Gaussian Elimination/Linear Algebra Problem?

Question

can someone check to make sure that I am finding the variable "a" correctly? I think b is correct, but I think a should be (2y-x)/3, whereas when I simplified it, I keep on getting (x+y)/3. Any help would be greatly appreciated. THANK YOU so much!

here is the problem

Here is my work

Answer 1

In your line where you have $$\left[\begin{array}{ c c | c} 1 & 2 & x\\ 0 & 2 & 2\left(\dfrac{y-2x}{-3}\right) \end{array}\right]$$ , you only add $-\left(\dfrac{y-2x}{-3}\right)$ to row $1$ instead of $-2\left(\dfrac{y-2x}{-3}\right)$. It should be $$\left[\begin{array}{ c c | c} 1 & 2 & x\\ 0 & 2 & 2\left(\dfrac{y-2x}{-3}\right) \end{array}\right]\to \left[\begin{array}{ c c | c} 1 & 0 & x-2\left(\dfrac{y-2x}{-3}\right)\\ 0 & 2 & 2\left(\dfrac{y-2x}{-3}\right) \end{array}\right]$$ $$\to \left[\begin{array}{ c c | c} 1 & 0 & \frac{-3x}{-3}+\left(\dfrac{-2y+4x}{-3}\right)\\ 0 & 1 & \left(\dfrac{y-2x}{-3}\right) \end{array}\right]$$ So, $a=\frac{-3x}{-3}+\left(\dfrac{-2y+4x}{-3}\right)=\dfrac{x-2y}{-3}=\boxed{\dfrac{2y-x}{3}}$