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Find the value of x and y : \[x-3y=3x-1=2x-y\].

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Hint: Let us assume \[x-3y=3x-1=2x-y=k\]. Now let us divide this equation into three parts. Now let us assume \[x-3y=k\] as equation (1). Now let us assume \[3x-1=k\] as equation (2). Now let us assume \[2x-y=k\] as equation (3). Now let us subtract equation (1) and equation (3), then we get a relation between x and y. Let us assume this as equation (4). Now let us subtract equation (1) and equation (2), then we get another relation between x and y. Let us assume this as equation (5). Now let us substitute equation (4) in equation (5), then we get the value of y. Now let us substitute equation (6) in equation (4), then we get the value of x.

Complete step-by-step answer:
Let us assume \[x-3y=3x-1=2x-y=k\].
From this, we can assume
\[\begin{align}
  & x-3y=k.....(1) \\
 & 3x-1=k......(2) \\
 & 2x-y=k.....(3) \\
\end{align}\]
Now let us subtract equation (1) and equation (3), then we get
\[\begin{align}
  & \Rightarrow \left( x-3y \right)-\left( 2x-y \right)=k-k \\
 & \Rightarrow x-3y-2x+y=0 \\
 & \Rightarrow -x-2y=0 \\
 & \Rightarrow x=-2y....(4) \\
\end{align}\]
Now let us subtract equation (1) and equation (2), then we get
\[\begin{align}
  & \Rightarrow \left( x-3y \right)-\left( 3x-1 \right)=k-k \\
 & \Rightarrow x-3x-3y+1=0 \\
 & \Rightarrow -2x-3y+1=0 \\
 & \Rightarrow 2x+3y-1=0....(5) \\
\end{align}\]
Now let us substitute equation (4) in equation (5), then we get
\[\begin{align}
  & \Rightarrow 2\left( -2y \right)+3y-1=0 \\
 & \Rightarrow -4y+3y-1=0 \\
 & \Rightarrow -y-1=0 \\
 & \Rightarrow y=-1....(6) \\
\end{align}\]
Now let us substitute equation (6) in equation (4), then we get
\[\begin{align}
  & \Rightarrow x=-2(-1) \\
 & \Rightarrow x=2.....(7) \\
\end{align}\]
So, from equation (6) and equation (7) it is clear that the value of x and y are 2 and -1.

Note: This problem can be alternatively.
Let us assume \[x-3y=3x-1=2x-y=k\].
From this, we can assume
\[\begin{align}
  & x-3y=k.....(1) \\
 & 3x-1=k......(2) \\
 & 2x-y=k.....(3) \\
\end{align}\]
Now let us subtract equation (1) and equation (2), then we get
\[\begin{align}
  & \Rightarrow \left( x-3y \right)-\left( 3x-1 \right)=k-k \\
 & \Rightarrow x-3x-3y+1=0 \\
 & \Rightarrow -2x-3y+1=0 \\
 & \Rightarrow 2x+3y-1=0....(4) \\
\end{align}\]
Now let us subtract equation (2) and equation (3), then we get
\[\begin{align}
  & \Rightarrow \left( 3x-1 \right)-\left( 2x-y \right)=k-k \\
 & \Rightarrow 3x-1-2x+y=0 \\
 & \Rightarrow x+y-1=0....(5) \\
\end{align}\]
Now let us subtract equation (4) and equation (5), then we get
\[\begin{align}
  & \Rightarrow \left( 2x+3y-1 \right)-\left( x+y-1 \right)=0 \\
 & \Rightarrow 2x+3y-1-x-y+1=0 \\
 & \Rightarrow x+2y=0....(6) \\
\end{align}\]
Now let us subtract equation (5) and equation (6), then we get
\[\begin{align}
  & \Rightarrow \left( x+y-1 \right)-\left( x+2y \right)=0 \\
 & \Rightarrow x+y-1-x-2y=0 \\
 & \Rightarrow -y-1=0 \\
 & \Rightarrow y=-1....(7) \\
\end{align}\]
Now let us substitute equation (7) in equation (6), then we get
\[\begin{align}
  & \Rightarrow x+2\left( -1 \right)=0 \\
 & \Rightarrow x-2=0 \\
 & \Rightarrow x=2.....(8) \\
\end{align}\]
So, from equation (7) and equation (8) it is clear that the value of x and y are 2 and -1.