75. Solve the system:

$$
\begin{aligned}
& x+y=2(z+u) \\
& x+z=3(y+u) \\
& x+u=4(y+z)
\end{aligned}
$$
75. Solution. Subtracting term by term from the 1st equation $2-\mathrm{e}$, we have:

$$
x-z=2 z+2 u-3 y-3 u, \text{ or } 4 y-3 z=-u
$$

Subtract the 3rd equation from the 1st: $y-u=2 z+2 u-4 y-4 z$, or $5 y+2 z=3 u$. Now we have the system:

$$
\begin{aligned}
4 y+u & =3 z \\
5 y-3 u & =-2 z
\end{aligned}
$$

9)

Equating the coefficients of $z$, we find $23 y-7 u=0$, from which $u=\frac{23}{7} y ;$ if $y=7$, then $u=23$ and $z=17$. From the 1st equation $\boldsymbol{x}=73$.