Solving linear system of equations when one variable cancels I have the following linear system of equations with two unknown variables $x$ and $y$.  There are two equations and two unknowns.  However, when the second equation is solved for $y$ and substituted into the first equation, the $x$ cancels.  Is there a way of re-writing this system or re-writing the problem so that I can solve for $x$ and $y$ using linear algebra or another type of numerical method?
$2.6513 = \frac{3}{2}y + \frac{x}{2}$
$1.7675 = y + \frac{x}{3}$
In the two equations above, $x=3$ and $y=0.7675$, but I want to solve for $x$ and $y$, given the system above.
If I subtract the second equation from the first, then:
$2.6513 - 1.7675 = \frac{3}{2}y - y + \frac{x}{2} - \frac{x}{3}$
Can the equation in this alternate form be useful in solving for $x$ and $y$?  Is there another procedure that I can use?
In this alternate form, would it be possible to limit $x$ and $y$ in some way so that a solution for $x$ and $y$ can be found by numerical optimization?

It appears that the system of equations is linearly dependent, since $\det(A) = 0$ using Cramer's rule and one equation can be transformed into the other by multiplication.  Since both equations are the same line, there is no intersection.
Perhaps there is a way to limit $x$ and $y$ so that some sort of optimization procedure could be used to determine $x$ and $y$.
