Question

$\left \{ {{x-5y=3} \atop {xy+3y=11}} \right.$Решите систему уравнений

Answer 1

Ответ:

Объяснение:

$\[\left\{\begin{gathered}x-5y=3\hfill\\xy+3y=11\hfill\\\end{gathered}\right.\Leftrightarrow\left\{\begin{gathered}x=5y+3\hfill\\xy+3y=11\hfill\\\end{gathered}\right.\Leftrightarrow\left\{\begin{gathered}x=5y+3\hfill\\(5y+3)\cdot y+3y=11\hfill\\\end{gathered}\right.\Leftrightarrow\]$

$\[\left\{\begin{gathered}x=5y+3\hfill\\5{y^2}+3y+3y-11=0\hfill\\\end{gathered}\right.\Leftrightarrow\left\{\begin{gathered}x=5y+3\hfill\\5{y^2}+6y-11=0\hfill\\\end{gathered}\right.\]$

Решим квадратное уравнение:

$\[\begin{gathered}5{y^2}+6y-11=0\hfill\\D={b^2}-4ac={6^2}-4\cdot 5\cdot (-11)=36+220=256\hfill\\{y_{1;2}}=\frac{{-b\pm\sqrt D}}{{2a}}=\frac{{-6\pm\sqrt {256}}}{{2\cdot 5}}=\frac{{-6\pm16}}{{10}}\hfill\\{y_1}=\frac{{-6+16}}{{10}}=\frac{{10}}{{10}}=1\hfill\\{y_2}=\frac{{-6-16}}{{10}}=\frac{{-22}}{{10}}=-2.2\hfill\\\end{gathered}\]$

Найдем x:

$\[x=5y+3\]$

При y₁ = 1

$\[{x_1}=5\cdot 1+3=8\]$

При y₂ = -2.2

$\[{x_2}=5\cdot (-2.2)+3=-8\]$

Ответ: x₁ = 8; y₁ = 1; x₂ = -8; y₂ = -2.2