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[Гост]

2x/3-y/5=2
(10x+3y)^2=100

[ammornil]

2x/3-y/5=2
(10x+3y)^2=100

[tex]\begin{array}{|l} \frac{\normalsize{2x}}{\normalsize{3}}-\frac{\normalsize{y}}{\normalsize{5}}=2 \\ (10x+3y)^{2}=100 \end{array} \Leftrightarrow \begin{array}{|l} 10x-3y=30 \\ (10x+3y)^{2}=100 \end{array} \Leftrightarrow \begin{array}{|l} 10x=30+3y \\ (30+3y+3y)^{2}=10^{2} \end{array} \Leftrightarrow \\ \begin{array}{lcl} \\ \phantom{q} \\ \begin{array}{|l} x=3+\frac{\normalsize{3y}}{\normalsize{10}} \\ 30+6y=10 \end{array} & \cup & \begin{array}{|l} x=3+\frac{\normalsize{3y}}{\normalsize{10}} \\ 30+6y=-10 \end{array} \\ \phantom{q} \\ \begin{array}{|l} x=3+\frac{\normalsize{3}}{\normalsize{10}}\cdot{\left( -\frac{10}{3} \right)} \\ y=-\frac{10}{3} \end{array} & \cup & \begin{array}{|l} x=3+\frac{\normalsize{3}}{\normalsize{10}}\cdot{\left( -\frac{20}{3} \right)} \\ y=-\frac{20}{3} \end{array} \\ \phantom{q} \\ \begin{array}{|l} x=2 \\ y=-\frac{10}{3} \end{array} & \cup & \begin{array}{|l} x=1 \\ y=-\frac{20}{3} \end{array} \end{array}[/tex]