Question

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

Answer 1

Ответ:

Объяснение:

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

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

Решим квадратное уравнение:

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

Найдем x:

$[x=5y+3]$

При y₁ = 1

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

При y₂ = -2.2

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

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