What is the value of the variables in the equation?

Question

$$x_1 + y_1 = 3$$ $$y_1 - x_3 = -1$$ $$x_3 + y_3 = 7$$ $$x_1 - y_3 = -3$$

Find the values of $x_1$, $x_3$, $y_1$ and $y_3$.

All I am getting is $x_1 + x_3 = 4$ and $y_1 + y_3 = 6$.

Answer 1

Let $x_1=a$, $y_1=b$, $x_3=c$ and $y_3=d$.

Thus, $b=3-a$ and from the second we obtain $c=4-a$, which from the third gives $d=3+a$ and we see that the fourth holds.

Id est, got infinitely many solutions: $$\{(a,3-a,4-a,3+a)\}$$

Answer 2

With this sort of system one should note that it can be represented as a matrix, where the elements are the coefficients for each variable: $$M = \left[\begin{array}{cccccc|c} 1 & 0 & 0 & 1 & 0 & 0 & 3\\ 0 & 0 & -1 & 1 & 0 & 0 & -1\\ 0 & 0 & 1 & 0 & 0 & 1 & 7\\ 1 & 0 & 0 & 0 & 0 & -1 & -3\end{array} \right]$$

Of course, to find a solution (if one exists), one must change the matrix $M$ into reduced row-echelon form, denoted here as $M_{rrf}$, which is as follows: $$M_{rrf} = \left[\begin{array}{cccccc|c} 1 & 0 & 0 & 0 & 0 & 0 & 3 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 6\end{array} \right] \\ \text{this was carried out by the following $\it row$ operations:} \\ -R_1 + R_4 \to R_4 \\ R_2 + R_3 \to R_3 \\R_4 + R_3 \to R_3 \\ -R_4 \to R_4 \\ -R_3 + R_2 \to R_2 \\ -R_3 + R_1 \to R_1 \\ -R_2 \to R_2 \\$$

Thus, since the 3rd row of $M_{rrf}$ is $\left[\begin{array}{cccccc|c} 0 & 0 & 0 & 1 & 0 & 0 & 0\end{array}\right]$, we have that $\text{dim kerM} = 1$ (recall that the dimension of the kernel is defined as the nullity, $\text{null M}$ of the matrix). Since, $null(M) = 1$, the system has infinitely many solutions.

Now, as shown by Michael's answer, we can parametrize our solution to create an expression that generates all of the solutions. Begin by defining the following: $$x_1 = 3 = a \\ x_3 = 1 = b \\ y_1 = 0 = c \\ y_3 = 6 = d$$ which gives us $$ a = a \\ a + c = 3 \\ -b + c = -1 \\ a - d = -3 \\ \text{which simplifies to:} \\ a = a \\ b = c + 1 = 4 - a \\ c = 3 - a \\ d = a + 3 $$

Therefore, the solution set for the system is $ \{a, 4-a, 3-a, 3+a \}$