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

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

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