Question

решить систему уравнений очень нужно xy/x-y=4, x+y=15

Answer 1

Відповідь:

( - 5 ; 20) или (12 ; 3)

Пояснення:

$\left \{ {{\frac{xy}{x-y}=4 } \atop {x+y=15}} \right. \\\left \{ {{\frac{xy}{x-y}=4 } \atop {x=15-y}} \right.\\\left \{ {{\frac{(15-y)y}{15-y-y}=4 } \atop {x=15-y}} \right.\\\left \{ {{\frac{15y-y^{2} }{15-2y}=4 } \atop {x=15-y}} \right.\\$

$\frac{15y-y^{2} }{15-2y}=4 \\\frac{15y-y^{2} }{15-2y}=\frac{4}{1} \\(15y-y^{2} )*1=(15-2y)*4\\15y-y^{2} =60-8y\\15y-y^{2} -60+8y=0\\-y^{2} +23y-60=0\\a=-1\\b=23\\c=-60\\D=b^{2} -4ac=23^{2} -4*(-1)*(-60)=529-240=2890\\y1=\frac{-b-\sqrt{D} }{2a} =\frac{-23-\sqrt{289} }{2*(-1)} =\frac{-23-17 }{-2}=\frac{-40 }{-2}=20\\y2=\frac{-b+\sqrt{D} }{2a} =\frac{-23+\sqrt{289} }{2*(-1)} =\frac{-23+17 }{-2}=\frac{-6 }{-2}=3$

$x=15-y\\x1=15-y1=15-20=-5\\x2=15-y2=15-3=12$