**Solution:**When a system of equations has infinite solutions this does not guarantee that any point of the plane is a solution, we can verify this fact by taking the arbitrary point $(2 ,1)$ which does not satisfy any of the equations, for this reason it can not be a system solution. Now, let's verify that the previous system of equations has infinite number of solutions, note that if we multiply the first equation by 3, we get the second equation.

$(3) \ast (x +2y =1)$ $ \Longrightarrow 3x +6y =3$

This says that the two equations are equivalent, we solve one of the two equations.

$x +2y =1 \Longrightarrow x =1 -2y$

We can affirm that the points of the plane that satisfy this relation are solutions of the system (if we work with the second equation we arrive at the same result). All the points $(\mathbf{x} ,\mathbf{y}) =(1 -2\mathbf{y} ,\mathbf{y})$ that have this form are solutions to the first equation (and also of the second because they are equivalent)

By assigning any real value to $y$, we obtain the corresponding value of $x$, and so we have a solution, we can give infinite values of $y$ and thus we get the value of $x$, there are clearly infinite solutions.

Example

$y =1 \Longrightarrow x = -1$ So the point $( -1 ,1)$ is a solution to the system of equations.

$y =2 \Longrightarrow x = -3$ So the point $( -3 ,2)$ is a solution to the system of equations.

Clearly the system has infinite solutions, but before that any arbitrary point like the point $(2 ,1)$, is not a solution to the system.