where $a+b+c\ne 0.$

Answer 1

Adding all the three equations, we get
$(x+y+z)(a+b+c)=0$,
Hence $x+y+z=0$,
Then from first of the given equations and this equation, we get
$(b+c)(-x)-ax=b-c$
or $x=\frac{c-b}{a+b+c}$.
Similarly $y=\frac{a-c}{a+b+c},z=\frac{b-a}{a+b+c}$.