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

Solve the following system of equations: ax+by=c, x+ay=1+c

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

Let a,b,c be real numbers with a^2 + b^2 + c^2 =1 . Show that the equation |[ax-by-c,bx+ay,cx+a],[bx+ay,-ax+by-c,cy+b],[cx+a,cy+b,-ax-by+c]|=0 represents a straight line.

Let a,b,c be real numbers with a^2 + b^2 + c^2 =1. Show that the equation |[ax-by-c,bx+ay,cx+a],[bx+ay,-ax+by-c,cy+b],[cx+a,cy+b,-ax-by+c]|=0 represents a straight line.

ax+by=2; bx+ay=3

ax+by=1 bx+ay=2

ax+by=5 bx+ay=3

Equations of the form ax+by=c and bx+ay=d where a!=b

Solve: ax - by = 1 bx + ay = 0

{:(ax + by = 1),(bx + ay = ((a + b)^(2))/(a^(2) + b^(2))-1):}