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

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