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A few hints:

• $Ax+b=y$ can be rewritten as $Ax=y-b$
• $x$ is the first vector of the canonical basis
• the image by $A$ of ith vector of the canonical basis is the ith column of A
  • hence any matrix whose first column is the vector $y-b = y+b = (0,1,1,0,,0,0,0,0)$ does the job

A few hints:

• $Ax+b=y$ can be rewritten as $Ax=y-b$
• $x$ is the first vector of the canonical basis
• the image by $A$ of the ith vector of the canonical basis is the ith column of A
  • hence any matrix whose first column is the vector $y-b = y+b = (0,1,1,0,,0,0,0,0)$ does the job

A few hints:

• $Ax+b=y$ can be rewritten as $Ax=y-b$
• $x$ is the first vector of the canonical basis
• the image by $A$ of the ith vector of the canonical basis is the ith column of A$A$
• hence any matrix whose first column is the vector $y-b = y+b = (0,1,1,0,,0,0,0,0)$ does the job

A few hints:

• $Ax+b=y$ can be rewritten as $Ax=y-b$
• $x$ is the first vector of the canonical basis
• the image by $A$ of the ith vector of the canonical basis is the ith column of $A$
• hence any matrix whose first column is the vector $y-b = y+b = (0,1,1,0,,0,0,0,0)$ does the job

EDIT Let me fix the question in comments:

• if $x$ is a (nonzero) vector, you can easily find an invertible matrix $B$ such that $x=Be0$ (where $e0 = (1,0,0,0,...)$)
• then any matrix $A$ such that $ABe0=y-b$ is a solution to your problem.
• by the previous part of the answer, the set of matrices $M$ such that $Me0=y-b$ is the $d(d-1)$ linear set $S$ of matrices whose first column is $y-b$ (and the other are the $d(d-1)$ free variables).
• then, the set of solutions of your original problem is the set $\{MB^-1 \mid M \in S\}$, which you can write in terms of the $d(d-1)$ free variables.

A few hints:

• $Ax+b=y$ can be rewritten as $Ax=y-b$
• $x$ is the first vector of the canonical basis
• the image by $A$ of the ith vector of the canonical basis is the ith column of $A$
• hence any matrix whose first column is the vector $y-b = y+b = (0,1,1,0,,0,0,0,0)$ does the job

EDIT Let me fix the question in comments:

• if $x$ is a (nonzero) vector, you can easily find an invertible matrix $B$ such that $x=Be0$ (where $e0 = (1,0,0,0,...)$)
• then any matrix $A$ such that $ABe0=y-b$ is a solution to your problem.
• by the previous part of the answer, the set of matrices $M$ such that $Me0=y-b$ is the $d(d-1)$ linear set $S$ of matrices whose first column is $y-b$ (and the other are the $d(d-1)$ free variables).
• then, the set of solutions of your original problem is the set $\{MB^-1 $\{MB^{-1} \mid M \in S\}$, which you can write in terms of the $d(d-1)$ free variables.

If you have problems in turning those hints into Sage code, do not hesitate to tell where you are locked.

A few hints:

• $Ax+b=y$ can be rewritten as $Ax=y-b$
• $x$ is the first vector of the canonical basis
• the image by $A$ of the ith vector of the canonical basis is the ith column of $A$
• hence any matrix whose first column is the vector $y-b = y+b = (0,1,1,0,,0,0,0,0)$ does the job

EDIT Let me fix the question in comments:

• if $x$ is a any (nonzero) vector, you can easily find an invertible matrix $B$ such that $x=Be0$ (where $e0 = (1,0,0,0,...)$)
• then any matrix $A$ such that $ABe0=y-b$ is a solution to your problem.
• by the previous part of the answer, the set of matrices $M$ such that $Me0=y-b$ is the $d(d-1)$ linear set $S$ of matrices whose first column is $y-b$ (and the other are the $d(d-1)$ free variables).
• then, the set of solutions of your original problem is the set $\{MB^{-1} \mid M \in S\}$, which you can write in terms of the $d(d-1)$ free variables.

If you have problems in turning those hints into Sage code, do not hesitate to tell where you are locked.