Question

Solve the following equations :
$x+y+z=a+b+c,\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=3,ax+by+cz=bc+ca+ab.$

Answer 1

We write the equations as
$(x-a)+(y-b)+(z-c)=0$
$(\frac{x}{a}-1)+(\frac{y}{b}-1)+(\frac{z}{c}-1)=0$
or $bc(x-a)+ca(y-b)+ab(z-c)=0$
and $ax+by+cz=bc+ca+ab,$
From $\left(1\right)$ and $\left(2\right)$, by cross-multiplication, we have
$\frac{x-a}{ab-ca}=\frac{y-b}{bc-ab}=\frac{z-c}{ca-bc}=k,$ say
Then $x=a+a(b-c)k,y=b+b(c-a)k,$$z=c+c(a-b)k$.
Substituting in $\left(3\right)$, we get
$a^2+b^2+c^2+[a^2(b-c)+b^2(c-a)+c^2(a-b)]k=bc+ca+ab$
or $a^2+b^2+c^2-(b-c)(c-a)(a-b)k=bc+ca+ab$
(Factorizing the coefficient of k)
or $k=\frac{a^2+b^2+c^2-bc-ca-ab}{(b-c)(c-a)(a-b)}$
Hence solution is given by $\left(4\right)$ where $k$ is given by $\left(5\right)$.