How do I prove that ax+by+cx=dx+ey?

Question

x,y,z ∈ ℝ³   and  x+y+3z=0

 

How do I prove ax+by+cz=dx+ey, if d and e are real numbers chosen to fit the equation? Thank you very much if you can help me! 

Answer 1

Since x+y+3z = 0, we can solve for z to get z = (x+y)/(-3).

 

So, ax+by+cz = ax+by-(1/3)(x+y)

 

                     = (a - (1/3))x + (b - 1/3))y 

 

                     = dx + ey, where d = a - (1/3) and e = b - (1/3)