Question

Помогите пожалуйста!!! На завтра надо!!!

Answer 1

...........................

Answer 2

Ответ:

1) $(1;-2)$
2) $(\frac{3}{10};-\frac{6}{5})=(0,3;-1,2)$

Объяснение:

1)
$\left \{ {{\frac{x+3}{4}-\frac{y-4}{8}=1\frac{3}{4} } \atop {\frac{x-4}{6}+\frac{y+2}{9}=-\frac{1}{2}}} \right.\\\left \{ {{\frac{x+3}{4}-\frac{y-4}{8}=\frac{7}{4} } \atop {\frac{x-4}{6}+\frac{y+2}{9}=-\frac{1}{2}}} \right.\\\left \{ {{\frac{8\cdot(x+3)}{4}-\frac{8\cdot(y-4)}{8}=\frac{8\cdot7}{4} } \atop {\frac{18\cdot(x-4)}{6}+\frac{18\cdot(y+2)}{9}=-\frac{18\cdot1}{2}}} \right.$
$\left \{{{2\cdot(x+3)-(y-4)=2\cdot7} \atop {3\cdot(x-4)+2\cdot(y+2)=-9}} \right.\\\left \{{{2x+6-y+4=14} \atop {3x-12+2y+4=-9}} \right.\\\left \{{{y=2x+6+4-14} \atop {3x+2y=-9+12-4}} \right.\\\left \{{{y=2x-4} \atop {3x+2y=-1}} \right.\\3x+2\cdot(2x-4)=-1\\3x+4x-8=-1\\7x=-1+8\\7x=7\\x=1\\y=2x-4=2\cdot1-4=2-4=-2\\\\(1;-2)$

2)
$\left \{ {{(x-4)^{2}-(x+4)^{2}=4y} \atop {(y+2)^{2}-(y+1)^{2}=2x}} \right. \\\left \{ {{x^{2}-8x+16-(x^{2}+8x+16)=4y} \atop {y^{2}+4y+4-(y^{2}+2y+1)=2x}} \right.\\\left \{ {{x^{2}-8x+16-x^{2}-8x-16=4y} \atop {y^{2}+4y+4-y^{2}-2y-1=2x}} \right.\\\left \{ {{-8x-8x=4y} \atop {4y+4-2y-1=2x}} \right.\\\left \{ {{-16x=4y} \atop {2y+3=2x}} \right.\\\left \{ {{-4x=y} \atop {2y+3=2x}} \right.\\2\cdot(-4x)+3=2x\\-8x+3=2x\\-8x-2x=-3\\-10x=-3\\x=\frac{-3}{-10}\\x=\frac{3}{10}=0,3$
$y=-4x=-4\cdot\frac{3}{10}=-\frac{4\cdot3}{10}=-\frac{12}{10}=-\frac{6}{5}=-1,2\\\\(\frac{3}{10};-\frac{6}{5})=(0,3;-1,2)$