Question

Помогите пожалуйста решить системы. В первом должен получиться ответ: (6;2);(-3 1/3;-7 1/3). А во втором: [1/7;2)

Answer 1

$\displaystyle\bf\\1)\\\\\left \{ {{x^{2} +2xy=60} \atop {x-y=4}} \right.\\\\\\\left \{ {{x^{2} +2x\cdot(x-4)-60=0} \atop {y=x-4}} \right. \\\\\\\left \{ {{x^{2} +2x^{2} -8x-60=0} \atop {y=x-4}} \right. \\\\\\\left \{ {{3x^{2} -8x-60=0} \atop {y=x-4}} \right.\\\\\\3x^{2} -8x-60=0\\\\D=(-8)^{2} -4\cdot 3\cdot(-60)=64+720=784=28^{2} \\\\\\x_{1} =\frac{8-28}{6} =\frac{-20}{6} =-3\frac{1}{3} \\\\\\x_{2} =\frac{8+28}{6} =\frac{36}{6}=6\\\\\\y_{1} =-3\frac{1}{3}-4=-7\frac{1}{3}$

$\displaystyle\bf\\y_{2} =6-4=2\\\\\\Otvet \ : \ \Big(-3\frac{1}{3} \ , \ -7\frac{1}{3} \Big) \ , \ \Big(6 \ , \ 2\Big)\\\\\\2)\\\\\left \{ {{2x-\dfrac{x+1}{2}\geq \dfrac{x-1}{3} } \atop {(x+5)(x-5)+37\leq (x-6)^{2} }} \right. \\\\\\\left \{ {{2x\cdot 6-\dfrac{x+1}{2}\cdot 6\geq \dfrac{x-1}{3}\cdot 6 } \atop {x^{2} -5^{2} +37\leq x^{2} -12x+36 }} \right. \\\\\\\left \{ {{12x-(x+1)\cdot 3\geq (x-1)\cdot 2} \atop {12x\leq 36-37+25}} \right.$

$\displaystyle\bf\\\left \{ {{12x-3x-3\geq 2x-2} \atop {12x\leq 24}} \right. \\\\\\\left \{ {{9x-2x\geq -2+3} \atop {x\leq 2}} \right. \\\\\\\left \{ {{7x\geq 1} \atop {x\leq 2}} \right.\\\\\\\left \{ {{x\geq \dfrac{1}{7} } \atop {x\leq 2}} \right.\\\\\\Otvet \ . \ \Big[\frac{1}{7} \ ; \ 2\Big]$