Question

A2=by +cz, b2=cz+ax,c2=ax+by

Answer 1

Answer:

Solution:

$x=\frac{b^2+c^2-a^2}{2a}$

$y=\frac{a^2-b^2+c^2}{2b}$

$z=\frac{a^2+b^2-c^2}{2c}$

Step-by-step explanation:

The given system of equations is solved by elimination method.

The system of equations is

$by+cz=a^2$...................(1)

$cz+ax=b^2$...................(2)

$ax+by=c^2$...................(3)

Adding these equations, we get

$2ax+2by+2cz=a^2+b^2+c^2$

$2(ax+by+cz)=a^2+b^2+c^2$

$ax+by+cz=\frac{a^2+b^2+c^2}{2}$....................(4)

(4)-(1) ⇒

$ax=\frac{a^2+b^2+c^2}{2}-a^2$

$ax=\frac{a^2+b^2+c^2-2a^2}{2}$

$ax=\frac{b^2+c^2-a^2}{2}$

$x=\frac{b^2+c^2-a^2}{2a}$

(4)-(2) ⇒

$by=\frac{a^2+b^2+c^2}{2}-b^2$

$by=\frac{a^2+b^2+c^2-2b^2}{2}$

$by=\frac{a^2-b^2+c^2}{2}$

$y=\frac{a^2-b^2+c^2}{2b}$

(4)-(3) ⇒

$cz=\frac{a^2+b^2+c^2}{2}-c^2$

$cz=\frac{a^2+b^2+c^2-2c^2}{2}$

$cz=\frac{a^2+b^2-c^2}{2}$

$z=\frac{a^2+b^2-c^2}{2c}$

Solution:

$x=\frac{b^2+c^2-a^2}{2a}$

$y=\frac{a^2-b^2+c^2}{2b}$

$z=\frac{a^2+b^2-c^2}{2c}$