[ax+by=c],[bx+az=1+c]

Solve by any method: ax+by=c, bx+ay=1+c

ax + by = c bx+ ay =1 +c

Prove the following: |[ax-by-cz,ay+bx,az+cx],[bx+ay,by-cz-ax,bz+cy],[cx+az,ay+bz,cz-ax-by]| = (a^2+b^2+c^2)(ax+by+cz)(x^2+y^2+z^2)

Solve the following linear equations: ax + by = c bx + ay = 1+c

Solve: ax+by=c,quad bx+ay=1+c

If a gt b gt c and the system of equtions ax +by +cz =0 , bx +cy+az=0 , cx+ay+bz=0 has a non-trivial solution then both the roots of the quadratic equation at^(2)+bt+c are

If a gt b gt c and the system of equtions ax +by +cz =0 , bx +cy+az=0 , cx+ay+bz=0 has a non-trivial solution then both the roots of the quadratic equation at^(2)+bt+c are

If a,b,c in R and a+b+c=0 and th esystem of equations ax+by+cz=0,bx+cy+az=0,cx+ay+bz=0 has a non-zero solution,then a:b:c is given by

If A=ax+by+cz , B=bx+cy+az , C=cx+ay+bz and a+b+c=0 , then prove that A^3+B^3+C^3=3ABC