A three-equation system

$\begin{cases} x + 2y + 3z = 10 & ...(1) \\ 3x + 6y + 10z = 31 & ...(2) \\ 5x + 11y + 15z = 52 & ...(3) \end{cases}$

From $(2) - 3 \times (1): \implies z = 1$ .

From $2\times (2) - (3):$

$\begin{aligned} x + y + 5z & = 10 & \small \color{#3D99F6} - 6z \text{ both sides} \\ x + y - z & = 10 - 6{\color{#3D99F6} z} = \boxed 4 & \small \color{#3D99F6} \text{Since }z = 1 \end{aligned}$

Triple equation (1) and subtract it from equation (2).This yields $z = 1$

Quintuple equation (1) and subtract it from equation (3). This yields $y = 2$

Substitute the values of $y$ and $z$ into equation (1). This yields $x = 3$

3 + 2 - 1 = 4

$x+y-z\text{/.}\, \text{Solve}\left[\left( \begin{array}{ccc} 1 & 2 & 3 \\ 3 & 6 & 10 \\ 5 & 11 & 15 \\ \end{array} \right).\{x,y,z\}=\{10,31,52\}\right] \Rightarrow 4$

The inverse of $\left( \begin{array}{ccc} 1 & 2 & 3 \\ 3 & 6 & 10 \\ 5 & 11 & 15 \\ \end{array} \right)$ is $\left( \begin{array}{ccc} 20 & -3 & -2 \\ -5 & 0 & 1 \\ -3 & 1 & 0 \\ \end{array} \right)$ . The matrix product of the inverse with $\begin{array}{c}10 \\ 31 \\ 52 \\ \end{array}$ yields $\begin{array}{c}3 \\ 2 \\ 1 \\ \end{array}$ , which yields the final result of $4$ .