Question

Розв'яжіть систему рівнянь 5y-x=4 x^2+3y^2=4

Answer 1

5y - x = 4
X^2 + 3y^2 = 4
••••••
X = 5y - 4
( 5y - 4 )^2 + 3y^2 = 4
25y^2 - 40y + 16 + 3y^2 - 4 = 0
28y^2 - 40y + 12 = 0
4( 7y^2 - 10y + 3 ) = 0
D = 100 - 84 = 16 = 4^2
y1 = ( 10 + 4 ) : 14 = 1
y2 = ( 10 - 6 ) : 14 = 2/7
X = 5y - 4
X1 = 5•1 - 4 = 1
X2 = 5•( 2/7 ) - 4 = ( 10/7 ) - 4 = 1 3/7 - 3 7/7 = - 2 4/7
ответ ( 1 ; 1 ) ; ( - 2 4/7 ; 2/7 )

Answer 2

$\left[\begin{array}{ccc}5y-x=4,\\x^2+3y^2=4\end{array}\right\left[\begin{array}{ccc}x=5y-4,\\x=\sqrt{4-3y^2}\end{array}\right\\5y-4=\sqrt{4-3y^2}\\25y^2-40y+16=4-3y^2\\7y^2-10y+3=0\\D=100-84=16=4^2\\y_1=\frac{10-4}{14}=\frac{3}{7}\\y_2=\frac{10+4}{14}=1\\\left[\begin{array}{ccc}x_1=5y_1-4\\x_2=5y_2-4\end{array}\right\left[\begin{array}{ccc}x_1=\frac{15}{7}-4=-\frac{13}{7}\\x_2=5-4=1\end{array}\right$

ответ: $(-1\frac{6}{7};\frac{3}{7});(1;1)$