Question

Розвяжіть систему рівнянь 5y-x=4 x2+3y2=4

Answer 1

$\left \{ {{5y-x=4} \atop {x^{2}+3y^{2}=4}} \right.\\\\\left \{ {{x=5y-4} \atop {(5y-4)^{2}+3y^{2}=4}} \right.\\\\\left \{ {{x=5y-4} \atop {25y^{2}-40y+16+3y^{2}-4=0}} \right.\\\\\left \{ {{x=5y-4} \atop {28y^{2}-40y+12=0 }} \right. \\\\\left \{ {{x=5y-4} \atop {7y^{2}-10y+3=0 }} \right.$

$7y^{2}-10y+3=0\\\\D=(-10)^{2}-4*7*3=100-84=16=4^{2}\\\\y_{1}=\frac{10+4}{14}=1\\\\y_{2} =\frac{10-4}{14}=\frac{3}{7}\\\\x_{1} =5*1-4=1\\\\x_{2}=5*\frac{3}{7}-4=\frac{15}{7}-4=2\frac{1}{7}-4=-1\frac{6}{7}\\\\Otvet:(1;1),(-1\frac{6}{7};\frac{3}{7} )$

Answer 2

х=5у-4

х²+3у²=4

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(5у-4)²+3у²-4=0

25у²-40у+16+3у²-4=0

28у²-40у+12=0

7у²-10у+3=0

D1=(b/2)²-ac=25-21=2²

у1=(-b/2+√D1)/a=(5+2)/7=1

y2=(5-2)/7=3/7

x1=5y-4=1

x2=15/7-4=2 1/7-4=-1 6/7

Ответ: (1;1) и (-1 6/7; 3/7)