I'm assuming you're referring to linear equations.
Although the linear system of equations $Ax = b$ might not have a solution when the system is overdetermined, you can always find a least-squares solution
$$\min_x \|Ax-b\|^2.$$
If the minimum value is $0$, a minimizer $x$ is a solution to the system. Otherwise, $x$ is the "closest possible" solution, in the sense of minimizing the residual error, to a system that has no solution.
To find a minimizer $x$, you take a derivative and set it equal to 0:
$$A^TAx -A^Tb = 0.$$
The matrix $A^TA$ might be singular, but $A^Tb$ always lies in its column space so this system always has a solution. You can find one such solution by calculating $x = A^{+}b$, where $A^{+}$ is the Moore-Penrose pseudoinverse of $A$.
So to find a solution to your overdetermined system, one approach is to compute(*) the pseudoinverse $A^{+}$, then calculate $x = A^{+}b$. You can check if $Ax=b$ to see if your overdetermined system did in fact have a solution.
(*): My favorite method, in terms of robustness and efficiency, for computing the pseudoinverse is to use the QR decomposition of $A$. The details are beyond the scope of this answer, but worth looking up if you're interested.
By looking carefully at this system we can answer the questions by inspection , hardly doing any arithmetic at all.
$ \ \ x + y = 2 \\
ax + y = b $
The coefficients of y are equal therefore there are two cases.
Case 1
If a = 1 then the left hand sides of the equations are identical. If $b \ne 2 $ then the lines are parallel (not coinciding) , therefore no solution. If b = 2 then the lines (are parallel) coincide , therefore infinite number of solutions.
Case 2
If $ a \ne 1 $ then there is no way to make the left hand sides identical (parallel) because of the y coefficients therefore there will be a unique solution.
Summary
(i) Unique solution for $a \ne 1 $
(ii) No solution for $ a = 1 $ AND $ b \ne 2 $
(iii) Infinite number of solutions for $ a = 1 $ AND $ b = 2 $
Best Answer
It does not have a unique solution, since, as you note, it has 6 unknowns and 4 equations. If every variable can be expressed as a function of $d$, we still have that $d$ can can take on any value in the domain. There would be as many solutions as there are values in the domain: infinitely many. Once $d$ would be determined, all other variables can then be determined.
Clearly, though, the solutions to the system of equations will depend on two free variables, not just $d$, since there are $6$ variables and $4$ equations. That is, four of the variables will be functions of one or both free variables.