Tentukan himpunan penyelesaian dari persamaan kuadrat .... ( 2x+3y=1...(1) (4x2+6xy+y2=-1...(2).... Question from @muhammadalfatih1

September 2018 2 75 Report

Tentukan himpunan penyelesaian dari persamaan kuadrat ....
( 2x+3y=1...(1)
(4x2+6xy+y2=-1...(2)

Takamori37 2x+3y = 1
3y = 1-2x
y = 1/3 (1-2x)
Substitusikan y pada persamaan 2:
$4x^2+6xy+y^2=1 \\ 4x^2+6x(\frac{1-2x}{3})+(\frac{1-2x}{3})^2=-1 \\ 4x^2+2x(1-2x)+\frac{1}{9}(1-2x)^2=-1 \\ 36x^2+18x(1-2x)+(1-2x)^2=-9 \\ 36x^2+18x-36x^2+1-4x+4x^2=-9 \\ 4x^2+14x+10=0 \\ 2x^2+7x+5=0 \\ (2x+5)(x+1)=0 \\ \\ $Untuk $x=-\frac{5}{2},\\ y=\frac{1-2x}{3}=\frac{1-2(-\frac{5}{2})}{3}=\frac{1-(-5)}{3}=\frac{6}{3}=2 \\ $Titik 1 : $(-\frac{5}{2},2) \\\\ $Untuk $x=-1 \\ y=\frac{1-2x}{3}=\frac{1-2(-1)}{3}=\frac{1+2}{3}=1 \\ $Titik 2 : $(-1,1) \\ \\ $Sehingga:$ \\ HP=\{(x,y)|(-\frac{5}{2},2),(-1,1)\}$

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AlfiantiRY A. misalkan persamaan 1 menjadi x
2x + 3y = 1 ⇔ x = $\frac{1-3y}{2}$

b. subtitusi x ke pers. 2
$4 x^{2} +6xy+ y^{2} =-1$
$4 ( \frac{1-3y}{2} )^{2}+6y( \frac{1-3y}{2} )+ y^{2} = -1$
$1-6y+9y^{2}+3y- 9y^{2}+ y^{2} =-1$
$y^{2} -3y+2=0$
$(y-2)(y-1)=0$
$y=2
$ ∨ $y=1$

u/ y=2 ⇒ $\frac{1-3.2}{2} = \frac{-5}{2}$
u/ y=1 ⇒ $\frac{1-3.1}{2} = \frac{-2}{2} = -1$

maka himpunan penyelesaiannya adalah { $( \frac{-5}{2} , 2), (-1 , 1)$ }

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