Question

Реши систему уравнений.

Answer 1

3x + y = a         x + 2y = b

$\left \{ {{a*b=\frac{238}{25}} \atop {\frac{a}{b}=\frac{17}{14}} \right. \\\\\left \{ {{a=\frac{17}{14}b } \atop {\frac{17}{14}b*b=\frac{238}{25}} \right. \\\\\left \{ {{a=\frac{17}{14}b } \atop {b^{2}=\frac{196}{25} }} \right.\\\\\left \{ {{a=\frac{17}{14}b } \atop {\left[\begin{array}{ccc}b_{1}=\frac{14}{5} \\b_{2}=-\frac{14}{5}\end{array}\right }} \right.$

$\left[\begin{array}{ccc}\left \{ {{a=\frac{17}{5} } \atop {b=\frac{14}{5} }} \right. \\\left \{ {{a=-\frac{17}{5} } \atop {b=-\frac{14}{5} }} \right. \end{array}\right \\\\\\\left[\begin{array}{ccc}\left \{ {{a=3,4 } \atop {b=2,8 }} \right. \\\left \{ {{a=-3,4 } \atop {b=-2,8 }} \right. \end{array}\right\\\\\\1)\left \{ {{3x+y=3,4} \atop {x+2y=2,8}} \right.\\\\+\left \{ {{3x+y=3,4} \atop {-0,5x-y=-1,4}} \right.\\--------\\2,5x=2\\\\x_{1} =0,8\\\\y_{1} =3,4-3*0,8=3,4-2,4=1$

$2)\left \{ {3x+y=-3,4} \atop {x+2y=-2,8}} \right.\\\\\left \{ {{3x+y=-3,4} \atop {-0,5-x=1,4}} \right. \\------\\2,5x=-2\\\\x_{2}=-0,8\\\\y_{2}=-3,4 -3*(-0,8)=-3,4+2,4=-1\\\\Otvet:\boxed{(x_{1}=\frac{4}{5};y_{1}=1) \ , \ (x_{2}=-\frac{4}{5};y_{2}=-1)}$