Решить систему уравнений 2u + v = 7 { |u - v| = 2

$\\ \left \{ {{2u+v=7} \atop {|u-v|}=2}} \right. \\ \\ \left \{ {{v=7-2u} \atop {|u-(7-2u)|}=2}} \right. \\ \left \{ {{v=7-2u} \atop {|u-7+2u|}=2}} \right. \\ \left \{ {{v=7-2u} \atop {|3u-7|}=2}} \right. \\ \\ |3u-7|}=2 \\ \ \\ 3u-7=2 \\ 3u=2+7 \\ 3u=9 \\ u= \frac{9}{3} \\ u=3 \\ \\ 3u-7=-2 \\ 3u=-2+7 \\ 3u=5 \\ u= \frac{5}{3} \\ u=1 \frac{2}{3} \\ u \approx1.(6) \\$

$\begin{cases}v=7-2u\\ u_1=3\\ u_2= \frac{5}{3}\end{cases} \\ \begin{cases}v_1=7-2*3\\ v_2=7-2* \frac{5}{3} \\ u_1=3\\ u_2= \frac{5}{3}\end{cases} \\ \begin{cases}v_1=7-6\\ v_2=7- \frac{10}{3} \\ u_1=3\\ u_2= \frac{5}{3}\end{cases} \\ \begin{cases}v_1=1\\ v_2=6 \frac{3}{3} - 3\frac{1}{3} \\ u_1=3\\ u_2= \frac{5}{3}\end{cases} \\ \begin{cases}v_1=1\\ v_2=3 \frac{2}{3} } \\ u_1=3\\ u_2=1 \frac{2}{3}\end{cases} \\ \begin{cases}v_1=1\\ v_2=3 ,(6) \\ u_1=3\\ u_2=1 ,(6)\end{cases} \\$

Ответ: v=1 u=3; v=3.(6) u=1.(6)