Matematika

Pertanyaan

SISTEM PERSAMAAN LINEAR UNTUK SEKOLAH MENENGAH ATAS
A. Hanya gunakan SUBTITUSI MURNI! (Tanpa campuran/eliminasi)
1. [tex] x^{2} [/tex] + [tex] y^{2} [/tex] = 25
y = 2x
2. [tex] x^{2} [/tex] + y = 9
x - y + 3 = 0
3. [tex] x^{2} [/tex] - y = 1
[tex] 2x^{2} [/tex] + 3y = 17

B. Hanya gunakan ELIMINASI MURNI (TANPA CAMPURAN/SUBTITUSI)
4. [tex] x^{2} [/tex] - 2y = 1
[tex] x^{2} [/tex] + 5y = 29
5. [tex] 3x^{2} [/tex] - [tex] y^{2} [/tex] = 11
[tex] x^{2} [/tex] + [tex] 4y^{2} [/tex] = 8
6. x - [tex] y^{2} [/tex] + 3 = 0
[tex] 2x^{2} [/tex] + [tex] y^{2} [/tex] - 4 = 0
7. [tex] 2x^{2} [/tex] + 4y = 13
[tex] x^{2} [/tex] - [tex] y^{2} [/tex] = [tex] y \frac{7}{2} ^{2} [/tex]


Soal : Tentukan X dan Y dari semua masing2 soal tersebut
Thanks kak :D

1 Jawaban

  • 1
    x²+y²=25
    y=2x

    x²+(2x)²=25
    x²+4x²=25
    5x²=25
    x²=5
    x=√5

    x²+y²=25
    (√5)²+y²=25
    5+y²=25
    y²=20
    y=√20
    y=2√5

    2)
    x²+y=9
    x-y+3=0

    x²+y=9
    x-y=-3
    y=x+3

    x²+y=9
    x²+x+3=9
    x²+x-6=0
    (x-2)(x+3)
    x1=2 V x2=-3

    y1=x1+3
    y1=2+3
    y1=5

    y2=x2+3
    y2=-3+3
    y2=0

    3)
    x²-y=1          x -3
    2x²+3y=17    x 1

    -3x²+3y=-3
    2x²+3y=17
    3y=17-2x²

    -3x²+17-2x²=-3
    -5x²=-20
    x²=4
    x=√4
    x=2

    x²-y=1
    2²-y=1
    4-y=1
    y=3

    4)
    x²-2y=1
    x²+5y=29 -
    -7y=-28
    y=4

    x²-2y=1
    x²-2(4)=1
    x²-8=1
    x²=9
    x=√9
    x=3

    5)
    3x²-y²=11  x 1
    x²+4y²=8  x 3

    3x²-y²=11
    3x²+12y²=24 -
    -13y²=-13
    y²=1
    y=√1
    y=1

    x²+4y²=8
    x²+4(1)²=8
    x²=4
    x=√4
    x=2

    6)
    x-y²+3=0
    2x²+y²-4=0

    x-y²=-3
    2x²+y²=4 +
    2x²+x=1
    2x²+x-1=0
    (x+2)(x-1)
    x1=-2 V x2=1

    x1-y1²+3=0
    -2-y1²+3=0
    -y1²=-1
    y1²=1
    y1=√1
    y1=1

    x2-y2²+3=0
    1-y2²+3=0
    -y2²=-4
    y2²=4
    y2=√4
    y2=2

    7)
    2x²+4y=13  x 1
    x²-y²=7/2    x 2

    2x²+4y=13
    2x²-2y²=7  -
    4y+2y²=6
    y²+2y-3=0
    (y+3)(y-1)
    y1=-3 V y2=1

    2x1²+4y1=13
    2x1²+4(-3)=13
    2x1²-12=13
    2x1²=25
    x1²=25/2
    x1=√25/2
    x1=5/√2
    x1=5√2/2

    2x2²+4y2=13
    2x2²+4(1)=13
    2x2²=9
    x2²=9/2
    x2=√9/2
    x2=3/√2
    x2=3√2/2

    Semoga Membantu

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