Question

Реши систему уравнений:
⎧⎩⎨14x−8y+x=2x4x−8y=−15
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Answer 1

$\left\{\begin{array}{ccc}\dfrac1{4x-8y}+x=2\\\dfrac x{4x-8y}=-15\end{array}\right.\\t=\dfrac1{4x-8y},~t\neq0\\\left\{\begin{array}{ccc}t+x=2\\tx=-15\end{array}\right.\\\left\{\begin{array}{ccc}t+x=2\\x=-\dfrac{15}t\end{array}\right.\\\left\{\begin{array}{ccc}t-\dfrac{15}t=2\\x=-\dfrac{15}t\end{array}\right.\\\left\{\begin{array}{ccc}t-2t-15=0\\x=-\dfrac{15}t\end{array}\right.\\\left\{\begin{array}{ccc}(t-1)^2=16\\x=-\dfrac{15}t\end{array}\right.$

$\left\{\begin{array}{ccc}t=1\pm4\\x=-\dfrac{15}t\end{array}\right.\\\left[\begin{array}{ccc}\left\{\begin{array}{ccc}t=-3\\x=5\end{array}\right.\\\left\{\begin{array}{ccc}t=5\\x=-3\end{array}\right.\end{array}\right.\\\left[\begin{array}{ccc}\left\{\begin{array}{ccc}\dfrac1{4(x-2y)}=-3\\x=5\end{array}\right.\\\left\{\begin{array}{ccc}\dfrac{1}{4(x-2y)}=5\\x=-3\end{array}\right.\end{array}\right.$

$\left[\begin{array}{ccc}\left\{\begin{array}{ccc}x-2y=-\dfrac1{12}\\x=5\end{array}\right.\\\left\{\begin{array}{ccc}x-2y=\dfrac1{20}\\x=-3\end{array}\right.\end{array}\right.\\\left[\begin{array}{ccc}\left\{\begin{array}{ccc}x=5\\y=2\dfrac{13}{24}\end{array}\right.\\\left\{\begin{array}{ccc}x=-3\\y=-1\dfrac{21}{40}\end{array}\right.\end{array}\right.$

$\left\{\begin{array}{ccc}x=-\fbox3\\y=-\fbox 1\dfrac{\fbox{21}}{\fbox{40}}\end{array}\right.$     $\left\{\begin{array}{ccc}x=\fbox5\\y=\fbox 2\dfrac{\fbox{13}}{\fbox{24}}\end{array}\right.$