Question

решить способом подстановки​

Answer 1

$2)\left \{ {{2x+3y=7} \atop {3x-y=10}} \right. \\\\\left \{ {{2x+3*(3x-10)=7} \atop {y=3x-10}} \right. \\\\\left \{ {{2x+9x-30=7} \atop {y=3x-10}} \right.\\\\\left \{ {{11x=37} \atop {y=3x-10}} \right. \\\\\left \{ {{x=3\frac{4}{11} } \atop {y=3*3\frac{4}{11}-10 }} \right.\\\\\left \{ {{x=3\frac{4}{11} } \atop {y=10\frac{1}{11 }-10} \right. \\\\\left \{ {{x=3\frac{4}{11} } \atop {y=\frac{1}{11} }} \right. \\\\Otvet:\boxed{(3\frac{4}{11} \ ; \ \frac{1}{11})}$

$3)\left \{ {{6(x-2y)=7-9y} \atop {8x+3y=5(2x+1)}} \right.\\\\\left \{ {{6x-12y=7-9y} \atop {8x+3y=10x+5}} \right.\\\\\left \{ {{6x-12y+9y=7} \atop {8x+3y-10x=5}} \right. \\\\\left \{ {{6x-3y=7} \atop {-2x+3y=5}} \right.\\\\\left \{ {{6x-3y=7} \atop {2x=3y-5}} \right.\\\\\left \{ {{6*(1,5y-2,5)-3y=7} \atop {x=1,5y-2,5}} \right.\\\\\left \{ {{9y-15-3y=7} \atop {x=1,5y-2,5}} \right. \\\\\left \{ {{6y=22} \atop {x=1,5y-2,5}} \right.$

$\left \{ {{y=3\frac{2}{3} } \atop {x=\frac{3}{2}*\frac{11}{3}-\frac{5}{2}}} \right.\\\\\left \{ {{y=3\frac{2}{3} } \atop {x=\frac{33}{6}-\frac{15}{6}}} \right.\\\\\left \{ {{y=3\frac{2}{3} } \atop {x=\frac{18}{6} }} \right.\\\\\left \{ {{y=3\frac{2}{3} } \atop {x=3}} \right.\\\\Otvet:\boxed{(3 \ ; \ 3\frac{2}{3})}$