Question

сочно даю 35 баллов решить способом подстановки


$\left \{ {x+y=3} \atop {x+3y=7}} \right.\\\lef \{ {{x+4y=5}\atop {x-2y=-3$

Answer 1

$\displaystyle\bf\\1)\\\\\left \{ {{x+y=3} \atop {x+3y=7}} \right. \\\\\\\left \{ {{x=3-y} \atop {3-y+3y=7}} \right. \\\\\\\left \{ {{x=3-y} \atop {2y=4}} \right. \\\\\\\left \{ {{x=3-2} \atop {y=2}} \right. \\\\\\\left \{ {{x=1} \atop {y=2}} \right. \\\\\\Otvet: \ (1 \ ; \ 2)$

$\displaystyle\bf\\2)\\\\\left \{ {{x+4y=5} \atop {x-2y=-3}} \right. \\\\\\\left \{ {{x=5-4y} \atop {5-4y-2y=-3}} \right. \\\\\\\left \{ {{x=5-4y} \atop {-6y=-8}} \right. \\\\\\\left \{ {{x=5-4\cdot\dfrac{4}{3} } \atop {y=1\dfrac{1}{3} }} \right. \\\\\\\left \{ {{x=5-5\dfrac{1}{3} } \atop {y=1\dfrac{1}{3} }} \right. \\\\\\\left \{ {{x=-\dfrac{1}{3} } \atop {y=1\dfrac{1}{3} }} \right. \\\\\\Otvet: \ \Big(-\frac{1}{3} \ ; \ 1\frac{1}{3} \Big)$