711.

Sistem jednačina

TEKST ZADATKA

Rešiti sistem jednačina.

x2+y2+x+y=3212(x+y)=7xyx^2+y^2+x+y=32 \\ 12(x+y)=7xy
REŠENJE ZADATKA

Iz druge jednačine izraziti x.x.

12(x+y)=7xy  ⟹  x=−12y12−7y12(x+y)=7xy \implies x=\frac{-12y}{12-7y}
DODATNO OBJAŠNJENJE

Uvrstiti xx u prvu jednačinu.

(−12y12−7y)2+y2+−12y12−7y+y=32144y2(12−7y)2+y2+−12y12−7y+y−32=0\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.

144y2(12−7y)2+y2(12−7y)2(12−7y)2+−12y(12−7y)(12−7y)2+y(12−7y)2(12−7y)2−32(12−7y)2(12−7y)2=0144y2+y2(144−168y+49y2)−12y(12−7y)+y(144−168y+49y2)−32(144−168y+49y2)(12−7y)2=0144y2+144y2−168y3+49y4−144y+84y2+144y−168y2+49y3−4608+5376y−1568y2(12−7y)2=049y4−119y3−1364y2+5376y−4608(12−7y)2=0∧y≠127\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 0,0, brojilac mora biti jednak 0.0. Jednačina se svodi na:

49y4−119y3−1364y2+5376y−4608=049y4−147y3+28y3−84y2−1280y2+3840y+1536y−4608=049y3(y−3)+28y2(y−3)−1280(y−3)+1536(y−3)=0(y−3)(49y3+28y2−1280y+1536)=0(y−3)(49y3−196y2+224y2−896y−384y+1536)=0(y−3)(49y2(y−4)+224(y−4)−384(y−4))=0(y−3)(y−4)(49y2+224y−384)=049y^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 yy su:

y−3=0∨y−4=0∨49y2+224y−384=0y1=3∨y2=4∨y3=−16+8107∨y4=−16−8107y-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}
DODATNO OBJAŠNJENJE

Uvrstiti y1=3,y_1=3, y2=4,y_2=4, y3=−16+8107y_3=\frac {-16+ 8\sqrt{10}} {7} i y4=−16+8107 y_4=\frac {-16+ 8\sqrt{10}} {7} u jednačinu x=−12y12−7y.x=\frac{-12y} {12-7y}.

x1=−12⋅312−7⋅3=−3612−21=−36−9=4x2=−12⋅412−7⋅4=−4812−28=−48−16=3x3=−12⋅−16+810712−7⋅−16+8107=−16−8107x4=−12⋅−16−810712−7⋅−16−8107=−16+8107x_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), (−16+8107, −16−8107) i (−16−8107, −16+8107)(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)

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