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?

$$\begin{equation*}
\left\{ 
\begin{array}{c}
2.6513=\frac{3}{2}y+\frac{x}{2} \\ 
1.7675=y+\frac{x}{3}
\end{array}
\right. 
\end{equation*}$$
If we multiply the first equation by $2$ and the second by $3$ we get
$$\begin{equation*}
\left\{ 
\begin{array}{c}
5.3026=3y+x \\ 
5.3025=3y+x
\end{array}
\right. 
\end{equation*}$$
This system has no solution because 
$$\begin{equation*}
5.3026\neq 5.3025
\end{equation*}$$
However if the number $2.6513$ resulted from rounding $2.65125$, then the
same computation yields
$$\begin{equation*}
\left\{ 
\begin{array}{c}
5.3025=3y+x \\ 
5.3025=3y+x
\end{array}
\right. 
\end{equation*}$$
which is satisfied by all $x,y$.
A system of the form
$$\begin{equation*}
\begin{pmatrix}
a_{11} & a_{12} \\ 
a_{21} & a_{22}
\end{pmatrix}
\begin{pmatrix}
x \\ 
y
\end{pmatrix}
=
\begin{pmatrix}
b_{1} \\ 
b_{2}
\end{pmatrix}
\end{equation*}$$
has the solution (Cramer's rule)
$$\begin{equation*}
\begin{pmatrix}
x \\ 
y
\end{pmatrix}
=\frac{1}{\det 
\begin{pmatrix}
a_{11} & a_{12} \\ 
a_{21} & a_{22}
\end{pmatrix}
}
\begin{pmatrix}
a_{22}b_{1}-a_{12}b_{2} \\ 
a_{11}b_{2}-a_{21}b_{1}
\end{pmatrix}
=
\begin{pmatrix}
\frac{a_{22}b_{1}-a_{12}b_{2}}{a_{11}a_{22}-a_{21}a_{12}} \\ 
\frac{a_{11}b_{2}-a_{21}b_{1}}{a_{11}a_{22}-a_{21}a_{12}}
\end{pmatrix}
\end{equation*}$$
if $\det 
\begin{pmatrix}
a_{11} & a_{12} \\ 
a_{21} & a_{22}%
\end{pmatrix}%
\neq 0$. 
In the present case, we have
$$\begin{equation*}
\begin{pmatrix}
\frac{1}{2} & \frac{3}{2} \\ 
\frac{1}{3} & 1
\end{pmatrix}
\begin{pmatrix}
x \\ 
y
\end{pmatrix}
=
\begin{pmatrix}
2.6513 \\ 
1.7675
\end{pmatrix}
\end{equation*}$$
and $$\det 
\begin{pmatrix}
\frac{1}{2} & \frac{3}{2} \\ 
\frac{1}{3} & 1
\end{pmatrix}
=0$$
