Question

Решить системы уравнений(где разделить - это дробь) 1)x+y=4 5/x-3/y=1 2)1/x+1/y=7/12 x+y=7

Answer 1

$\left \{ {{x+y=4} \atop { \frac{5}{x} - \frac{3}{y} =1}} \right.$, $\left \{ {{x=4-y} \atop { \frac{5}{4-y} - \frac{3}{y} =1}} \right.$, $\frac{5y-3y+12}{y*(y-4)} =1$, $\frac{2y+12}{y(y-4)} =1$, $2y+12= y^{2} -4y$, $y^{2}-6y-12=0$, $D=36-4*(-12)=84$, $y_{1} = \frac{6- 2\sqrt{21} }{2} =3- \sqrt{21}$, $y_{2} = \frac{6+2 \sqrt{21} }{2} =3+ \sqrt{21}$, $x_{1} =4-(3- \sqrt{21)} =1+ \sqrt{21}$, $x_{2}=4-(3+ \sqrt{21})=1- \sqrt{21}$.
$\left \{ {{ \frac{1}{x}+ \frac{1}{y}= \frac{7}{12} } \atop {x+y=7}} \right.$, $\left \{ {{ \frac{1}{x}+ \frac{1}{y} = \frac{7}{12} } \atop {x=7-y}} \right.$, $\frac{1}{7-y} + \frac{1}{y} = \frac{7}{12}$, $\frac{y+7-y}{y*(7-y)} = \frac{7}{12}$, $\frac{7}{y*(7-y)} = \frac{7}{12}$, $12=7y- y^{2}$, $y^{2}-7y+12=0$, $D=49-4*12=1$, $y_{1} = \frac{7-1}{2} =3$, $y_{2} = \frac{7+1}{2} =4$, $x_{1} =7-3=4$, $x_{2} =7-4=3$.