Сложим эти уравнения:

[tex](x + y)^{2} = 64[/tex]

Тогда:

[tex] |x + y| = 8[/tex]

Verified answer

Ответ:  8 .

Можно выразить сумму (х+у) из каждого уравнения .

[tex]\left\{\begin{array}{l}x\cdot (x+y)=29\\y\cdot (x+y)=35\end{array}\right\ \ \left\{\begin{array}{l}\ \ (x+y)=\dfrac{29}{x}\\(x+y)=\dfrac{35}{y}\end{array}\right\ \ \left\{\begin{array}{l}\dfrac{29}{x}=\dfrac{35}{y}\\y\cdot (x+y)=35\end{array}\right[/tex]  

[tex]\left\{\begin{array}{l}\dfrac{y}{x}=\dfrac{35}{29}\\y\cdot (x+y)=35\end{array}\right\ \ \left\{\begin{array}{l}x=\dfrac{29\cdot y}{35}\\y\cdot (\dfrac{29\, y}{35}+y)=35\end{array}\right\ \ \left\{\begin{array}{l}x=\dfrac{29\cdot y}{35}\\\dfrac{64\, y^2}{35}=35\end{array}\right\ \ \left\{\begin{array}{l}x=\dfrac{29\cdot y}{35}\\\ y^2=\dfrac{35^2}{64}\end{array}\right[/tex]

[tex]\left\{\begin{array}{l}x=\dfrac{29\cdot y}{35}\\\ \ y=\pm \dfrac{35}{8}\end{array}\right\ \ \left\{\begin{array}{l}x=\pm \dfrac{29}{8}\\\ \ y=\pm \dfrac{35}{8}\end{array}\right[/tex]

[tex]|x+y|=\Big|\dfrac{29}{8}+\dfrac{35}{8} \Big|=\Big|\dfrac{64}{8}\Big|=|8|=8\\\\\\ili\ \ \ |x+y|=\Big|-\dfrac{29}{8}-\dfrac{35}{8} \Big|=\Big|-\dfrac{64}{8}\Big|=|-8|=8[/tex]