# How to find solution of these equations?

$$\begin{cases}x_1+y_1=a\\ x_1-y_2=b\\ y_1-x_2=c\\ x_2+y_2=d \end{cases}$$

How to find a solution for these equations because from elimination and substitution method I end up with 2 equations with totally same variable part on L.H.S like below:

$$\begin{cases}y_1+y_1=A\\ y_1+y_2=B\end{cases}$$

• So, it must be $A=B$ and there are infinitely many solutions.
– mfl
Feb 10, 2019 at 18:03
• Label the equations (1),(2),(3),(4). Then (1)-(2) gives $y_1+y_2=a-b$ and (3)+(4) gives $y_1+y_2=c+d$. If $a-b\ne c+d$, then the equations are inconsistent and have no solutions. If $a-b=c+d$, then one equation is redundant and you have infinitely many solutions. Feb 10, 2019 at 18:30

Writing the augmented matrix of this linear system in reduced row echelon form, you obtain that either there are no solutions, or the solutions make an affine line in $$K^4$$. I ordered the unknowns as $$(x_1,x_2,y_1,y_2)$$:
\begin{align} &\left[\begin{array}{cccc|c} 1&0&1&0&a \\ 1&0&0&-1&b\\ 0&-1& 1&0&c \\ 0&1&0&1&d \end{array}\right]\rightsquigarrow \left[\begin{array}{cccc|c} 1&0&1&0&a \\ 0&1&0&1&d \\ 1&0&0&-1&b\\ 0&-1& 1&0&c \\ \end{array}\right]\rightsquigarrow \left[\begin{array}{cccc|c} 1&0&1&0&a \\ 0&1&0&1&d \\ 0&0&-1&-1&b-a\\ 0&-1& 1&0&c \\ \end{array}\right]\rightsquigarrow \\ \rightsquigarrow&\left[\begin{array}{cccc|c} 1&0&1&0&a \\ 0&1&0&1&d \\ 0&0&-1&-1&b-a\\ 0&0& 1&1&c+d \\ \end{array}\right]\rightsquigarrow \left[\begin{array}{cccc|c} 1&0&0&-1&b \\ 0&1&0&1&d \\ 0&0&1&1&a-b\\ 0&0& 0&0&-a+b+c+d \\ \end{array}\right]. \end{align}