Question

Помогите решить систему лилейных уравнений
3х+10у=-12
8х+4у=-7

Answer 1

$\left \{ {{3x+10y=-12} \atop {8x+4y=-7}} \right.$

$\left \{ {{3x+10y=-12} \atop {-20x-10y=17.5}} \right.$

$\left \{ {{-17x=5.5} \atop {3x+10y=-12}} \right.$

$\left \{ {{x=5.5:(-17)} \atop {10y=-12-3x}} \right.$

$\left \{ {{x= \frac{55}{10} :(-17)} \atop {10y=-12-3x}} \right.$

$\left \{ {{x= \frac{11}{2} *(- \frac{1}{17}) } \atop {10y=-12-3x}} \right.$

$\left \{ {{x=- \frac{11}{34} } \atop {10y=-12-3*(- \frac{11}{34})$

$\left \{ {{x=- \frac{11}{34} } \atop {10y=-12+ \frac{33}{34}$

$\left \{ {{x=- \frac{11}{34} } \atop {10y=-11 \frac{1}{34}$

$\left \{ {{x=- \frac{11}{34} } \atop {y=- \frac{375}{34}* \frac{1}{10}$

$\left \{ {{x=- \frac{11}{34} } \atop {y=- \frac{375}{340}$

$\left \{ {{x=- \frac{11}{34} } \atop {y=- 1 \frac{7}{68}$

Ответ: $(- \frac{11}{34} ;-1 \frac{7}{68})$