Sistem: zbir kvadrata 32 i uslov 12(x + y) = 7xy

Iz druge jednačine izraziti $x.$

12(x+y)=7xy \implies x=\frac{-12y}{12-7y}

Uvrstiti $x$ u prvu jednačinu.

\bigg(\frac{-12y}{12-7y}\bigg)^2+y^2+\frac{-12y}{12-7y}+y=32 \\ \frac{144y^2} {(12-7y)^2}+y^2+\frac{-12y}{12-7y}+y-32=0

Svesti na zajednički imenilac i srediti izraz.

\frac{144y^2} {(12-7y)^2}+\frac{y^2(12-7y)^2}{(12-7y)^2}+\frac{-12y(12-7y)}{(12-7y)^2}+\frac{y(12-7y)^2} {(12-7y)^2}-\frac{32(12-7y)^2} {(12-7y)^2}=0 \\ \frac{144y^2+y^2(144-168y+49y^2)-12y(12-7y)+y(144-168y+49y^2)-32(144-168y+49y^2)} {(12-7y)^2}=0 \\ \frac{144y^2+144y^2-168y^3+49y^4-144y+84y^2+144y-168y^2+49y^3-4608+5376y-1568y^2} {(12-7y)^2}=0 \\ \frac{49y^4-119y^3-1364y^2+5376y-4608} {(12-7y)^2}=0 \quad\land\quad y\not=\frac{12}7

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

49y^4-119y^3-1364y^2+5376y-4608=0 \\ 49y^4-147y^3+28y^3-84y^2-1280y^2+3840y+1536y-4608=0 \\ 49y^3(y-3)+28y^2(y-3)-1280(y-3)+1536(y-3)=0 \\ (y-3)(49y^3+28y^2-1280y+1536)=0 \\ (y-3)(49y^3-196y^2+224y^2-896y-384y+1536)=0 \\ (y-3)(49y^2(y-4)+224(y-4)-384(y-4))=0 \\ (y-3)(y-4)(49y^2+224y-384)=0

Rešenja za $y$ su:

y-3=0 \quad\lor\quad y-4=0 \quad\lor\quad 49y^2+224y-384=0\\ y_1=3 \quad\lor\quad y_2=4 \quad\lor\quad y_3=\frac {-16+ 8\sqrt{10}} {7} \quad\lor\quad y_4=\frac {-16-8\sqrt{10}} {7}

Uvrstiti $y_1=3,$ $y_2=4,$ $y_3=\frac {-16+ 8\sqrt{10}} {7}$ i $y_4=\frac {-16+ 8\sqrt{10}} {7}$ u jednačinu $x=\frac{-12y} {12-7y}.$

x_1=\frac{-12\cdot3}{12-7\cdot3}=\frac{-36}{12-21}=\frac{-36}{-9}=4 \\ x_2=\frac{-12\cdot4}{12-7\cdot4}=\frac{-48}{12-28}=\frac{-48}{-16}=3 \\ x_3=\frac{-12\cdot\frac {-16+ 8\sqrt{10}} {7}}{12-7\cdot\frac {-16+ 8\sqrt{10}} {7}}=\frac {-16- 8\sqrt{10}} {7} \\ x_4=\frac{-12\cdot\frac {-16- 8\sqrt{10}} {7}}{12-7\cdot\frac {-16- 8\sqrt{10}} {7}}=\frac {-16+ 8\sqrt{10}} {7}

Rešenje sistema je skup uređenih parova:

(4, 3) , \ (3,4), \ \bigg(\frac {-16+ 8\sqrt{10}} {7} , \ \frac {-16- 8\sqrt{10}} {7} \bigg) \ \text{i} \ \bigg(\frac {-16- 8\sqrt{10}} {7} , \ \frac {-16+ 8\sqrt{10}} {7} \bigg)

$x^2+y^2+x+y=32, \ 12(x+y)=7xy$