Question

Решите систему уравнений
$\left \{ {{\frac{4}{x+y} +\frac{3}{x-y} =5} \atop {\frac{2}{x+y}+\frac{9}{y-x} =-1}} \right.$

Answer 1

Ответ:

(2; –1)

Объяснение:

Сделаем замену:

$\displaystyle\frac{1}{{x + y}} = u,\ \displaystyle\frac{1}{{x - y}} = v,$

тогда

$\left\{ \begin{array}{l}4u + 3v = 5,\\2u - 9v = - 1.\end{array} \right.$

Домножим первое уравнение на 3 и сложим со вторым:

$14u = 14;\\\\u = 1.$

Тогда из первого уравнения

$3v = 5 - 4u;\\\\3v = 1;\\\\v = \displaystyle\frac{1}{3}.$

Делая обратную замену, получаем:

$\left\{ \begin{array}{l}\displaystyle\frac{1}{{x + y}} = 1,\\\displaystyle\frac{1}{{x - y}} = \displaystyle\frac{1}{3};\end{array} \right.\\\\\left\{ \begin{array}{l}x + y = 1,\\x - y = 3.\end{array} \right.$

Складывая оба уравнения, получаем

$2x = 4;\\\\x = 2,$

тогда из второго уравнения $y = x - 3 = 2 - 3 = - 1.$

Answer 2

$\displaystyle\bf\\\left \{ {{\dfrac{4}{x+y} +\dfrac{3}{x-y} =5} \atop {\dfrac{2}{x+y}+\dfrac{9}{y-x}=-1 }} \right. \\\\\\\left \{ {{\dfrac{4}{x+y} +\dfrac{3}{x-y} =5} \atop {\dfrac{2}{x+y}-\dfrac{9}{x-y}=-1 }} \right. \\\\\\\frac{2}{x+y} =m \ \ ; \ \ \frac{3}{x-y} =n\\\\\\\left \{ {{2m+n=5} \ \atop {m-3n=-1} \ |\cdot(-2)} \right. \\\\\\+\left \{ {{2m+n=5} \atop {-2m+6n=2}} \right.\\ ----------\\7n=7\\\\n=1\\\\2m=5-n=5-1=4\\\\m=2$

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