Sistem: x na 2 + xy + y na 2 = 4 i x + xy + y = 2

Iz druge jednačine izraziti $x.$

x+xy+y=2 \implies x=\frac{2-y} {1+y}

Uvrstiti $x$ u prvu jednačinu.

\bigg(\frac{2-y} {1+y}\bigg)^2+\frac{2-y} {1+y}\cdot y+y^2=4 \\ \frac {4-4y+y^2} {(1+y)^2}+\frac {2y-y^2} {1+y}+y^2-4=0

Svesti na zajednički imenilac i srediti izraz.

\frac {4-4y+y^2} {(1+y)^2}+\frac {(2y-y^2)(1+y)} {(1+y)^2}+\frac{y^2(1+y)^2}{(1+y)^2}-\frac{4(1+y)^2} {(1+y)^2}=0 \\ \frac {4-4y+y^2+(2y-y^2)(1+y)+y^2(1+y)^2-4(1+y)^2} {(1+y)^2}=0 \\ \frac {4-4y+y^2+2y+2y^2-y^2-y^3+y^2(1+2y+y^2)-4(1+2y+y^2)} {(1+y)^2}=0 \\ \frac {4-4y+y^2+2y+2y^2-y^2-y^3+y^2+2y^3+y^4-4-8y-4y^2} {(1+y)^2}=0 \\ \frac {-10y-y^2+y^3+y^4} {(1+y)^2}=0 \quad\land\quad y\not=-1

Da bi izraz bio jednak $0,$ brojilac mora biti jednak $0.$ Jednačina se svodi na:

-10y-y^2+y^3+y^4=0 \\ -y(10+y-y^2-y^3)=0 \\ -y(-y^3+2y^2-3y^2+6y-5y+10)=0 \\ -y(-y^2(y-2)-3y(y-2)-5(y-2))=0 \\ -y(-(y-2)(y^2+3y+5))=0\\ y(y-2)(y^2+3y+5)=0

Rešenja za $y$ su:

y=0 \quad\lor\quad y-2=0 \quad\lor\quad y^2+3y+5=0 \\ y_1=0 \quad\lor\quad y_2=2 \quad\lor\quad y_3=\frac {-3+ i\sqrt{11}} {2} \quad\lor\quad y_4=\frac {-3-i\sqrt{11}} {2}

Uvrstiti $y_1=0,$ $y_2=2,$ $y_3=\frac {-3+ i\sqrt{11}} {2}$ i $y_4=\frac {-3-i\sqrt{11}} {2}$ u jednačinu $x=\frac{2-y} {1+y}.$

x_1=\frac{2-0} {1+0}=\frac 21=2 \\ x_2=\frac{2-2} {1+2}=\frac03=0 \\ x_3=\frac{2-\frac {-3+ i\sqrt{11}} {2}} {1+\frac {-3+ i\sqrt{11}} {2}}=\frac{\frac {4+3- i\sqrt{11}} {2}} {\frac {2-3+ i\sqrt{11}} {2}}=\frac{7- i\sqrt{11}} {-1+ i\sqrt{11}}=\frac {-3-i\sqrt{11}} 2 \\ x_4=\frac{2-\frac {-3- i\sqrt{11}} {2}} {1+\frac {-3- i\sqrt{11}} {2}}=\frac{\frac {4+3+ i\sqrt{11}} {2}} {\frac {2-3- i\sqrt{11}} {2}}=\frac{7+i\sqrt{11}} {-1- i\sqrt{11}}=\frac {-3+i\sqrt{11}} 2

Rešenje sistema je skup uređenih parova:

(2,0), \ (0,2) , \bigg(\frac {-3-i\sqrt{11}} 2, \ \frac {-3+i\sqrt{11}} 2 \bigg)\ \text{ i }\ \bigg(\frac {-3+i\sqrt{11}} 2, \ \frac {-3-i\sqrt{11}} 2 \bigg)

$x^2+xy+y^2=4, \ x+xy+y=2$