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Question
\\(\
$$\begin{cases} 6x + 3y = 5 \\\\ 2x + y = 1 \\end{cases}$$
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Step1: Multiply the second equation
We multiply the second equation \(2x + y = 1\) by 3 to make the coefficients of \(y\) the same in both equations. So we get \(3\times(2x + y)=3\times1\), which simplifies to \(6x + 3y = 3\).
Step2: Subtract the equations
Now we have the first equation \(6x + 3y = 5\) and the new second equation \(6x + 3y = 3\). If we subtract the second from the first: \((6x + 3y)-(6x + 3y)=5 - 3\). Simplifying the left side gives \(0\) and the right side gives \(2\), so \(0 = 2\), which is a contradiction.
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The system of equations has no solution.