If $a$ and $b$ are given integers, prove that the systems of equations, $x + y + 2z + 2t = a$ and $2x - 2y + z - t = b$ has a solution in integers $x, y, z, t$.
I tried adding both equations and got $3x-y+3z-t=a+b$ but I couldn't figure out how to prove this.
Including the RHS of the reduced system we get \begin{equation} \begin{bmatrix} 1 & 1 & 2 & 2 & a \\ 0 & -4 & -3 & -5 & b-2a\\ \end{bmatrix} \end{equation} So $b-2a+3z+5t$ must be divisible by 4. Since $3 \cdot 2+5(-1)=1$ we find $z=2k, t=-k$ such that $b-2a+3z+5t=4u$ for some integer $u$. With $y=-u$ the second equation of the reduced system is satisfied. Finally $x$ may be calculated from the first one. Then the solution has only integer components.
Write it in matrix form and show that it has full row rank, which implies that there exists at least one solution. The coefficient matrix is as follows:
\begin{equation} \begin{bmatrix} 1 & 1 & 2 & 2 \\ 2 & -2 & 1 & -1 \\ \end{bmatrix} \end{equation}
We can row reduce it once and find the following
\begin{equation} \begin{bmatrix} 1 & 1 & 2 & 2 \\ 0 & -4 & -3 & -5 \\ \end{bmatrix} \end{equation}
We see that both $1$ and $-4$ are pivots, meaning that this $2\times 4$ matrix has 2 pivots (full row rank). From here, it is a theorem that full row rank would imply existence of solution(s). Proof can be found in any typical linear algebra textbook, or this question.
See this wiki link for more about ranks.
