Решить систему 1/x+1/y=3/4 и (x^2+y^2)/(x+y)=10/3

Question 1

Question 2

Решите задачу:

$\left \{ {{ \frac{1}{x}+\frac{1}{y}=\frac{3}{4} } \atop { \frac{x^2+y^2}{x+y} = \frac{10}{3} }} \right. \\ \left \{ {{ \frac{x+y}{xy}=\frac{3}{4} } \atop { \frac{x^2+y^2+2xy-2xy}{x+y} = \frac{10}{3} }} \right. \\ \left \{ {{ \frac{xy}{x+y}=\frac{4}{3} } \atop { \frac{(x+y)^2-2xy}{x+y} = \frac{10}{3} }} \right. \\ \left \{ {{ \frac{xy}{x+y}=\frac{4}{3} } \atop { \frac{(x+y)^2}{x+y}-\frac{2xy}{x+y} = \frac{10}{3} }} \right.$
$\left \{ {{ \frac{xy}{x+y}=\frac{4}{3} } \atop { (x+y)-2\frac{xy}{x+y} = \frac{10}{3} }} \right. \\ \left \{ {{ \frac{xy}{x+y}=\frac{4}{3} } \atop { (x+y)-2*\frac{4}{3} = \frac{10}{3} }} \right. \\ \left \{ {{ \frac{xy}{x+y}=\frac{4}{3} } \atop { (x+y)-\frac{8}{3} = \frac{10}{3} }} \right. \\ \left \{ {{ \frac{xy}{x+y}=\frac{4}{3} } \atop { x+y=\frac{10}{3}+\frac{8}{3} }} \right. \\ \left \{ {{ \frac{xy}{x+y}=\frac{4}{3} } \atop { x+y=\frac{18}{3} }} \right.$
$\left \{ {{ \frac{xy}{x+y}=\frac{4}{3} } \atop { x+y=6 }} \right. \\ \left \{ {{ \frac{xy}{6}=\frac{4}{3} } \atop { x+y=6 }} \right. \\ \left \{ {{ xy=\frac{4}{3}*6 } \atop { x+y=6 }} \right. \\ \left \{ {{ xy=8 } \atop { x+y=6 }} \right. \\ \left \{ {{ x(6-x)=8 } \atop { y=6-x }} \right. \\ \left \{ {{ x6-x^2=8 } \atop { y=6-x }} \right. \\ \left \{ {{ x_{1}=2; \ \ x_{2}=4 } \atop { y=6-x }} \right. \\ \left \{ {{ x_{1}=2; \ \ x_{2}=4 } \atop { y_{1}=4; \ \ y_{2}=2 }} \right. \\$