Question

Rozwiaz układ równań:
xy=360
(x-4)(y+3)=360

Answer 1

Xy=360
(x-4)(y+3)=360

xy=360
xy+3x-4y-12=360

xy=360
3x-4y-12=0     /:3

xy=360
x-4/3y-4=0   wyznaczamy x

xy=360
x= 4/3y+4   podstawiamy do pierwszego rownania 

x= 4/3y+4 
(4/3y+4)y=360

x= 4/3y+4 
4/3y^2 + 4y = 360   /-360

x= 4/3y+4 
4/3y^2+4y-360=0    
rozwiazujemy rownanie kwadratowe 
Δ = 4^2 - (-360*4/3*4) = 16 + 1920 = 1936 > 0 wiec dwa rozwiazania
√Δ = √1936 = 44
y₁ = (-b-√Δ)/2a = (-4-44)/2*(4/3) = -48 : 8/3 = -48*3/8 = - 18
y₂ = (-b+√Δ)/2a = (-4+44) / 2*(4/3) = 40 : 8/3 = 40 * 3/8 = 15 
Podstawiamy kolejno do rownania x= 4/3y+4

y₁ = (-b-√Δ)/2a = (-4-44)/2*(4/3) = -48 : 8/3 = -48*3/8 = - 18
x₁ = 4/3*(-18) + 4 

y₁ = (-b-√Δ)/2a = (-4-44)/2*(4/3) = -48 : 8/3 = -48*3/8 = - 18
x₁ = -24 + 4 = -20

Teraz obliczamy dla y₂;
y₂ = (-b+√Δ)/2a = (-4+44) / 2*(4/3) = 40 : 8/3 = 40 * 3/8 = 15 
x₂ = 4/3*15+4 

y₂ = (-b+√Δ)/2a = (-4+44) / 2*(4/3) = 40 : 8/3 = 40 * 3/8 = 15 
x₂ = 20+4

y₂ = (-b+√Δ)/2a = (-4+44) / 2*(4/3) = 40 : 8/3 = 40 * 3/8 = 15 
x₂ = 24 

Rozwiązanie to:
(-20, -18) lub (24, 15)