# Equation system with multiple unknown real numbers

Have been having problems with this equation system for a while, $$\begin{array}{l} x - y - az = 1\\ ax + y + az = a\\ ax + 3y + 3z = -1 \end{array}$$ where I need to find all the values of $$a\in \mathbb{R}$$.

I have tried to solve the system with elimination, by subtracting the first line multiplied by $$(-a)$$ with the second and third line and so on. I have found $$z$$ to be $$\frac{-1 -a}{3+3a}$$ but after integrating it and solving for $$y$$ I'm lost.

Any help is appreciated, thanks in advance!

• Welcome to Maths.SX! Are you sure of the value found for $z$? I obtain $\frac{-1-a}{3\color{red}-3a}$. – Bernard Feb 25 at 20:31
• – mfl Feb 25 at 20:36
• Sorry for a late response. @Bernad Yes your value of z is correct. I missed a minus there thanks :) – Irrumasti Feb 26 at 14:45
• @mfl I usually use matrixes when solving on paper but decided not to use it here as I already have the code used written in a latex document. Thanks anyways though. – Irrumasti Feb 26 at 14:48

Hint: Adding equation 1 and 2, we get $$x(a+1)=a+1$$ so $$(a+1)(x-1)=0$$ Can you proceed?
• If $$x=1$$ so you will get $$z(a-1)=\frac{1}{3}$$ and so $$z=\frac{1}{3(a-1)}$$ for $$a\neq 1$$ – Dr. Sonnhard Graubner Feb 26 at 15:04
• I $$a=-1$$ then there are infinity many solutions. – Dr. Sonnhard Graubner Feb 26 at 15:06